Discussion:
Diophantine equation : x^{2n} + y^{2m} = z^2
bassam king karzeddin
2016-08-22 14:31:12 UTC
Conjecture: If (x, y, z) are nonzero co prime integers, and (n, m) are positive integers > 1, then this Diophantine equation :

x^{2n} + y^{2m} = z^2

Doesn't have any integer solution
bassam king karzeddin
2016-08-22 14:46:33 UTC
Post by bassam king karzeddin
x^{2n} + y^{2m} = z^2
Doesn't have any integer solution
Proof is left as an easy exercise to a talented school students!
Simon
2016-08-24 14:45:29 UTC
Post by bassam king karzeddin
Post by bassam king karzeddin
x^{2n} + y^{2m} = z^2
Doesn't have any integer solution
Proof is left as an easy exercise to a talented school students!
Isn't there a formula for generating all Pythagorean triples? That formula may show ...
Jerry Kraus
2016-08-24 14:52:51 UTC
Post by Simon
Post by bassam king karzeddin
Post by bassam king karzeddin
x^{2n} + y^{2m} = z^2
Doesn't have any integer solution
Proof is left as an easy exercise to a talented school students!
Isn't there a formula for generating all Pythagorean triples? That formula may show ...
Here's a general solution method for x^2 + y^2 = z^2

http://fermatslasttheorem.blogspot.com/2005/05/pythagorean-triples-solution.html

z = d[p^2 + q^2]
y = d[p^2 - q^2]
x = d[2pq]

Clearly, x,y,z are only coprime in the case of d =1, by definition. Hence, 3,4,5 are the only coprime solution of x^{2n} + y^{2m} = z^2. So, we can limit the exceptions to my rule not merely to n=m=1, but to a single value for x,y,z where n=m=1.

The general idea is that coprimes have no common factors/divisors, and that for two squared numbers to be equal, they must have common factors.
Peter Percival
2016-08-24 15:16:40 UTC
Post by Jerry Kraus
Post by Simon
Isn't there a formula for generating all Pythagorean triples? That formula may show ...
Here's a general solution method for x^2 + y^2 = z^2
http://fermatslasttheorem.blogspot.com/2005/05/pythagorean-triples-solution.html
z = d[p^2 + q^2]
y = d[p^2 - q^2]
x = d[2pq]
Clearly, x,y,z are only coprime in the case of d =1, by definition. Hence, 3,4,5 are the only coprime solution of x^{2n} + y^{2m} = z^2. So, we can limit the exceptions to my rule not merely to n=m=1, but to a single value for x,y,z where n=m=1.
Hence?
Post by Jerry Kraus
The general idea is that coprimes have no common factors/divisors, and that for two squared numbers to be equal, they must have common factors.
--
Do, as a concession to my poor wits, Lord Darlington, just explain
to me what you really mean.
I think I had better not, Duchess. Nowadays to be intelligible is
to be found out. -- Oscar Wilde, Lady Windermere's Fan
Jerry Kraus
2016-08-24 15:01:56 UTC
Post by Simon
Post by bassam king karzeddin
Post by bassam king karzeddin
x^{2n} + y^{2m} = z^2
Doesn't have any integer solution
Proof is left as an easy exercise to a talented school students!
Isn't there a formula for generating all Pythagorean triples? That formula may show ...
The general idea is that coprimes have no common factors/divisors, and that for two squared numbers to be equal, they must have common factors.

http://fermatslasttheorem.blogspot.com/2005/05/pythagorean-triples-solution.html

This seems quite consistent with the general solution provided, in the link, above
quasi
2016-08-25 21:54:16 UTC
Post by bassam king karzeddin
Post by bassam king karzeddin
Conjecture: If (x, y, z) are nonzero co prime integers, and
(n, m) are positive integers > 1, then this Diophantine
x^{2n} + y^{2m} = z^2
doesn't have any integer solution
Proof is left as an easy exercise to a talented school students!
Do you actually have a solution?

If so, I'd like to see it.

quasi
quasi
2016-08-27 07:52:14 UTC
Post by quasi
Post by bassam king karzeddin
Post by bassam king karzeddin
Conjecture: If (x, y, z) are nonzero co prime integers, and
(n, m) are positive integers > 1, then this Diophantine
x^{2n} + y^{2m} = z^2
doesn't have any integer solution
Proof is left as an easy exercise to a talented school students!
Do you actually have a solution?
If so, I'd like to see it.
Based on the lack of response, I suspect that the OP doesn't
actually have a proof of his claim.

Even for the special cases (m,n) = (2,3) or (m,n) = (3,3),
I suspect there may be no known resolution. Explicitly:

* Do there exist integers x,y,z with gcd(x,y,z) = 1 such that
x^6 + y^4 = z^2?

* Do there exist integers x,y,z with gcd(x,y,z) = 1 such that
x^6 + y^6 = z^2?

Of the above two problems, the second one looks more open to
attack, but the attacks I tried didn't lead to a complete
resolution.

So at this point, I'll restate my challenge to bassam king
karzeddin (or anyone else):

Is there _any_ pair of integers m,n > 1, not both even, for
which you can _prove_ that there is no triple x,y,z of integers
with gcd(x,y,z) = 1 such that x^(2n) + y^(2m) = z^2?

quasi
quasi
2016-08-27 07:58:01 UTC
Post by quasi
Post by quasi
Post by bassam king karzeddin
Post by bassam king karzeddin
Conjecture: If (x, y, z) are nonzero co prime integers, and
(n, m) are positive integers > 1, then this Diophantine
x^{2n} + y^{2m} = z^2
doesn't have any integer solution
Proof is left as an easy exercise to a talented school students!
Do you actually have a solution?
If so, I'd like to see it.
Based on the lack of response, I suspect that the OP doesn't
actually have a proof of his claim.
Even for the special cases (m,n) = (2,3) or (m,n) = (3,3),
* Do there exist integers x,y,z
I meant: Do there exist nonzero integers x,y,z
Post by quasi
with gcd(x,y,z) = 1 such that x^6 + y^4 = z^2?
* Do there exist integers x,y,z
I meant: Do there exist nonzero integers x,y,z
Post by quasi
with gcd(x,y,z) = 1 such that x^6 + y^6 = z^2?
Of the above two problems, the second one looks more open to
attack, but the attacks I tried didn't lead to a complete
resolution.
So at this point, I'll restate my challenge to bassam king
Is there _any_ pair of integers m,n > 1, not both even, for
which you can _prove_ that there is no triple x,y,z of integers
I meant: no triple x,y,z of nonzero integers
Post by quasi
with gcd(x,y,z) = 1 such that x^(2n) + y^(2m) = z^2?
quasi
quasi
2016-08-27 11:16:59 UTC
Post by quasi
Post by quasi
Post by quasi
Post by bassam king karzeddin
Post by bassam king karzeddin
Conjecture: If (x, y, z) are nonzero co prime integers, and
(n, m) are positive integers > 1, then this Diophantine
x^{2n} + y^{2m} = z^2
doesn't have any integer solution
Proof is left as an easy exercise to a talented school students!
Do you actually have a solution?
If so, I'd like to see it.
Based on the lack of response, I suspect that the OP doesn't
actually have a proof of his claim.
Even for the special cases (m,n) = (2,3) or (m,n) = (3,3),
* Do there exist integers x,y,z
I meant: Do there exist nonzero integers x,y,z
Post by quasi
with gcd(x,y,z) = 1 such that x^6 + y^4 = z^2?
* Do there exist integers x,y,z
I meant: Do there exist nonzero integers x,y,z
Post by quasi
with gcd(x,y,z) = 1 such that x^6 + y^6 = z^2?
Of the above two problems, the second one looks more open to
attack, but the attacks I tried didn't lead to a complete
resolution.
So at this point, I'll restate my challenge to bassam king
Is there _any_ pair of integers m,n > 1, not both even, for
which you can _prove_ that there is no triple x,y,z of integers
I meant: no triple x,y,z of nonzero integers
Post by quasi
with gcd(x,y,z) = 1 such that x^(2n) + y^(2m) = z^2?
Update:

I can prove the claim for the case where gcd(m,n) > 1.

In particular, there do not exist nonzero integers x,y,z
with gcd(x,y,z) = 1 such that x^6 + y^6 = z^2.

But I'm not yet able to resolve the test case (m,n) = (2,3),
or explicitly, the question:

* Do there exist nonzero integers x,y,z
with gcd(x,y,z) = 1 such that x^6 + y^4 = z^2?

quasi
Jerry Kraus
2016-08-23 13:20:49 UTC
Post by bassam king karzeddin
x^{2n} + y^{2m} = z^2
Doesn't have any integer solution
Conjecture: If (x, y, z) are nonzero co prime integers, and (n, m) are positive integers > 1, then this Diophantine equation :

x^{2n} + y^{2m} = z^2

Doesn't have any integer solution

x,y,z are coprime
Therefore, x,y,z are all odd, or 1 of them is even, and the other two are odd.

If all three are odd, there can be no integer solution to

x^{2n} + y^{2m} = z^2

because x^{2n} + y^{2m} is always even
z^2 is always odd
and the sqaure root of an odd number cannot be even.

If one of x,y,z is even, the other two are odd:

z = { x^{2n} + y^{2m} }^(1/2)

But, again, x,y,z are coprime.

z = { x^{2n} + y^{2m} }^(1/2)

directly implies

For some value v
v1 + v2 ... + va = v1 + v2 ... + va

Where z = v1 + v2 ... + va
{ x^{2n} + y^{2m} }^(1/2) = v1 + v2 ... + va

But, since x,y,z are coprime, no such value v exists.

Q.E.D.
quasi
2016-08-23 16:51:08 UTC
Post by bassam king karzeddin
Post by bassam king karzeddin
Conjecture: If (x, y, z) are nonzero co prime integers, and
(n, m) are positive integers > 1, then this Diophantine
x^{2n} + y^{2m} = z^2
Doesn't have any integer solution
Conjecture: If (x, y, z) are nonzero co prime integers, and
(n, m) are positive integers > 1, then this Diophantine
x^{2n} + y^{2m} = z^2
Doesn't have any integer solution
x,y,z are coprime
Therefore, x,y,z are all odd, or 1 of them is even, and the
other two are odd.
If all three are odd, there can be no integer solution to
x^{2n} + y^{2m} = z^2
because x^{2n} + y^{2m} is always even
z^2 is always odd
and the sqaure root of an odd number cannot be even.
z = { x^{2n} + y^{2m} }^(1/2)
But, again, x,y,z are coprime.
z = { x^{2n} + y^{2m} }^(1/2)
directly implies
For some value v
v1 + v2 ... + va = v1 + v2 ... + va
??
Post by bassam king karzeddin
Where z = v1 + v2 ... + va
{ x^{2n} + y^{2m} }^(1/2) = v1 + v2 ... + va
??
Post by bassam king karzeddin
But, since x,y,z are coprime, no such value v exists.
Q.E.D.
How are you using the information that m,n > 1?

If m,n = 1, there are lots of solutions -- for example
x = 3, y = 4, z = 5.

quasi
Jerry Kraus
2016-08-23 18:32:30 UTC
Post by quasi
Post by bassam king karzeddin
Post by bassam king karzeddin
Conjecture: If (x, y, z) are nonzero co prime integers, and
(n, m) are positive integers > 1, then this Diophantine
x^{2n} + y^{2m} = z^2
Doesn't have any integer solution
Conjecture: If (x, y, z) are nonzero co prime integers, and
(n, m) are positive integers > 1, then this Diophantine
x^{2n} + y^{2m} = z^2
Doesn't have any integer solution
x,y,z are coprime
Therefore, x,y,z are all odd, or 1 of them is even, and the
other two are odd.
If all three are odd, there can be no integer solution to
x^{2n} + y^{2m} = z^2
because x^{2n} + y^{2m} is always even
z^2 is always odd
and the sqaure root of an odd number cannot be even.
z = { x^{2n} + y^{2m} }^(1/2)
But, again, x,y,z are coprime.
z = { x^{2n} + y^{2m} }^(1/2)
directly implies
For some value v
v1 + v2 ... + va = v1 + v2 ... + va
??
Post by bassam king karzeddin
Where z = v1 + v2 ... + va
{ x^{2n} + y^{2m} }^(1/2) = v1 + v2 ... + va
??
Post by bassam king karzeddin
But, since x,y,z are coprime, no such value v exists.
Q.E.D.
How are you using the information that m,n > 1?
If m,n = 1, there are lots of solutions -- for example
x = 3, y = 4, z = 5.
quasi
Sorry, quasi, you're using one of the earlier versions, which were in error.

Try this one, which is the latest, as posted on Google Groups.

z = { x^{2n} + y^{2m} }^(1/2)

directly implies

For some value v
v1 + v2 ... + vv = v1 + v2 ... + vv

Where z^2 = v1 + v2 ... + vv
{ x^{2n} + y^{2m} } = v1 + v2 ... + vv

But, since x,y,z are coprime, no such value v exists.

--------------------------------------------------------------------------

m,n are specified as being greater than 1, Quasi, in the original post, so best not to use counter-examples that involve m,n =1.

The general idea is that coprimes have no common factors/divisors, and that for two squared numbers to be equal, they must have common factors.
Arturo Magidin
2016-08-23 18:39:36 UTC
Post by Jerry Kraus
Post by quasi
Post by bassam king karzeddin
Post by bassam king karzeddin
Conjecture: If (x, y, z) are nonzero co prime integers, and
(n, m) are positive integers > 1, then this Diophantine
x^{2n} + y^{2m} = z^2
Doesn't have any integer solution
Conjecture: If (x, y, z) are nonzero co prime integers, and
(n, m) are positive integers > 1, then this Diophantine
x^{2n} + y^{2m} = z^2
Doesn't have any integer solution
x,y,z are coprime
Therefore, x,y,z are all odd, or 1 of them is even, and the
other two are odd.
If all three are odd, there can be no integer solution to
x^{2n} + y^{2m} = z^2
because x^{2n} + y^{2m} is always even
z^2 is always odd
and the sqaure root of an odd number cannot be even.
z = { x^{2n} + y^{2m} }^(1/2)
But, again, x,y,z are coprime.
z = { x^{2n} + y^{2m} }^(1/2)
directly implies
For some value v
v1 + v2 ... + va = v1 + v2 ... + va
??
Post by bassam king karzeddin
Where z = v1 + v2 ... + va
{ x^{2n} + y^{2m} }^(1/2) = v1 + v2 ... + va
??
Post by bassam king karzeddin
But, since x,y,z are coprime, no such value v exists.
Q.E.D.
How are you using the information that m,n > 1?
If m,n = 1, there are lots of solutions -- for example
x = 3, y = 4, z = 5.
quasi
Sorry, quasi, you're using one of the earlier versions, which were in error.
Try this one, which is the latest, as posted on Google Groups.
z = { x^{2n} + y^{2m} }^(1/2)
directly implies
For some value v
v1 + v2 ... + vv = v1 + v2 ... + vv
Where z^2 = v1 + v2 ... + vv
{ x^{2n} + y^{2m} } = v1 + v2 ... + vv
But, since x,y,z are coprime, no such value v exists.
--------------------------------------------------------------------------
m,n are specified as being greater than 1, Quasi, in the original post, so best not to use counter-examples that involve m,n =1.
The general idea is that coprimes have no common factors/divisors, and that for two squared numbers to be equal, they must have common factors.
It would seem to me that your argument can be repeated to "show" that no solutions exist to

x^2 + y^2 = z^2

with x, y, z coprime integers. Where did you use that n and m are larger than 1? Nowhere that I can spot. However, we *know* solutions exist to this equation. So, try running your argument with n=m=1 and see if it seems to hold. If it does seem to hold, then you're wrong.

(The reason I suggest this is that, like quasi, I cannot make heads or tails of what it is you are trying to do; it looks like a random collection of symbols in vaguely mathematical shape)
--
Arturo Magidin
Jerry Kraus
2016-08-23 19:14:04 UTC
Post by Arturo Magidin
Post by Jerry Kraus
Post by quasi
Post by bassam king karzeddin
Post by bassam king karzeddin
Conjecture: If (x, y, z) are nonzero co prime integers, and
(n, m) are positive integers > 1, then this Diophantine
x^{2n} + y^{2m} = z^2
Doesn't have any integer solution
Conjecture: If (x, y, z) are nonzero co prime integers, and
(n, m) are positive integers > 1, then this Diophantine
x^{2n} + y^{2m} = z^2
Doesn't have any integer solution
x,y,z are coprime
Therefore, x,y,z are all odd, or 1 of them is even, and the other two are odd.
If all three are odd, there can be no integer solution to
x^{2n} + y^{2m} = z^2
because x^{2n} + y^{2m} is always even
z^2 is always odd
and the sqaure root of an odd number cannot be even.
z = { x^{2n} + y^{2m} }^(1/2)
But, again, x,y,z are coprime.
z = { x^{2n} + y^{2m} }^(1/2)
directly implies
For some value v
v1 + v2 ... + va = v1 + v2 ... + va
??
Post by bassam king karzeddin
Where z = v1 + v2 ... + va
{ x^{2n} + y^{2m} }^(1/2) = v1 + v2 ... + va
??
Post by bassam king karzeddin
But, since x,y,z are coprime, no such value v exists.
Q.E.D.
How are you using the information that m,n > 1?
If m,n = 1, there are lots of solutions -- for example
x = 3, y = 4, z = 5.
quasi
Sorry, quasi, you're using one of the earlier versions, which were in error.
Try this one, which is the latest, as posted on Google Groups.
z = { x^{2n} + y^{2m} }^(1/2)
directly implies
For some value v
v1 + v2 ... + vv = v1 + v2 ... + vv
Where z^2 = v1 + v2 ... + vv
{ x^{2n} + y^{2m} } = v1 + v2 ... + vv
But, since x,y,z are coprime, no such value v exists.
--------------------------------------------------------------------------
m,n are specified as being greater than 1, Quasi, in the original post, so best not to use counter-examples that involve m,n =1.
The general idea is that coprimes have no common factors/divisors, and that for two squared numbers to be equal, they must have common factors.
It would seem to me that your argument can be repeated to "show" that no solutions exist to
x^2 + y^2 = z^2
with x, y, z coprime integers. Where did you use that n and m are larger than 1? Nowhere that I can spot. However, we *know* solutions exist to this equation. So, try running your argument with n=m=1 and see if it seems to hold. If it does seem to hold, then you're wrong.
(The reason I suggest this is that, like quasi, I cannot make heads or tails of what it is you are trying to do; it looks like a random collection of symbols in vaguely mathematical shape)
--
Arturo Magidin
Obviously, m,n =1 represent a special case, which is why they are excepted. Your objection that I have not specifically dealt with the special case is legitimate, if somewhat overstated.

For the case of m=n =1 we actually have the equation of a cone, a standard geometrical figure, and cones can indeed have integer values for their x,y, and z coordinates.

Hence the critical point is that the exponent of x and y must be even, but greater than z, for the rule to apply.

Why, exactly? I shall have to cogitate further, but, I'll probably have an explanation for you, fairly shortly. Say, within 24 hours.
Arturo Magidin
2016-08-23 19:21:29 UTC
Post by Jerry Kraus
Post by Arturo Magidin
Post by Jerry Kraus
Post by quasi
Post by bassam king karzeddin
Post by bassam king karzeddin
Conjecture: If (x, y, z) are nonzero co prime integers, and
(n, m) are positive integers > 1, then this Diophantine
x^{2n} + y^{2m} = z^2
Doesn't have any integer solution
Conjecture: If (x, y, z) are nonzero co prime integers, and
(n, m) are positive integers > 1, then this Diophantine
x^{2n} + y^{2m} = z^2
Doesn't have any integer solution
x,y,z are coprime
Therefore, x,y,z are all odd, or 1 of them is even, and the
other two are odd.
If all three are odd, there can be no integer solution to
x^{2n} + y^{2m} = z^2
because x^{2n} + y^{2m} is always even
z^2 is always odd
and the sqaure root of an odd number cannot be even.
z = { x^{2n} + y^{2m} }^(1/2)
But, again, x,y,z are coprime.
z = { x^{2n} + y^{2m} }^(1/2)
directly implies
For some value v
v1 + v2 ... + va = v1 + v2 ... + va
??
Post by bassam king karzeddin
Where z = v1 + v2 ... + va
{ x^{2n} + y^{2m} }^(1/2) = v1 + v2 ... + va
??
Post by bassam king karzeddin
But, since x,y,z are coprime, no such value v exists.
Q.E.D.
How are you using the information that m,n > 1?
If m,n = 1, there are lots of solutions -- for example
x = 3, y = 4, z = 5.
quasi
Sorry, quasi, you're using one of the earlier versions, which were in error.
Try this one, which is the latest, as posted on Google Groups.
z = { x^{2n} + y^{2m} }^(1/2)
directly implies
For some value v
v1 + v2 ... + vv = v1 + v2 ... + vv
Where z^2 = v1 + v2 ... + vv
{ x^{2n} + y^{2m} } = v1 + v2 ... + vv
But, since x,y,z are coprime, no such value v exists.
--------------------------------------------------------------------------
m,n are specified as being greater than 1, Quasi, in the original post, so best not to use counter-examples that involve m,n =1.
The general idea is that coprimes have no common factors/divisors, and that for two squared numbers to be equal, they must have common factors.
It would seem to me that your argument can be repeated to "show" that no solutions exist to
x^2 + y^2 = z^2
with x, y, z coprime integers. Where did you use that n and m are larger than 1? Nowhere that I can spot. However, we *know* solutions exist to this equation. So, try running your argument with n=m=1 and see if it seems to hold. If it does seem to hold, then you're wrong.
(The reason I suggest this is that, like quasi, I cannot make heads or tails of what it is you are trying to do; it looks like a random collection of symbols in vaguely mathematical shape)
--
Arturo Magidin
Obviously, m,n =1 represent a special case, which is why they are excepted. Your objection that I have not specifically dealt with the special case is legitimate, if somewhat overstated.
That is not what I meant.

The point is that, even though the assumption is that n and m are larger than 1, there does not seem to be any point in your argument where you *use* the fact that n and m are larger than one. If you do indeed never *use* the fact that n and m are larger than one, and your argument were correct, then the argument would *also* establish the assertion for the case when n and m are equal to 1.

For instance: suppose you want to prove "If n>2, then n<n+1". And you argue as follows: we know that 0<1. Adding the same quantity to both sides respects the inequality, hence n+0 < n+1; therefore, n<n+1, which is what we wanted to prove." Now, we never used the fact that n>2, so even though we are allowed to assume that n>2, the actual argument given is more general: it establishes the "larger" fact that for ALL n (and not just n>2), n<n+1.

This is an important and useful exercise when you are looking at a proof: are all hypotheses actually being used? If not, is a hypothesis superfluous? Can we establish a more general statement? Etc.

So, while you are allowed to assume that m>1 and n>1, you don't seem to actually use these facts (in so far as I can understand your argument, which to be honest, is not that far).
Post by Jerry Kraus
For the case of m=n =1 we actually have the equation of a cone,
a standard geometrical figure, and cones can indeed have integer values for their x,y, and z coordinates.
Or, it's well known that there exist pythagorean triples: integers a, b, and c, pairwise relatively prime, all nonzero, such that a^2+b^2=c^2. We can even describe all such triples.
Post by Jerry Kraus
Hence the critical point is that the exponent of x and y must be even, but greater than z, for the rule to apply.
This is equally nonsensical to me. Very little of what you have written makes any sense to me so far.
Post by Jerry Kraus
Why, exactly? I shall have to cogitate further, but, I'll probably have an explanation for you, fairly shortly. Say, within 24 hours.
Take your time. I'm in no hurry.
--
Arturo Magidin
Jerry Kraus
2016-08-23 19:30:25 UTC
Post by Arturo Magidin
Post by Jerry Kraus
Post by Arturo Magidin
Post by Jerry Kraus
Post by quasi
Post by bassam king karzeddin
Post by bassam king karzeddin
Conjecture: If (x, y, z) are nonzero co prime integers, and
(n, m) are positive integers > 1, then this Diophantine
x^{2n} + y^{2m} = z^2
Doesn't have any integer solution
Conjecture: If (x, y, z) are nonzero co prime integers, and
(n, m) are positive integers > 1, then this Diophantine
x^{2n} + y^{2m} = z^2
Doesn't have any integer solution
x,y,z are coprime
Therefore, x,y,z are all odd, or 1 of them is even, and the
other two are odd.
If all three are odd, there can be no integer solution to
x^{2n} + y^{2m} = z^2
because x^{2n} + y^{2m} is always even
z^2 is always odd
and the sqaure root of an odd number cannot be even.
z = { x^{2n} + y^{2m} }^(1/2)
But, again, x,y,z are coprime.
z = { x^{2n} + y^{2m} }^(1/2)
directly implies
For some value v
v1 + v2 ... + va = v1 + v2 ... + va
??
Post by bassam king karzeddin
Where z = v1 + v2 ... + va
{ x^{2n} + y^{2m} }^(1/2) = v1 + v2 ... + va
??
Post by bassam king karzeddin
But, since x,y,z are coprime, no such value v exists.
Q.E.D.
How are you using the information that m,n > 1?
If m,n = 1, there are lots of solutions -- for example
x = 3, y = 4, z = 5.
quasi
Sorry, quasi, you're using one of the earlier versions, which were in error.
Try this one, which is the latest, as posted on Google Groups.
z = { x^{2n} + y^{2m} }^(1/2)
directly implies
For some value v
v1 + v2 ... + vv = v1 + v2 ... + vv
Where z^2 = v1 + v2 ... + vv
{ x^{2n} + y^{2m} } = v1 + v2 ... + vv
But, since x,y,z are coprime, no such value v exists.
--------------------------------------------------------------------------
m,n are specified as being greater than 1, Quasi, in the original post, so best not to use counter-examples that involve m,n =1.
The general idea is that coprimes have no common factors/divisors, and that for two squared numbers to be equal, they must have common factors.
It would seem to me that your argument can be repeated to "show" that no solutions exist to
x^2 + y^2 = z^2
with x, y, z coprime integers. Where did you use that n and m are larger than 1? Nowhere that I can spot. However, we *know* solutions exist to this equation. So, try running your argument with n=m=1 and see if it seems to hold. If it does seem to hold, then you're wrong.
(The reason I suggest this is that, like quasi, I cannot make heads or tails of what it is you are trying to do; it looks like a random collection of symbols in vaguely mathematical shape)
--
Arturo Magidin
Obviously, m,n =1 represent a special case, which is why they are excepted. Your objection that I have not specifically dealt with the special case is legitimate, if somewhat overstated.
That is not what I meant.
The point is that, even though the assumption is that n and m are larger than 1, there does not seem to be any point in your argument where you *use* the fact that n and m are larger than one. If you do indeed never *use* the fact that n and m are larger than one, and your argument were correct, then the argument would *also* establish the assertion for the case when n and m are equal to 1.
For instance: suppose you want to prove "If n>2, then n<n+1". And you argue as follows: we know that 0<1. Adding the same quantity to both sides respects the inequality, hence n+0 < n+1; therefore, n<n+1, which is what we wanted to prove." Now, we never used the fact that n>2, so even though we are allowed to assume that n>2, the actual argument given is more general: it establishes the "larger" fact that for ALL n (and not just n>2), n<n+1.
This is an important and useful exercise when you are looking at a proof: are all hypotheses actually being used? If not, is a hypothesis superfluous? Can we establish a more general statement? Etc.
So, while you are allowed to assume that m>1 and n>1, you don't seem to actually use these facts (in so far as I can understand your argument, which to be honest, is not that far).
Post by Jerry Kraus
For the case of m=n =1 we actually have the equation of a cone,
a standard geometrical figure, and cones can indeed have integer values for their x,y, and z coordinates.
Or, it's well known that there exist pythagorean triples: integers a, b, and c, pairwise relatively prime, all nonzero, such that a^2+b^2=c^2. We can even describe all such triples.
Post by Jerry Kraus
Hence the critical point is that the exponent of x and y must be even, but greater than z, for the rule to apply.
This is equally nonsensical to me. Very little of what you have written makes any sense to me so far.
Post by Jerry Kraus
Why, exactly? I shall have to cogitate further, but, I'll probably have an explanation for you, fairly shortly. Say, within 24 hours.
Take your time. I'm in no hurry.
--
Arturo Magidin
Well, Arturo, I see a couple of possibilities, here,

1. The general argument holds, but, there are special grounds why it's perfectly logical it wouldn't hold for the special case of m,n =1.

2. The general argument can be modified somewhat to hold for all m,n, including m,n =1.

As I've said, I'll have to go off and cogitate a bit on the matter, but I'll probably have some more ideas in a day or so.
Arturo Magidin
2016-08-23 19:33:03 UTC
Post by Jerry Kraus
Post by Arturo Magidin
Post by Jerry Kraus
Post by Arturo Magidin
Post by Jerry Kraus
Post by quasi
Post by bassam king karzeddin
Post by bassam king karzeddin
Conjecture: If (x, y, z) are nonzero co prime integers, and
(n, m) are positive integers > 1, then this Diophantine
x^{2n} + y^{2m} = z^2
Doesn't have any integer solution
Conjecture: If (x, y, z) are nonzero co prime integers, and
(n, m) are positive integers > 1, then this Diophantine
x^{2n} + y^{2m} = z^2
Doesn't have any integer solution
x,y,z are coprime
Therefore, x,y,z are all odd, or 1 of them is even, and the
other two are odd.
If all three are odd, there can be no integer solution to
x^{2n} + y^{2m} = z^2
because x^{2n} + y^{2m} is always even
z^2 is always odd
and the sqaure root of an odd number cannot be even.
z = { x^{2n} + y^{2m} }^(1/2)
But, again, x,y,z are coprime.
z = { x^{2n} + y^{2m} }^(1/2)
directly implies
For some value v
v1 + v2 ... + va = v1 + v2 ... + va
??
Post by bassam king karzeddin
Where z = v1 + v2 ... + va
{ x^{2n} + y^{2m} }^(1/2) = v1 + v2 ... + va
??
Post by bassam king karzeddin
But, since x,y,z are coprime, no such value v exists.
Q.E.D.
How are you using the information that m,n > 1?
If m,n = 1, there are lots of solutions -- for example
x = 3, y = 4, z = 5.
quasi
Sorry, quasi, you're using one of the earlier versions, which were in error.
Try this one, which is the latest, as posted on Google Groups.
z = { x^{2n} + y^{2m} }^(1/2)
directly implies
For some value v
v1 + v2 ... + vv = v1 + v2 ... + vv
Where z^2 = v1 + v2 ... + vv
{ x^{2n} + y^{2m} } = v1 + v2 ... + vv
But, since x,y,z are coprime, no such value v exists.
--------------------------------------------------------------------------
m,n are specified as being greater than 1, Quasi, in the original post, so best not to use counter-examples that involve m,n =1.
The general idea is that coprimes have no common factors/divisors, and that for two squared numbers to be equal, they must have common factors.
It would seem to me that your argument can be repeated to "show" that no solutions exist to
x^2 + y^2 = z^2
with x, y, z coprime integers. Where did you use that n and m are larger than 1? Nowhere that I can spot. However, we *know* solutions exist to this equation. So, try running your argument with n=m=1 and see if it seems to hold. If it does seem to hold, then you're wrong.
(The reason I suggest this is that, like quasi, I cannot make heads or tails of what it is you are trying to do; it looks like a random collection of symbols in vaguely mathematical shape)
--
Arturo Magidin
Obviously, m,n =1 represent a special case, which is why they are excepted. Your objection that I have not specifically dealt with the special case is legitimate, if somewhat overstated.
That is not what I meant.
The point is that, even though the assumption is that n and m are larger than 1, there does not seem to be any point in your argument where you *use* the fact that n and m are larger than one. If you do indeed never *use* the fact that n and m are larger than one, and your argument were correct, then the argument would *also* establish the assertion for the case when n and m are equal to 1.
For instance: suppose you want to prove "If n>2, then n<n+1". And you argue as follows: we know that 0<1. Adding the same quantity to both sides respects the inequality, hence n+0 < n+1; therefore, n<n+1, which is what we wanted to prove." Now, we never used the fact that n>2, so even though we are allowed to assume that n>2, the actual argument given is more general: it establishes the "larger" fact that for ALL n (and not just n>2), n<n+1.
This is an important and useful exercise when you are looking at a proof: are all hypotheses actually being used? If not, is a hypothesis superfluous? Can we establish a more general statement? Etc.
So, while you are allowed to assume that m>1 and n>1, you don't seem to actually use these facts (in so far as I can understand your argument, which to be honest, is not that far).
Post by Jerry Kraus
For the case of m=n =1 we actually have the equation of a cone,
a standard geometrical figure, and cones can indeed have integer values for their x,y, and z coordinates.
Or, it's well known that there exist pythagorean triples: integers a, b, and c, pairwise relatively prime, all nonzero, such that a^2+b^2=c^2. We can even describe all such triples.
Post by Jerry Kraus
Hence the critical point is that the exponent of x and y must be even, but greater than z, for the rule to apply.
This is equally nonsensical to me. Very little of what you have written makes any sense to me so far.
Post by Jerry Kraus
Why, exactly? I shall have to cogitate further, but, I'll probably have an explanation for you, fairly shortly. Say, within 24 hours.
Take your time. I'm in no hurry.
--
Arturo Magidin
Well, Arturo, I see a couple of possibilities, here,
1. The general argument holds, but, there are special grounds why it's perfectly logical it wouldn't hold for the special case of m,n =1.
2. The general argument can be modified somewhat to hold for all m,n, including m,n =1.
But the problem is that if the general argument "hold for all m,n, including m,n=1", then the argument must be WRONG (because the conclusion you get when m=n=1 is known to be FALSE).

Unless you are claiming that you have discovered an inconsistency in basic mathematics.

That's the point: if your argument does not actually require that m and n be greater than 1, if you never actually use this fact, then we don't need to go any further: the argument *must be incorrect*. Because your argument could then be used to establish a false statement, and therefore must be invalid. I don't need to find exactly which step is invalid, I would know that it must be invalid for it would otherwise establish a provably false conclusion.
--
Arturo Magidin
Jerry Kraus
2016-08-23 19:40:09 UTC
Post by Arturo Magidin
Post by Jerry Kraus
Post by Arturo Magidin
Post by Jerry Kraus
Post by Arturo Magidin
Post by Jerry Kraus
Post by quasi
Post by bassam king karzeddin
Post by bassam king karzeddin
Conjecture: If (x, y, z) are nonzero co prime integers, and
(n, m) are positive integers > 1, then this Diophantine
x^{2n} + y^{2m} = z^2
Doesn't have any integer solution
Conjecture: If (x, y, z) are nonzero co prime integers, and
(n, m) are positive integers > 1, then this Diophantine
x^{2n} + y^{2m} = z^2
Doesn't have any integer solution
x,y,z are coprime
Therefore, x,y,z are all odd, or 1 of them is even, and the
other two are odd.
If all three are odd, there can be no integer solution to
x^{2n} + y^{2m} = z^2
because x^{2n} + y^{2m} is always even
z^2 is always odd
and the sqaure root of an odd number cannot be even.
z = { x^{2n} + y^{2m} }^(1/2)
But, again, x,y,z are coprime.
z = { x^{2n} + y^{2m} }^(1/2)
directly implies
For some value v
v1 + v2 ... + va = v1 + v2 ... + va
??
Post by bassam king karzeddin
Where z = v1 + v2 ... + va
{ x^{2n} + y^{2m} }^(1/2) = v1 + v2 ... + va
??
Post by bassam king karzeddin
But, since x,y,z are coprime, no such value v exists.
Q.E.D.
How are you using the information that m,n > 1?
If m,n = 1, there are lots of solutions -- for example
x = 3, y = 4, z = 5.
quasi
Sorry, quasi, you're using one of the earlier versions, which were in error.
Try this one, which is the latest, as posted on Google Groups.
z = { x^{2n} + y^{2m} }^(1/2)
directly implies
For some value v
v1 + v2 ... + vv = v1 + v2 ... + vv
Where z^2 = v1 + v2 ... + vv
{ x^{2n} + y^{2m} } = v1 + v2 ... + vv
But, since x,y,z are coprime, no such value v exists.
--------------------------------------------------------------------------
m,n are specified as being greater than 1, Quasi, in the original post, so best not to use counter-examples that involve m,n =1.
The general idea is that coprimes have no common factors/divisors, and that for two squared numbers to be equal, they must have common factors.
It would seem to me that your argument can be repeated to "show" that no solutions exist to
x^2 + y^2 = z^2
with x, y, z coprime integers. Where did you use that n and m are larger than 1? Nowhere that I can spot. However, we *know* solutions exist to this equation. So, try running your argument with n=m=1 and see if it seems to hold. If it does seem to hold, then you're wrong.
(The reason I suggest this is that, like quasi, I cannot make heads or tails of what it is you are trying to do; it looks like a random collection of symbols in vaguely mathematical shape)
--
Arturo Magidin
Obviously, m,n =1 represent a special case, which is why they are excepted. Your objection that I have not specifically dealt with the special case is legitimate, if somewhat overstated.
That is not what I meant.
The point is that, even though the assumption is that n and m are larger than 1, there does not seem to be any point in your argument where you *use* the fact that n and m are larger than one. If you do indeed never *use* the fact that n and m are larger than one, and your argument were correct, then the argument would *also* establish the assertion for the case when n and m are equal to 1.
For instance: suppose you want to prove "If n>2, then n<n+1". And you argue as follows: we know that 0<1. Adding the same quantity to both sides respects the inequality, hence n+0 < n+1; therefore, n<n+1, which is what we wanted to prove." Now, we never used the fact that n>2, so even though we are allowed to assume that n>2, the actual argument given is more general: it establishes the "larger" fact that for ALL n (and not just n>2), n<n+1.
This is an important and useful exercise when you are looking at a proof: are all hypotheses actually being used? If not, is a hypothesis superfluous? Can we establish a more general statement? Etc.
So, while you are allowed to assume that m>1 and n>1, you don't seem to actually use these facts (in so far as I can understand your argument, which to be honest, is not that far).
Post by Jerry Kraus
For the case of m=n =1 we actually have the equation of a cone,
a standard geometrical figure, and cones can indeed have integer values for their x,y, and z coordinates.
Or, it's well known that there exist pythagorean triples: integers a, b, and c, pairwise relatively prime, all nonzero, such that a^2+b^2=c^2. We can even describe all such triples.
Post by Jerry Kraus
Hence the critical point is that the exponent of x and y must be even, but greater than z, for the rule to apply.
This is equally nonsensical to me. Very little of what you have written makes any sense to me so far.
Post by Jerry Kraus
Why, exactly? I shall have to cogitate further, but, I'll probably have an explanation for you, fairly shortly. Say, within 24 hours.
Take your time. I'm in no hurry.
--
Arturo Magidin
Well, Arturo, I see a couple of possibilities, here,
1. The general argument holds, but, there are special grounds why it's perfectly logical it wouldn't hold for the special case of m,n =1.
2. The general argument can be modified somewhat to hold for all m,n, including m,n =1.
But the problem is that if the general argument "hold for all m,n, including m,n=1", then the argument must be WRONG (because the conclusion you get when m=n=1 is known to be FALSE).
Unless you are claiming that you have discovered an inconsistency in basic mathematics.
That's the point: if your argument does not actually require that m and n be greater than 1, if you never actually use this fact, then we don't need to go any further: the argument *must be incorrect*. Because your argument could then be used to establish a false statement, and therefore must be invalid. I don't need to find exactly which step is invalid, I would know that it must be invalid for it would otherwise establish a provably false conclusion.
--
Arturo Magidin
Your point is that my argument can be applied to m,n =1 -- although these, are specifically excepted -- but doesn't hold for m,n =1, as currently stated. Fair enough. I will attempt to justify the exception, or modify the argument. I will, however, have to think about it.
Peter Percival
2016-08-23 19:16:05 UTC
Post by Jerry Kraus
Post by quasi
Post by bassam king karzeddin
Post by bassam king karzeddin
Conjecture: If (x, y, z) are nonzero co prime integers, and
(n, m) are positive integers > 1, then this Diophantine
x^{2n} + y^{2m} = z^2
Doesn't have any integer solution
Conjecture: If (x, y, z) are nonzero co prime integers, and
(n, m) are positive integers > 1, then this Diophantine
x^{2n} + y^{2m} = z^2
Doesn't have any integer solution
x,y,z are coprime
Therefore, x,y,z are all odd, or 1 of them is even, and the
other two are odd.
If all three are odd, there can be no integer solution to
x^{2n} + y^{2m} = z^2
because x^{2n} + y^{2m} is always even
z^2 is always odd
and the sqaure root of an odd number cannot be even.
z = { x^{2n} + y^{2m} }^(1/2)
But, again, x,y,z are coprime.
z = { x^{2n} + y^{2m} }^(1/2)
directly implies
For some value v
v1 + v2 ... + va = v1 + v2 ... + va
??
Post by bassam king karzeddin
Where z = v1 + v2 ... + va
{ x^{2n} + y^{2m} }^(1/2) = v1 + v2 ... + va
??
Post by bassam king karzeddin
But, since x,y,z are coprime, no such value v exists.
Q.E.D.
How are you using the information that m,n > 1?
If m,n = 1, there are lots of solutions -- for example
x = 3, y = 4, z = 5.
quasi
Sorry, quasi, you're using one of the earlier versions, which were in error.
Try this one, which is the latest, as posted on Google Groups.
z = { x^{2n} + y^{2m} }^(1/2)
directly implies
For some value v
v1 + v2 ... + vv = v1 + v2 ... + vv
What does that mean? Surely v1 + v2 ... + vv = v1 + v2 ... + vv
whatever z equals?
Post by Jerry Kraus
Where z^2 = v1 + v2 ... + vv
{ x^{2n} + y^{2m} } = v1 + v2 ... + vv
But, since x,y,z are coprime, no such value v exists.
--------------------------------------------------------------------------
m,n are specified as being greater than 1, Quasi, in the original post, so best not to use counter-examples that involve m,n =1.
The general idea is that coprimes have no common factors/divisors, and that for two squared numbers to be equal, they must have common factors.
--
I have had a tremor of bliss, a wink of heaven, a whisper,
And I would no longer be denied; all things
Proceed to a joyful consummation.
Becket through the pen of Eliot, /Murder in the Cathedral/
Jerry Kraus
2016-08-23 19:24:08 UTC
Post by Peter Percival
Post by Jerry Kraus
Post by quasi
Post by bassam king karzeddin
Post by bassam king karzeddin
Conjecture: If (x, y, z) are nonzero co prime integers, and
(n, m) are positive integers > 1, then this Diophantine
x^{2n} + y^{2m} = z^2
Doesn't have any integer solution
Conjecture: If (x, y, z) are nonzero co prime integers, and
(n, m) are positive integers > 1, then this Diophantine
x^{2n} + y^{2m} = z^2
Doesn't have any integer solution
x,y,z are coprime
Therefore, x,y,z are all odd, or 1 of them is even, and the other two are odd.
If all three are odd, there can be no integer solution to
x^{2n} + y^{2m} = z^2
because x^{2n} + y^{2m} is always even
z^2 is always odd
and the sqaure root of an odd number cannot be even.
z = { x^{2n} + y^{2m} }^(1/2)
But, again, x,y,z are coprime.
z = { x^{2n} + y^{2m} }^(1/2)
directly implies
For some value v
v1 + v2 ... + va = v1 + v2 ... + va
??
Post by bassam king karzeddin
Where z = v1 + v2 ... + va
{ x^{2n} + y^{2m} }^(1/2) = v1 + v2 ... + va
??
Post by bassam king karzeddin
But, since x,y,z are coprime, no such value v exists.
Q.E.D.
How are you using the information that m,n > 1?
If m,n = 1, there are lots of solutions -- for example
x = 3, y = 4, z = 5.
quasi
Sorry, quasi, you're using one of the earlier versions, which were in error.
Try this one, which is the latest, as posted on Google Groups.
z = { x^{2n} + y^{2m} }^(1/2)
directly implies
For some value v
v1 + v2 ... + vv = v1 + v2 ... + vv
What does that mean? Surely v1 + v2 ... + vv = v1 + v2 ... + vv
whatever z equals?
It's just another way of saying/representing that z^2 = x^{2n} + y^{2m}, but emphasizing the sums of factors to the value of the square.
Arturo Magidin
2016-08-23 19:30:26 UTC
Post by Jerry Kraus
Post by Peter Percival
Post by Jerry Kraus
Post by quasi
Post by bassam king karzeddin
Post by bassam king karzeddin
Conjecture: If (x, y, z) are nonzero co prime integers, and
(n, m) are positive integers > 1, then this Diophantine
x^{2n} + y^{2m} = z^2
Doesn't have any integer solution
Conjecture: If (x, y, z) are nonzero co prime integers, and
(n, m) are positive integers > 1, then this Diophantine
x^{2n} + y^{2m} = z^2
Doesn't have any integer solution
x,y,z are coprime
Therefore, x,y,z are all odd, or 1 of them is even, and the other two are odd.
If all three are odd, there can be no integer solution to
x^{2n} + y^{2m} = z^2
because x^{2n} + y^{2m} is always even
z^2 is always odd
and the sqaure root of an odd number cannot be even.
z = { x^{2n} + y^{2m} }^(1/2)
But, again, x,y,z are coprime.
z = { x^{2n} + y^{2m} }^(1/2)
directly implies
For some value v
v1 + v2 ... + va = v1 + v2 ... + va
??
Post by bassam king karzeddin
Where z = v1 + v2 ... + va
{ x^{2n} + y^{2m} }^(1/2) = v1 + v2 ... + va
??
Post by bassam king karzeddin
But, since x,y,z are coprime, no such value v exists.
Q.E.D.
How are you using the information that m,n > 1?
If m,n = 1, there are lots of solutions -- for example
x = 3, y = 4, z = 5.
quasi
Sorry, quasi, you're using one of the earlier versions, which were in error.
Try this one, which is the latest, as posted on Google Groups.
z = { x^{2n} + y^{2m} }^(1/2)
directly implies
For some value v
v1 + v2 ... + vv = v1 + v2 ... + vv
What does that mean? Surely v1 + v2 ... + vv = v1 + v2 ... + vv
whatever z equals?
It's just another way of saying/representing that z^2 = x^{2n} + y^{2m}, but emphasizing the sums of factors to the value of the square.
Let me tell you all the ways this does not make sense to me:

1. You only specify v; you are also using v for two things: as the generic name for the summands, and as a value of the index. That's *BAD* notation!

2. You don't specify what kind of animals v1, v2, etc. are. Integers? Real numbers? Positive rationals? What?

3. You don't specify how you are picking these objects, or what role they are supposed to play. How do they relate to x and y? To z?

4. Are these just random numbers that add up to z^2, and hence to x^{2n}+y^{2m}? Can we pick any we want?

And then you say "since x,y,z are coprime, no such value v exists", which seems like a nonsequitur given how little information there is about what v is or does, or how it relates to anything at hand.
--
Arturo Magidin
Jerry Kraus
2016-08-23 19:35:28 UTC
Post by Arturo Magidin
Post by Jerry Kraus
Post by Peter Percival
Post by Jerry Kraus
Post by quasi
Post by bassam king karzeddin
Post by bassam king karzeddin
Conjecture: If (x, y, z) are nonzero co prime integers, and
(n, m) are positive integers > 1, then this Diophantine
x^{2n} + y^{2m} = z^2
Doesn't have any integer solution
Conjecture: If (x, y, z) are nonzero co prime integers, and
(n, m) are positive integers > 1, then this Diophantine
x^{2n} + y^{2m} = z^2
Doesn't have any integer solution
x,y,z are coprime
Therefore, x,y,z are all odd, or 1 of them is even, and the
other two are odd.
If all three are odd, there can be no integer solution to
x^{2n} + y^{2m} = z^2
because x^{2n} + y^{2m} is always even
z^2 is always odd
and the sqaure root of an odd number cannot be even.
z = { x^{2n} + y^{2m} }^(1/2)
But, again, x,y,z are coprime.
z = { x^{2n} + y^{2m} }^(1/2)
directly implies
For some value v
v1 + v2 ... + va = v1 + v2 ... + va
??
Post by bassam king karzeddin
Where z = v1 + v2 ... + va
{ x^{2n} + y^{2m} }^(1/2) = v1 + v2 ... + va
??
Post by bassam king karzeddin
But, since x,y,z are coprime, no such value v exists.
Q.E.D.
How are you using the information that m,n > 1?
If m,n = 1, there are lots of solutions -- for example
x = 3, y = 4, z = 5.
quasi
Sorry, quasi, you're using one of the earlier versions, which were in error.
Try this one, which is the latest, as posted on Google Groups.
z = { x^{2n} + y^{2m} }^(1/2)
directly implies
For some value v
v1 + v2 ... + vv = v1 + v2 ... + vv
What does that mean? Surely v1 + v2 ... + vv = v1 + v2 ... + vv
whatever z equals?
It's just another way of saying/representing that z^2 = x^{2n} + y^{2m}, but emphasizing the sums of factors to the value of the square.
1. You only specify v; you are also using v for two things: as the generic name for the summands, and as a value of the index. That's *BAD* notation!
2. You don't specify what kind of animals v1, v2, etc. are. Integers? Real numbers? Positive rationals? What?
3. You don't specify how you are picking these objects, or what role they are supposed to play. How do they relate to x and y? To z?
4. Are these just random numbers that add up to z^2, and hence to x^{2n}+y^{2m}? Can we pick any we want?
And then you say "since x,y,z are coprime, no such value v exists", which seems like a nonsequitur given how little information there is about what v is or does, or how it relates to anything at hand.
--
Arturo Magidin
The values of v are integers, as the problem pertains to integers. They are factors of x, y, z, insofar as this is, or is not possible. Coprimes, by definition, have no common factors/divisors, hence the relevance of the argument regarding coprimes to the problem. If we have no common factors, we cannot sum factors to the value of a square, to equate values. That's a general sketch of the concepts involved.
Arturo Magidin
2016-08-23 19:44:27 UTC
Post by Jerry Kraus
Post by Arturo Magidin
Post by Jerry Kraus
Post by Peter Percival
Post by Jerry Kraus
Post by quasi
Post by bassam king karzeddin
Post by bassam king karzeddin
Conjecture: If (x, y, z) are nonzero co prime integers, and
(n, m) are positive integers > 1, then this Diophantine
x^{2n} + y^{2m} = z^2
Doesn't have any integer solution
Conjecture: If (x, y, z) are nonzero co prime integers, and
(n, m) are positive integers > 1, then this Diophantine
x^{2n} + y^{2m} = z^2
Doesn't have any integer solution
x,y,z are coprime
Therefore, x,y,z are all odd, or 1 of them is even, and the
other two are odd.
If all three are odd, there can be no integer solution to
x^{2n} + y^{2m} = z^2
because x^{2n} + y^{2m} is always even
z^2 is always odd
and the sqaure root of an odd number cannot be even.
z = { x^{2n} + y^{2m} }^(1/2)
But, again, x,y,z are coprime.
z = { x^{2n} + y^{2m} }^(1/2)
directly implies
For some value v
v1 + v2 ... + va = v1 + v2 ... + va
??
Post by bassam king karzeddin
Where z = v1 + v2 ... + va
{ x^{2n} + y^{2m} }^(1/2) = v1 + v2 ... + va
??
Post by bassam king karzeddin
But, since x,y,z are coprime, no such value v exists.
Q.E.D.
How are you using the information that m,n > 1?
If m,n = 1, there are lots of solutions -- for example
x = 3, y = 4, z = 5.
quasi
Sorry, quasi, you're using one of the earlier versions, which were in error.
Try this one, which is the latest, as posted on Google Groups.
z = { x^{2n} + y^{2m} }^(1/2)
directly implies
For some value v
v1 + v2 ... + vv = v1 + v2 ... + vv
What does that mean? Surely v1 + v2 ... + vv = v1 + v2 ... + vv
whatever z equals?
It's just another way of saying/representing that z^2 = x^{2n} + y^{2m}, but emphasizing the sums of factors to the value of the square.
1. You only specify v; you are also using v for two things: as the generic name for the summands, and as a value of the index. That's *BAD* notation!
2. You don't specify what kind of animals v1, v2, etc. are. Integers? Real numbers? Positive rationals? What?
3. You don't specify how you are picking these objects, or what role they are supposed to play. How do they relate to x and y? To z?
4. Are these just random numbers that add up to z^2, and hence to x^{2n}+y^{2m}? Can we pick any we want?
And then you say "since x,y,z are coprime, no such value v exists", which seems like a nonsequitur given how little information there is about what v is or does, or how it relates to anything at hand.
--
Arturo Magidin
The values of v are integers, as the problem pertains to integers.
Again: you are using v to denote TWO DIFFERENT THINGS. That's bad. If v is the index, and only the index, then you should pick a different letter to denote the values. E.g., w. Are you saying the value of v, the INDEX, in an integer, or are you saying that the SUMMANDS are integers? Or both?
Post by Jerry Kraus
They are factors of x, y, z, insofar as this is, or is not possible.
If they are factors (and hence they act MULTIPLICATIVELY), then why are you writing that they ADD UP to z?

That is, if w_1, w_2, ...., w_v are the (prime? some other type of) FACTORS of z, then it is not the case the z^2 = w_1 + w_2 + ... + w_v (as you write); rather, you would have to have something like z=(w_1)*(w_2)*...*(w_v). Multiplication, not addition.

And if they are factors of x and/or y, then they are not factors of z. So this does not make any sense to me either.
Post by Jerry Kraus
Coprimes, by definition, have no common factors/divisors,
except for 1 and -1.
Post by Jerry Kraus
hence the relevance of the argument regarding coprimes to the problem.
But you are using addition, not multiplication!

If w_1,...,w_v are the (prime, possibly repeated) factors of z, then

z=(w_1)*...*(w_v).

You then have that x^{2n}+y^{2m} = (w_1)^2*...*(w_v)^2, and (perhaps) that
(x^{2n}+y^{2m})^{1/2} = (w_1)*...*(w_v).

But here's the problem: you have a sum on the left and a product on the right. The factors of a sum do not need to be factors of either summand. For example, 2 is a factor of 3 + 5, but is not a factor of either 3 or 5; no problem there! So the fact that your "v_i" are factors of z, and therefore cannot be factors of either x or y, does not preclude them from being factors of x^{2n}+y^{2m}. Because here we have a sum, not a product.
Post by Jerry Kraus
If we have no common factors, we cannot sum factors to the value of a square, to equate values. That's a general sketch of the concepts involved.
I think this is confused nonsense. You are adding factors instead of multiplying them.

For example, with n=m=1, your argument would seem to be that since the only factor of 5 is 5, then (x^2+y^2)^(1/2) = 5 requires 5 to be a factor of x and y, which is impossible because we are assuming that x, y, and z=5 are coprime. But this is not true, since x=3 and y=4, the first one having factor 3 and the second having prime factorization 2*2, still satisfy (x^2+y^2)^{1/2} = 5 and hence x^2+y^2 = z^2. We have the same assumptions (coprimeness, a square root, the factors of z and z^2). But an obviously false conclusion.
--
Arturo Magidin
--
Arturo Magidin
Jerry Kraus
2016-08-23 19:55:57 UTC
Post by Arturo Magidin
Post by Jerry Kraus
Post by Arturo Magidin
Post by Jerry Kraus
Post by Peter Percival
Post by Jerry Kraus
Post by quasi
Post by bassam king karzeddin
Post by bassam king karzeddin
Conjecture: If (x, y, z) are nonzero co prime integers, and
(n, m) are positive integers > 1, then this Diophantine
x^{2n} + y^{2m} = z^2
Doesn't have any integer solution
Conjecture: If (x, y, z) are nonzero co prime integers, and
(n, m) are positive integers > 1, then this Diophantine
x^{2n} + y^{2m} = z^2
Doesn't have any integer solution
x,y,z are coprime
Therefore, x,y,z are all odd, or 1 of them is even, and the
other two are odd.
If all three are odd, there can be no integer solution to
x^{2n} + y^{2m} = z^2
because x^{2n} + y^{2m} is always even
z^2 is always odd
and the sqaure root of an odd number cannot be even.
z = { x^{2n} + y^{2m} }^(1/2)
But, again, x,y,z are coprime.
z = { x^{2n} + y^{2m} }^(1/2)
directly implies
For some value v
v1 + v2 ... + va = v1 + v2 ... + va
??
Post by bassam king karzeddin
Where z = v1 + v2 ... + va
{ x^{2n} + y^{2m} }^(1/2) = v1 + v2 ... + va
??
Post by bassam king karzeddin
But, since x,y,z are coprime, no such value v exists.
Q.E.D.
How are you using the information that m,n > 1?
If m,n = 1, there are lots of solutions -- for example
x = 3, y = 4, z = 5.
quasi
Sorry, quasi, you're using one of the earlier versions, which were in error.
Try this one, which is the latest, as posted on Google Groups.
z = { x^{2n} + y^{2m} }^(1/2)
directly implies
For some value v
v1 + v2 ... + vv = v1 + v2 ... + vv
What does that mean? Surely v1 + v2 ... + vv = v1 + v2 ... + vv
whatever z equals?
It's just another way of saying/representing that z^2 = x^{2n} + y^{2m}, but emphasizing the sums of factors to the value of the square.
1. You only specify v; you are also using v for two things: as the generic name for the summands, and as a value of the index. That's *BAD* notation!
2. You don't specify what kind of animals v1, v2, etc. are. Integers? Real numbers? Positive rationals? What?
3. You don't specify how you are picking these objects, or what role they are supposed to play. How do they relate to x and y? To z?
4. Are these just random numbers that add up to z^2, and hence to x^{2n}+y^{2m}? Can we pick any we want?
And then you say "since x,y,z are coprime, no such value v exists", which seems like a nonsequitur given how little information there is about what v is or does, or how it relates to anything at hand.
--
Arturo Magidin
The values of v are integers, as the problem pertains to integers.
Again: you are using v to denote TWO DIFFERENT THINGS. That's bad. If v is the index, and only the index, then you should pick a different letter to denote the values. E.g., w. Are you saying the value of v, the INDEX, in an integer, or are you saying that the SUMMANDS are integers? Or both?
Post by Jerry Kraus
They are factors of x, y, z, insofar as this is, or is not possible.
If they are factors (and hence they act MULTIPLICATIVELY), then why are you writing that they ADD UP to z?
That is, if w_1, w_2, ...., w_v are the (prime? some other type of) FACTORS of z, then it is not the case the z^2 = w_1 + w_2 + ... + w_v (as you write); rather, you would have to have something like z=(w_1)*(w_2)*...*(w_v). Multiplication, not addition.
And if they are factors of x and/or y, then they are not factors of z. So this does not make any sense to me either.
Post by Jerry Kraus
Coprimes, by definition, have no common factors/divisors,
except for 1 and -1.
Post by Jerry Kraus
hence the relevance of the argument regarding coprimes to the problem.
But you are using addition, not multiplication!
If w_1,...,w_v are the (prime, possibly repeated) factors of z, then
z=(w_1)*...*(w_v).
You then have that x^{2n}+y^{2m} = (w_1)^2*...*(w_v)^2, and (perhaps) that
(x^{2n}+y^{2m})^{1/2} = (w_1)*...*(w_v).
But here's the problem: you have a sum on the left and a product on the right. The factors of a sum do not need to be factors of either summand. For example, 2 is a factor of 3 + 5, but is not a factor of either 3 or 5; no problem there! So the fact that your "v_i" are factors of z, and therefore cannot be factors of either x or y, does not preclude them from being factors of x^{2n}+y^{2m}. Because here we have a sum, not a product.
Post by Jerry Kraus
If we have no common factors, we cannot sum factors to the value of a square, to equate values. That's a general sketch of the concepts involved.
I think this is confused nonsense. You are adding factors instead of multiplying them.
For example, with n=m=1, your argument would seem to be that since the only factor of 5 is 5, then (x^2+y^2)^(1/2) = 5 requires 5 to be a factor of x and y, which is impossible because we are assuming that x, y, and z=5 are coprime. But this is not true, since x=3 and y=4, the first one having factor 3 and the second having prime factorization 2*2, still satisfy (x^2+y^2)^{1/2} = 5 and hence x^2+y^2 = z^2. We have the same assumptions (coprimeness, a square root, the factors of z and z^2). But an obviously false conclusion.
--
Arturo Magidin
--
Arturo Magidin
I think one of the points I'm exploring here is the relationship between multiplication and addition, actually. After all, x^2 can simply be represented as the sum of x, x's -- x1 + x2 + x3 ... all the way up to the x'th x. That's why I think the coprime aspect is quite relevant to
x^{2n} + y^{2m} = z^2. The summation of x and y doesn't entirely eliminate the significance of the coprime aspect. But, as we agree, for m,n =1 there seems to be an exception, whose significance I am currently contemplating.
Arturo Magidin
2016-08-23 20:05:51 UTC
Post by Jerry Kraus
Post by Arturo Magidin
Post by Jerry Kraus
Post by Arturo Magidin
Post by Jerry Kraus
Post by Peter Percival
Post by Jerry Kraus
Post by quasi
Post by bassam king karzeddin
Post by bassam king karzeddin
Conjecture: If (x, y, z) are nonzero co prime integers, and
(n, m) are positive integers > 1, then this Diophantine
x^{2n} + y^{2m} = z^2
Doesn't have any integer solution
Conjecture: If (x, y, z) are nonzero co prime integers, and
(n, m) are positive integers > 1, then this Diophantine
x^{2n} + y^{2m} = z^2
Doesn't have any integer solution
x,y,z are coprime
Therefore, x,y,z are all odd, or 1 of them is even, and the
other two are odd.
If all three are odd, there can be no integer solution to
x^{2n} + y^{2m} = z^2
because x^{2n} + y^{2m} is always even
z^2 is always odd
and the sqaure root of an odd number cannot be even.
z = { x^{2n} + y^{2m} }^(1/2)
But, again, x,y,z are coprime.
z = { x^{2n} + y^{2m} }^(1/2)
directly implies
For some value v
v1 + v2 ... + va = v1 + v2 ... + va
??
Post by bassam king karzeddin
Where z = v1 + v2 ... + va
{ x^{2n} + y^{2m} }^(1/2) = v1 + v2 ... + va
??
Post by bassam king karzeddin
But, since x,y,z are coprime, no such value v exists.
Q.E.D.
How are you using the information that m,n > 1?
If m,n = 1, there are lots of solutions -- for example
x = 3, y = 4, z = 5.
quasi
Sorry, quasi, you're using one of the earlier versions, which were in error.
Try this one, which is the latest, as posted on Google Groups.
z = { x^{2n} + y^{2m} }^(1/2)
directly implies
For some value v
v1 + v2 ... + vv = v1 + v2 ... + vv
What does that mean? Surely v1 + v2 ... + vv = v1 + v2 ... + vv
whatever z equals?
It's just another way of saying/representing that z^2 = x^{2n} + y^{2m}, but emphasizing the sums of factors to the value of the square.
1. You only specify v; you are also using v for two things: as the generic name for the summands, and as a value of the index. That's *BAD* notation!
2. You don't specify what kind of animals v1, v2, etc. are. Integers? Real numbers? Positive rationals? What?
3. You don't specify how you are picking these objects, or what role they are supposed to play. How do they relate to x and y? To z?
4. Are these just random numbers that add up to z^2, and hence to x^{2n}+y^{2m}? Can we pick any we want?
And then you say "since x,y,z are coprime, no such value v exists", which seems like a nonsequitur given how little information there is about what v is or does, or how it relates to anything at hand.
--
Arturo Magidin
The values of v are integers, as the problem pertains to integers.
Again: you are using v to denote TWO DIFFERENT THINGS. That's bad. If v is the index, and only the index, then you should pick a different letter to denote the values. E.g., w. Are you saying the value of v, the INDEX, in an integer, or are you saying that the SUMMANDS are integers? Or both?
Post by Jerry Kraus
They are factors of x, y, z, insofar as this is, or is not possible.
If they are factors (and hence they act MULTIPLICATIVELY), then why are you writing that they ADD UP to z?
That is, if w_1, w_2, ...., w_v are the (prime? some other type of) FACTORS of z, then it is not the case the z^2 = w_1 + w_2 + ... + w_v (as you write); rather, you would have to have something like z=(w_1)*(w_2)*...*(w_v). Multiplication, not addition.
And if they are factors of x and/or y, then they are not factors of z. So this does not make any sense to me either.
Post by Jerry Kraus
Coprimes, by definition, have no common factors/divisors,
except for 1 and -1.
Post by Jerry Kraus
hence the relevance of the argument regarding coprimes to the problem.
But you are using addition, not multiplication!
If w_1,...,w_v are the (prime, possibly repeated) factors of z, then
z=(w_1)*...*(w_v).
You then have that x^{2n}+y^{2m} = (w_1)^2*...*(w_v)^2, and (perhaps) that
(x^{2n}+y^{2m})^{1/2} = (w_1)*...*(w_v).
But here's the problem: you have a sum on the left and a product on the right. The factors of a sum do not need to be factors of either summand. For example, 2 is a factor of 3 + 5, but is not a factor of either 3 or 5; no problem there! So the fact that your "v_i" are factors of z, and therefore cannot be factors of either x or y, does not preclude them from being factors of x^{2n}+y^{2m}. Because here we have a sum, not a product.
Post by Jerry Kraus
If we have no common factors, we cannot sum factors to the value of a square, to equate values. That's a general sketch of the concepts involved.
I think this is confused nonsense. You are adding factors instead of multiplying them.
For example, with n=m=1, your argument would seem to be that since the only factor of 5 is 5, then (x^2+y^2)^(1/2) = 5 requires 5 to be a factor of x and y, which is impossible because we are assuming that x, y, and z=5 are coprime. But this is not true, since x=3 and y=4, the first one having factor 3 and the second having prime factorization 2*2, still satisfy (x^2+y^2)^{1/2} = 5 and hence x^2+y^2 = z^2. We have the same assumptions (coprimeness, a square root, the factors of z and z^2). But an obviously false conclusion.
--
Arturo Magidin
--
Arturo Magidin
I think one of the points I'm exploring here is the relationship between multiplication and addition, actually. After all, x^2 can simply be represented as the sum of x, x's -- x1 + x2 + x3 ... all the way up to the x'th x.
And there you go again: WHY are you trying to force x to play two different roles? NEVER DO THAT! x should either be the generic name of the summands, OR the number of the summands. It should NOT be both. You wouldn't talk about "the fourth four" unless you actually means that you had four fours. So don't talk about "the xth x". Use different names.

True, numbers can be represented as sums. But in general, expressions of numbers as sums of positive integers ("partitions") have little to do with expressions of numbers as products of primes ("factorizations"). We can express 9 as 2+7, but this has little to do with the fact that 9 =3*3.
Post by Jerry Kraus
That's why I think the coprime aspect is quite relevant to
x^{2n} + y^{2m} = z^2.
And I don't think so. 9 and 8 are coprime, but I can write them using common summands: 9 = 1+3+5, and 8=3+5. Common summands, even though 9 and 8 are coprime.
Post by Jerry Kraus
The summation of x and y doesn't entirely eliminate the significance of the coprime aspect.
Frankly, either I have no idea what you are thinking, or your thinking is utterly muddled and confused (and so *you* don't really know what you are thinking).

Partitions of integers seldom have anything to do with factorizations of integers. The number of ways of expressing a positive integer as a sum grows with the integer; the number of ways of expressing a positive integer as a product of primes is constant (always equal to 1), and the number of factors does not change monotonically. Expressing a sum of the form A+B as a product just tells you that for each prime factor of the sum, either both A and B are divisible by the prime, or neither is. But it is entirely possible that A+B = q1*...*qn, but A and B have little or nothing to do with any of the q_i, and that the prime factors of A and B are all distinct from each other and from q_1,...,q_n.
--
Arturo Magidin
Peter Percival
2016-08-23 20:20:27 UTC
Post by Jerry Kraus
I think one of the points I'm exploring here is the relationship
between multiplication and addition, actually. After all, x^2 can
simply be represented as the sum of x, x's -- x1 + x2 + x3 ... all
the way up to the x'th x.
Could you give an example of a sum of x's adding up to x^2?

If it's just a matter of, e.g. with x=3, 2 + 2 + 5 = 9, then there will
be many such sequences, won't there?
--
I have had a tremor of bliss, a wink of heaven, a whisper,
And I would no longer be denied; all things
Proceed to a joyful consummation.
Becket through the pen of Eliot, /Murder in the Cathedral/
Arturo Magidin
2016-08-23 20:27:59 UTC
Post by Peter Percival
Post by Jerry Kraus
I think one of the points I'm exploring here is the relationship
between multiplication and addition, actually. After all, x^2 can
simply be represented as the sum of x, x's -- x1 + x2 + x3 ... all
the way up to the x'th x.
Could you give an example of a sum of x's adding up to x^2?
If it's just a matter of, e.g. with x=3, 2 + 2 + 5 = 9, then there will
be many such sequences, won't there?
I *think* he is thinking of multiplication as repeated addition. So 9 = 3*3 is a sum of 3 with itself, three summands: 9 = 3+3+3.

So he really meant

x^2 = x + x + ... + x (x summands)

which works for positive integer values of x.
--
Arturo Magidin
Jerry Kraus
2016-08-24 14:02:55 UTC
Post by Arturo Magidin
Post by Peter Percival
Post by Jerry Kraus
I think one of the points I'm exploring here is the relationship
between multiplication and addition, actually. After all, x^2 can
simply be represented as the sum of x, x's -- x1 + x2 + x3 ... all
the way up to the x'th x.
Could you give an example of a sum of x's adding up to x^2?
If it's just a matter of, e.g. with x=3, 2 + 2 + 5 = 9, then there will
be many such sequences, won't there?
I *think* he is thinking of multiplication as repeated addition. So 9 = 3*3 is a sum of 3 with itself, three summands: 9 = 3+3+3.
So he really meant
x^2 = x + x + ... + x (x summands)
which works for positive integer values of x.
--
Arturo Magidin
Yes, that's correct, Arturo.

By the way, are there are wide range of integer solutions, for x,y,z, other than 3,4,5 for x^{2n} + y^{2m} = z^2? That's the only one I'm well familiar with.

Here's a general solution method for x^2 + y^2 = z^2

http://fermatslasttheorem.blogspot.com/2005/05/pythagorean-triples-solution.html

z = d[p^2 + q^2]
y = d[p^2 - q^2]
x = d[2pq]

Clearly, x,y,z are only coprime in the case of d =1, by definition. Hence, 3,4,5 are the only coprime solution of x^{2n} + y^{2m} = z^2. So, we can limit the exceptions to my rule not merely to n=m=1, but to a single value for x,y,z where n=m=1.
Peter Percival
2016-08-24 14:09:12 UTC
Post by Jerry Kraus
Post by Arturo Magidin
Post by Peter Percival
Post by Jerry Kraus
I think one of the points I'm exploring here is the relationship
between multiplication and addition, actually. After all, x^2 can
simply be represented as the sum of x, x's -- x1 + x2 + x3 ... all
the way up to the x'th x.
Could you give an example of a sum of x's adding up to x^2?
If it's just a matter of, e.g. with x=3, 2 + 2 + 5 = 9, then there will
be many such sequences, won't there?
I *think* he is thinking of multiplication as repeated addition. So 9 = 3*3 is a sum of 3 with itself, three summands: 9 = 3+3+3.
So he really meant
x^2 = x + x + ... + x (x summands)
which works for positive integer values of x.
--
Arturo Magidin
Yes, that's correct, Arturo.
By the way, are there are wide range of integer solutions, for x,y,z, other than 3,4,5 for x^{2n} + y^{2m} = z^2?
Are you still requiring n,m > 1 ? Because if so x,y,z = 3,4,5 isn't a soln.
Post by Jerry Kraus
That's the only one I'm well familiar with.
Here's a general solution method for x^2 + y^2 = z^2
http://fermatslasttheorem.blogspot.com/2005/05/pythagorean-triples-solution.html
z = d[p^2 + q^2]
y = d[p^2 - q^2]
x = d[2pq]
Clearly, x,y,z are only coprime in the case of d =1, by definition. Hence, 3,4,5 are the only coprime solution of x^{2n} + y^{2m} = z^2. So, we can limit the exceptions to my rule not merely to n=m=1, but to a single value for x,y,z where n=m=1.
--
Do, as a concession to my poor wits, Lord Darlington, just explain
to me what you really mean.
I think I had better not, Duchess. Nowadays to be intelligible is
to be found out. -- Oscar Wilde, Lady Windermere's Fan
Jerry Kraus
2016-08-24 14:37:03 UTC
Post by Peter Percival
Post by Jerry Kraus
Post by Arturo Magidin
Post by Peter Percival
Post by Jerry Kraus
I think one of the points I'm exploring here is the relationship
between multiplication and addition, actually. After all, x^2 can
simply be represented as the sum of x, x's -- x1 + x2 + x3 ... all
the way up to the x'th x.
Could you give an example of a sum of x's adding up to x^2?
If it's just a matter of, e.g. with x=3, 2 + 2 + 5 = 9, then there will
be many such sequences, won't there?
I *think* he is thinking of multiplication as repeated addition. So 9 = 3*3 is a sum of 3 with itself, three summands: 9 = 3+3+3.
So he really meant
x^2 = x + x + ... + x (x summands)
which works for positive integer values of x.
--
Arturo Magidin
Yes, that's correct, Arturo.
By the way, are there are wide range of integer solutions, for x,y,z, other than 3,4,5 for x^{2n} + y^{2m} = z^2?
Are you still requiring n,m > 1 ? Because if so x,y,z = 3,4,5 isn't a soln.
Post by Jerry Kraus
That's the only one I'm well familiar with.
Here's a general solution method for x^2 + y^2 = z^2
http://fermatslasttheorem.blogspot.com/2005/05/pythagorean-triples-solution.html
z = d[p^2 + q^2]
y = d[p^2 - q^2]
x = d[2pq]
Clearly, x,y,z are only coprime in the case of d =1, by definition. Hence, 3,4,5 are the only coprime solution of x^{2n} + y^{2m} = z^2. So, we can limit the exceptions to my rule not merely to n=m=1, but to a single value for x,y,z where n=m=1.
--
Do, as a concession to my poor wits, Lord Darlington, just explain
to me what you really mean.
I think I had better not, Duchess. Nowadays to be intelligible is
to be found out. -- Oscar Wilde, Lady Windermere's Fan
Since there are no integer solutions where n,m >1, obviously I am discussing the case in which n,m = 1.

What I'm emphasizing, and proving here, is that there are actually no coprime integer solutions other than 3,4,5, even with n=m=1, for x^{2n} + y^{2m} = z^2.

Hence, just a single,unique numerical exception to my general rule:

The general idea is that coprimes have no common factors/divisors, and that for two squared numbers to be equal, they must have common factors.
Peter Percival
2016-08-24 15:11:30 UTC
Post by Jerry Kraus
What I'm emphasizing, and proving here, is that there are actually no coprime integer solutions other than 3,4,5, even with n=m=1, for x^{2n} + y^{2m} = z^2.
5,12,13 etc etc
--
Do, as a concession to my poor wits, Lord Darlington, just explain
to me what you really mean.
I think I had better not, Duchess. Nowadays to be intelligible is
to be found out. -- Oscar Wilde, Lady Windermere's Fan
Jerry Kraus
2016-08-24 15:26:30 UTC
Post by Peter Percival
Post by Jerry Kraus
What I'm emphasizing, and proving here, is that there are actually no coprime integer solutions other than 3,4,5, even with n=m=1, for x^{2n} + y^{2m} = z^2.
5,12,13 etc etc
--
Do, as a concession to my poor wits, Lord Darlington, just explain
to me what you really mean.
I think I had better not, Duchess. Nowadays to be intelligible is
to be found out. -- Oscar Wilde, Lady Windermere's Fan
Sorry, Peter, you're right, of course. Still

z = d[p^2 + q^2]
y = d[p^2 - q^2]
x = d[2pq]

does seem to work, as a general solution.
Arturo Magidin
2016-08-24 18:29:08 UTC
Post by Jerry Kraus
Post by Peter Percival
Post by Jerry Kraus
What I'm emphasizing, and proving here, is that there are actually no coprime integer solutions other than 3,4,5, even with n=m=1, for x^{2n} + y^{2m} = z^2.
5,12,13 etc etc
--
Do, as a concession to my poor wits, Lord Darlington, just explain
to me what you really mean.
I think I had better not, Duchess. Nowadays to be intelligible is
to be found out. -- Oscar Wilde, Lady Windermere's Fan
Sorry, Peter, you're right, of course. Still
z = d[p^2 + q^2]
y = d[p^2 - q^2]
x = d[2pq]
does seem to work, as a general solution.
They *are* a general solution. Any choice of p and q that satisfy:

1) p>q
2) p and q are of opposite parity
3) p and q are relatively prime

will yield a solution to a^2+b^2=c^2 in which x,y, and z are pairwise relatively prime. This has been known for centuries.
--
Arturo Magidin
Jerry Kraus
2016-08-24 18:35:45 UTC
Post by Arturo Magidin
Post by Jerry Kraus
Post by Peter Percival
Post by Jerry Kraus
What I'm emphasizing, and proving here, is that there are actually no coprime integer solutions other than 3,4,5, even with n=m=1, for x^{2n} + y^{2m} = z^2.
5,12,13 etc etc
--
Do, as a concession to my poor wits, Lord Darlington, just explain
to me what you really mean.
I think I had better not, Duchess. Nowadays to be intelligible is
to be found out. -- Oscar Wilde, Lady Windermere's Fan
Sorry, Peter, you're right, of course. Still
z = d[p^2 + q^2]
y = d[p^2 - q^2]
x = d[2pq]
does seem to work, as a general solution.
1) p>q
2) p and q are of opposite parity
3) p and q are relatively prime
will yield a solution to a^2+b^2=c^2 in which x,y, and z are pairwise relatively prime. This has been known for centuries.
--
Arturo Magidin
So, Arturo, given the following general proof regarding x^n + y^n = z^n:

http://mathrefresher.blogspot.com/2005/05/coprime-numbers-xn-yn-zn.html

is the original post conjecture regarding x^{2n} + y^{2m} = z^2 having no coprime integer solutions for n,m >1, likely to be accurate, or provable? I'm having my doubts.
Arturo Magidin
2016-08-24 18:48:05 UTC
Post by Jerry Kraus
Post by Arturo Magidin
Post by Jerry Kraus
Post by Peter Percival
Post by Jerry Kraus
What I'm emphasizing, and proving here, is that there are actually no coprime integer solutions other than 3,4,5, even with n=m=1, for x^{2n} + y^{2m} = z^2.
5,12,13 etc etc
--
Do, as a concession to my poor wits, Lord Darlington, just explain
to me what you really mean.
I think I had better not, Duchess. Nowadays to be intelligible is
to be found out. -- Oscar Wilde, Lady Windermere's Fan
Sorry, Peter, you're right, of course. Still
z = d[p^2 + q^2]
y = d[p^2 - q^2]
x = d[2pq]
does seem to work, as a general solution.
1) p>q
2) p and q are of opposite parity
3) p and q are relatively prime
will yield a solution to a^2+b^2=c^2 in which x,y, and z are pairwise relatively prime. This has been known for centuries.
--
Arturo Magidin
http://mathrefresher.blogspot.com/2005/05/coprime-numbers-xn-yn-zn.html
is the original post conjecture regarding x^{2n} + y^{2m} = z^2 having no coprime integer solutions for n,m >1, likely to be accurate, or provable? I'm having my doubts.
As I said elsewhere, that "general proof" (a pretty darn standard argument in number theory), I see absolutely no change about the original equation. In short, I think that page is completely and totally irrelevant to the query at hand.

While the case n=m allows for a proof of the implication: "IF there is a solution, THEN there is a solution in coprime integers" (which means that if you can prove that there are no coprime solutions then there are no solutions at all), when n=/=m I do not see how to prove that. But all that this means is that a proof that there are no coprime solutions does not, in and of itself, imply that there are no solutions at all.

So, to me, this is irrelevant. I also see you have failed to address any of the issues of your meanderings from yesterday, so at this point, I am beginning to wonder what the point is of engaging with you.
--
Arturo Magidin
Arturo Magidin
2016-08-24 18:24:10 UTC
Post by Jerry Kraus
Post by Arturo Magidin
Post by Peter Percival
Post by Jerry Kraus
I think one of the points I'm exploring here is the relationship
between multiplication and addition, actually. After all, x^2 can
simply be represented as the sum of x, x's -- x1 + x2 + x3 ... all
the way up to the x'th x.
Could you give an example of a sum of x's adding up to x^2?
If it's just a matter of, e.g. with x=3, 2 + 2 + 5 = 9, then there will
be many such sequences, won't there?
I *think* he is thinking of multiplication as repeated addition. So 9 = 3*3 is a sum of 3 with itself, three summands: 9 = 3+3+3.
So he really meant
x^2 = x + x + ... + x (x summands)
which works for positive integer values of x.
--
Arturo Magidin
Yes, that's correct, Arturo.
By the way, are there are wide range of integer solutions, for x,y,z, other than 3,4,5 for x^{2n} + y^{2m} = z^2? That's the only one I'm well familiar with.
Here's a general solution method for x^2 + y^2 = z^2
http://fermatslasttheorem.blogspot.com/2005/05/pythagorean-triples-solution.html
z = d[p^2 + q^2]
y = d[p^2 - q^2]
x = d[2pq]
Clearly, x,y,z are only coprime in the case of d =1, by definition. Hence, 3,4,5 are the only coprime solution of x^{2n} + y^{2m} = z^2. So, we can limit the exceptions to my rule not merely to n=m=1, but to a single value for x,y,z where n=m=1.
Any choice of p>q that are relatively and of opposite parity will yield a primitive pythagorean triple (with d=1). The (3,4,5) trile is obtained by taking p=2, q=1. Taking, for example, p=6, q=5 gives x=60, y=11, z=61, which satisfy x^2 + y^2 = 3600 + 121 = 3721 = 61^2 = z^2.

Or p=11, q=6, which gives x=132, y=85, z=157. Etc.
--
Arturo Magidin
Peter Percival
2016-08-23 19:51:34 UTC
Post by Jerry Kraus
Post by Arturo Magidin
Post by Jerry Kraus
Post by Peter Percival
Post by Jerry Kraus
Post by quasi
Post by bassam king karzeddin
Post by bassam king karzeddin
Conjecture: If (x, y, z) are nonzero co prime integers, and
(n, m) are positive integers > 1, then this Diophantine
x^{2n} + y^{2m} = z^2
Doesn't have any integer solution
Conjecture: If (x, y, z) are nonzero co prime integers, and
(n, m) are positive integers > 1, then this Diophantine
x^{2n} + y^{2m} = z^2
Doesn't have any integer solution
x,y,z are coprime
Therefore, x,y,z are all odd, or 1 of them is even, and the other two are odd.
If all three are odd, there can be no integer solution to
x^{2n} + y^{2m} = z^2
because x^{2n} + y^{2m} is always even
z^2 is always odd
and the sqaure root of an odd number cannot be even.
z = { x^{2n} + y^{2m} }^(1/2)
But, again, x,y,z are coprime.
z = { x^{2n} + y^{2m} }^(1/2)
directly implies
For some value v
v1 + v2 ... + va = v1 + v2 ... + va
??
Post by bassam king karzeddin
Where z = v1 + v2 ... + va
{ x^{2n} + y^{2m} }^(1/2) = v1 + v2 ... + va
??
Post by bassam king karzeddin
But, since x,y,z are coprime, no such value v exists.
Q.E.D.
How are you using the information that m,n > 1?
If m,n = 1, there are lots of solutions -- for example
x = 3, y = 4, z = 5.
quasi
Sorry, quasi, you're using one of the earlier versions, which were in error.
Try this one, which is the latest, as posted on Google Groups.
z = { x^{2n} + y^{2m} }^(1/2)
directly implies
For some value v
v1 + v2 ... + vv = v1 + v2 ... + vv
What does that mean? Surely v1 + v2 ... + vv = v1 + v2 ... + vv
whatever z equals?
It's just another way of saying/representing that z^2 = x^{2n} + y^{2m}, but emphasizing the sums of factors to the value of the square.
1. You only specify v; you are also using v for two things: as the generic name for the summands, and as a value of the index. That's *BAD* notation!
2. You don't specify what kind of animals v1, v2, etc. are. Integers? Real numbers? Positive rationals? What?
3. You don't specify how you are picking these objects, or what role they are supposed to play. How do they relate to x and y? To z?
4. Are these just random numbers that add up to z^2, and hence to x^{2n}+y^{2m}? Can we pick any we want?
And then you say "since x,y,z are coprime, no such value v exists", which seems like a nonsequitur given how little information there is about what v is or does, or how it relates to anything at hand.
--
Arturo Magidin
The values of v are integers,
So v1 + v2 ... + vv is 1 or 6 or 18 or ... according as v is 1 or 2 or 3
...?
Post by Jerry Kraus
as the problem pertains to integers. They are factors of x, y, z, insofar as this is, or is not possible. Coprimes, by definition, have no common factors/divisors, hence the relevance of the argument regarding coprimes to the problem. If we have no common factors, we cannot sum factors to the value of a square, to equate values. That's a general sketch of the concepts involved.
--
I have had a tremor of bliss, a wink of heaven, a whisper,
And I would no longer be denied; all things
Proceed to a joyful consummation.
Becket through the pen of Eliot, /Murder in the Cathedral/
Jerry Kraus
2016-08-23 19:58:01 UTC
Post by Peter Percival
Post by Jerry Kraus
Post by Arturo Magidin
Post by Jerry Kraus
Post by Peter Percival
Post by Jerry Kraus
Post by quasi
Post by bassam king karzeddin
Post by bassam king karzeddin
Conjecture: If (x, y, z) are nonzero co prime integers, and
(n, m) are positive integers > 1, then this Diophantine
x^{2n} + y^{2m} = z^2
Doesn't have any integer solution
Conjecture: If (x, y, z) are nonzero co prime integers, and
(n, m) are positive integers > 1, then this Diophantine
x^{2n} + y^{2m} = z^2
Doesn't have any integer solution
x,y,z are coprime
Therefore, x,y,z are all odd, or 1 of them is even, and the
other two are odd.
If all three are odd, there can be no integer solution to
x^{2n} + y^{2m} = z^2
because x^{2n} + y^{2m} is always even
z^2 is always odd
and the sqaure root of an odd number cannot be even.
z = { x^{2n} + y^{2m} }^(1/2)
But, again, x,y,z are coprime.
z = { x^{2n} + y^{2m} }^(1/2)
directly implies
For some value v
v1 + v2 ... + va = v1 + v2 ... + va
??
Post by bassam king karzeddin
Where z = v1 + v2 ... + va
{ x^{2n} + y^{2m} }^(1/2) = v1 + v2 ... + va
??
Post by bassam king karzeddin
But, since x,y,z are coprime, no such value v exists.
Q.E.D.
How are you using the information that m,n > 1?
If m,n = 1, there are lots of solutions -- for example
x = 3, y = 4, z = 5.
quasi
Sorry, quasi, you're using one of the earlier versions, which were in error.
Try this one, which is the latest, as posted on Google Groups.
z = { x^{2n} + y^{2m} }^(1/2)
directly implies
For some value v
v1 + v2 ... + vv = v1 + v2 ... + vv
What does that mean? Surely v1 + v2 ... + vv = v1 + v2 ... + vv
whatever z equals?
It's just another way of saying/representing that z^2 = x^{2n} + y^{2m}, but emphasizing the sums of factors to the value of the square.
1. You only specify v; you are also using v for two things: as the generic name for the summands, and as a value of the index. That's *BAD* notation!
2. You don't specify what kind of animals v1, v2, etc. are. Integers? Real numbers? Positive rationals? What?
3. You don't specify how you are picking these objects, or what role they are supposed to play. How do they relate to x and y? To z?
4. Are these just random numbers that add up to z^2, and hence to x^{2n}+y^{2m}? Can we pick any we want?
And then you say "since x,y,z are coprime, no such value v exists", which seems like a nonsequitur given how little information there is about what v is or does, or how it relates to anything at hand.
--
Arturo Magidin
The values of v are integers,
So v1 + v2 ... + vv is 1 or 6 or 18 or ... according as v is 1 or 2 or 3
...?
Post by Jerry Kraus
as the problem pertains to integers. They are factors of x, y, z, insofar as this is, or is not possible. Coprimes, by definition, have no common factors/divisors, hence the relevance of the argument regarding coprimes to the problem. If we have no common factors, we cannot sum factors to the value of a square, to equate values. That's a general sketch of the concepts involved.
--
I have had a tremor of bliss, a wink of heaven, a whisper,
And I would no longer be denied; all things
Proceed to a joyful consummation.
Becket through the pen of Eliot, /Murder in the Cathedral/
v is a single value summed v times to achieve the value of v^2. Is that reasonably clear? As I've been saying, I think the relationship between addition, multiplication and exponentiation is quite relevant to this particular problem.
Arturo Magidin
2016-08-23 20:09:47 UTC
Post by Jerry Kraus
Post by Peter Percival
Post by Jerry Kraus
Post by Arturo Magidin
Post by Jerry Kraus
Post by Peter Percival
Post by Jerry Kraus
Post by quasi
Post by bassam king karzeddin
Post by bassam king karzeddin
Conjecture: If (x, y, z) are nonzero co prime integers, and
(n, m) are positive integers > 1, then this Diophantine
x^{2n} + y^{2m} = z^2
Doesn't have any integer solution
Conjecture: If (x, y, z) are nonzero co prime integers, and
(n, m) are positive integers > 1, then this Diophantine
x^{2n} + y^{2m} = z^2
Doesn't have any integer solution
x,y,z are coprime
Therefore, x,y,z are all odd, or 1 of them is even, and the
other two are odd.
If all three are odd, there can be no integer solution to
x^{2n} + y^{2m} = z^2
because x^{2n} + y^{2m} is always even
z^2 is always odd
and the sqaure root of an odd number cannot be even.
z = { x^{2n} + y^{2m} }^(1/2)
But, again, x,y,z are coprime.
z = { x^{2n} + y^{2m} }^(1/2)
directly implies
For some value v
v1 + v2 ... + va = v1 + v2 ... + va
??
Post by bassam king karzeddin
Where z = v1 + v2 ... + va
{ x^{2n} + y^{2m} }^(1/2) = v1 + v2 ... + va
??
Post by bassam king karzeddin
But, since x,y,z are coprime, no such value v exists.
Q.E.D.
How are you using the information that m,n > 1?
If m,n = 1, there are lots of solutions -- for example
x = 3, y = 4, z = 5.
quasi
Sorry, quasi, you're using one of the earlier versions, which were in error.
Try this one, which is the latest, as posted on Google Groups.
z = { x^{2n} + y^{2m} }^(1/2)
directly implies
For some value v
v1 + v2 ... + vv = v1 + v2 ... + vv
What does that mean? Surely v1 + v2 ... + vv = v1 + v2 ... + vv
whatever z equals?
It's just another way of saying/representing that z^2 = x^{2n} + y^{2m}, but emphasizing the sums of factors to the value of the square.
1. You only specify v; you are also using v for two things: as the generic name for the summands, and as a value of the index. That's *BAD* notation!
2. You don't specify what kind of animals v1, v2, etc. are. Integers? Real numbers? Positive rationals? What?
3. You don't specify how you are picking these objects, or what role they are supposed to play. How do they relate to x and y? To z?
4. Are these just random numbers that add up to z^2, and hence to x^{2n}+y^{2m}? Can we pick any we want?
And then you say "since x,y,z are coprime, no such value v exists", which seems like a nonsequitur given how little information there is about what v is or does, or how it relates to anything at hand.
--
Arturo Magidin
The values of v are integers,
So v1 + v2 ... + vv is 1 or 6 or 18 or ... according as v is 1 or 2 or 3
...?
Post by Jerry Kraus
as the problem pertains to integers. They are factors of x, y, z, insofar as this is, or is not possible. Coprimes, by definition, have no common factors/divisors, hence the relevance of the argument regarding coprimes to the problem. If we have no common factors, we cannot sum factors to the value of a square, to equate values. That's a general sketch of the concepts involved.
--
I have had a tremor of bliss, a wink of heaven, a whisper,
And I would no longer be denied; all things
Proceed to a joyful consummation.
Becket through the pen of Eliot, /Murder in the Cathedral/
v is a single value summed v times to achieve the value of v^2.
Ouch. If you have a single value v summed v times, then you have vv=v+...+v (v summands)= v^2, NOT "v1+v2+...+vv".

But you had z="v1+v2+...+vv", which apparently meant just meant z=vv=v^2. But we DO NOT know that z itself is a square; this assumption is unwarranted.

Moreover, again, the fact that (x^{2n}+y^{2m})^{1/2} = vv (assuming that were established) does not tell you anything about how the value of v is related to the factors of x^{2n} and y^{2m}. A sum may have factors that have NOTHING to do with the factors of the summand. 3 and 22 add up to 5*5, but 5 has nothing to do with the factors of 3 or of 22.
Post by Jerry Kraus
Is that reasonably clear? As I've been saying, I think the relationship between addition, multiplication and exponentiation is quite relevant to this particular problem.
But just because something is relevant does not mean you have been saying anything coherent about them. You haven't. And just saying something does not actually make it so.
--
Arturo Magidin
Peter Percival
2016-08-23 20:23:25 UTC
Post by Jerry Kraus
Post by Peter Percival
So v1 + v2 ... + vv is 1 or 6 or 18 or ... according as v is 1 or 2 or 3
...?
[...]
Post by Jerry Kraus
v is a single value summed v times to achieve the value of v^2. Is that reasonably clear? As I've been saying, I think the relationship between addition, multiplication and exponentiation is quite relevant to this particular problem.
So with v = 4, we have 4 (that's v1) + 4 (that's v2) + 4 (that's v3) + 4
(that's v4) = 16? You are just saying that v*v = v^2.
--
I have had a tremor of bliss, a wink of heaven, a whisper,
And I would no longer be denied; all things
Proceed to a joyful consummation.
Becket through the pen of Eliot, /Murder in the Cathedral/
Peter Percival
2016-08-23 19:36:23 UTC
Post by Arturo Magidin
Post by Jerry Kraus
Post by Peter Percival
Post by Jerry Kraus
Post by quasi
Post by bassam king karzeddin
Post by bassam king karzeddin
Conjecture: If (x, y, z) are nonzero co prime integers, and
(n, m) are positive integers > 1, then this Diophantine
x^{2n} + y^{2m} = z^2
Doesn't have any integer solution
Conjecture: If (x, y, z) are nonzero co prime integers, and
(n, m) are positive integers > 1, then this Diophantine
x^{2n} + y^{2m} = z^2
Doesn't have any integer solution
x,y,z are coprime
Therefore, x,y,z are all odd, or 1 of them is even, and the other two are odd.
If all three are odd, there can be no integer solution to
x^{2n} + y^{2m} = z^2
because x^{2n} + y^{2m} is always even
z^2 is always odd
and the sqaure root of an odd number cannot be even.
z = { x^{2n} + y^{2m} }^(1/2)
But, again, x,y,z are coprime.
z = { x^{2n} + y^{2m} }^(1/2)
directly implies
For some value v
v1 + v2 ... + va = v1 + v2 ... + va
??
Post by bassam king karzeddin
Where z = v1 + v2 ... + va
{ x^{2n} + y^{2m} }^(1/2) = v1 + v2 ... + va
??
Post by bassam king karzeddin
But, since x,y,z are coprime, no such value v exists.
Q.E.D.
How are you using the information that m,n > 1?
If m,n = 1, there are lots of solutions -- for example
x = 3, y = 4, z = 5.
quasi
Sorry, quasi, you're using one of the earlier versions, which were in error.
Try this one, which is the latest, as posted on Google Groups.
z = { x^{2n} + y^{2m} }^(1/2)
directly implies
For some value v
v1 + v2 ... + vv = v1 + v2 ... + vv
What does that mean? Surely v1 + v2 ... + vv = v1 + v2 ... + vv
whatever z equals?
It's just another way of saying/representing that z^2 = x^{2n} + y^{2m}, but emphasizing the sums of factors to the value of the square.
1. You only specify v; you are also using v for two things: as the generic name for the summands, and as a value of the index. That's *BAD* notation!
Indeed so. When I saw v1 + v2 I thought v_1 + v_2, and when I saw vv I
thought v*v. I decided in the end that v*1 + v*2 ... + v*v was meant.
Post by Arturo Magidin
2. You don't specify what kind of animals v1, v2, etc. are. Integers? Real numbers? Positive rationals? What?
3. You don't specify how you are picking these objects, or what role they are supposed to play. How do they relate to x and y? To z?
4. Are these just random numbers that add up to z^2, and hence to x^{2n}+y^{2m}? Can we pick any we want?
And then you say "since x,y,z are coprime, no such value v exists", which seems like a nonsequitur given how little information there is about what v is or does, or how it relates to anything at hand.
--
I have had a tremor of bliss, a wink of heaven, a whisper,
And I would no longer be denied; all things
Proceed to a joyful consummation.
Becket through the pen of Eliot, /Murder in the Cathedral/
Jerry Kraus
2016-08-23 13:24:33 UTC
Post by bassam king karzeddin
x^{2n} + y^{2m} = z^2
Doesn't have any integer solution
Conjecture: If (x, y, z) are nonzero co prime integers, and (n, m) are positive integers > 1, then this Diophantine equation :

x^{2n} + y^{2m} = z^2

Doesn't have any integer solution

x,y,z are coprime
Therefore, x,y,z are all odd, or 1 of them is even, and the other two are odd.

If all three are odd, there can be no integer solution to

x^{2n} + y^{2m} = z^2

because x^{2n} + y^{2m} is always even
z^2 is always odd
and the sqaure root of an odd number cannot be even.

If one of x,y,z is even, the other two are odd:

z = { x^{2n} + y^{2m} }^(1/2)

But, again, x,y,z are coprime.

z = { x^{2n} + y^{2m} }^(1/2)

directly implies

For some value v
v1 + v2 ... + vv = v1 + v2 ... + vv

Where z = v1 + v2 ... + vv
{ x^{2n} + y^{2m} }^(1/2) = v1 + v2 ... + vv

But, since x,y,z are coprime, no such value v exists.

Q.E.D.
quasi
2016-08-23 17:01:01 UTC
Post by bassam king karzeddin
Post by bassam king karzeddin
Conjecture: If (x, y, z) are nonzero co prime integers,
and (n, m) are positive integers > 1, then this Diophantine
x^{2n} + y^{2m} = z^2
Doesn't have any integer solution
Conjecture: If (x, y, z) are nonzero co prime integers,
and (n, m) are positive integers > 1, then this Diophantine
x^{2n} + y^{2m} = z^2
Doesn't have any integer solution
x,y,z are coprime
Therefore, x,y,z are all odd, or 1 of them is even, and the
other two are odd.
If all three are odd, there can be no integer solution to
x^{2n} + y^{2m} = z^2
because x^{2n} + y^{2m} is always even
z^2 is always odd
and the sqaure root of an odd number cannot be even.
z = { x^{2n} + y^{2m} }^(1/2)
But, again, x,y,z are coprime.
z = { x^{2n} + y^{2m} }^(1/2)
directly implies
For some value v
v1 + v2 ... + vv = v1 + v2 ... + vv
??
Post by bassam king karzeddin
Where z = v1 + v2 ... + vv
{ x^{2n} + y^{2m} }^(1/2) = v1 + v2 ... + vv
??
Post by bassam king karzeddin
But, since x,y,z are coprime, no such value v exists.
Q.E.D.
Looks like nonsense to me.

quasi
Jerry Kraus
2016-08-23 14:04:00 UTC
Post by bassam king karzeddin
x^{2n} + y^{2m} = z^2
Doesn't have any integer solution
Conjecture: If (x, y, z) are nonzero co prime integers, and (n, m) are positive integers > 1, then this Diophantine equation :

x^{2n} + y^{2m} = z^2

Doesn't have any integer solution

x,y,z are coprime
Therefore, x,y,z are all odd, or 1 of them is even, and the other two are odd.

If all three are odd, there can be no integer solution to

x^{2n} + y^{2m} = z^2

because x^{2n} + y^{2m} is always even
z^2 is always odd
and the sqaure root of an odd number cannot be even.

If one of x,y,z is even, the other two are odd:

z = { x^{2n} + y^{2m} }^(1/2)

But, again, x,y,z are coprime.

z = { x^{2n} + y^{2m} }^(1/2)

directly implies

For some value v
v1 + v2 ... + vv = v1 + v2 ... + vv

Where z^2 = v1 + v2 ... + vv
{ x^{2n} + y^{2m} } = v1 + v2 ... + vv

But, since x,y,z are coprime, no such value v exists.

Q.E.D.

I keep finding mistakes in this proof, and keep deleting it from the thread. I think I've got it right, this time, though. Well, hopefully, anyway.
quasi
2016-08-23 17:07:17 UTC
Post by bassam king karzeddin
Post by bassam king karzeddin
Conjecture: If (x, y, z) are nonzero co prime integers,
and (n, m) are positive integers > 1, then this Diophantine
x^{2n} + y^{2m} = z^2
Doesn't have any integer solution
Conjecture: If (x, y, z) are nonzero co prime integers,
and (n, m) are positive integers > 1, then this Diophantine
x^{2n} + y^{2m} = z^2
Doesn't have any integer solution
x,y,z are coprime
Therefore, x,y,z are all odd, or 1 of them is even, and the
other two are odd.
If all three are odd, there can be no integer solution to
x^{2n} + y^{2m} = z^2
because x^{2n} + y^{2m} is always even
z^2 is always odd
and the sqaure root of an odd number cannot be even.
z = { x^{2n} + y^{2m} }^(1/2)
But, again, x,y,z are coprime.
z = { x^{2n} + y^{2m} }^(1/2)
directly implies
For some value v
v1 + v2 ... + vv = v1 + v2 ... + vv
??
Post by bassam king karzeddin
Where z^2 = v1 + v2 ... + vv
{ x^{2n} + y^{2m} } = v1 + v2 ... + vv
??
Post by bassam king karzeddin
But, since x,y,z are coprime, no such value v exists.
Q.E.D.
I keep finding mistakes in this proof, and keep deleting it
Deletes don't work on most modern usenet servers.
Post by bassam king karzeddin
I think I've got it right, this time, though.
No.
Post by bassam king karzeddin
Well, hopefully, anyway.
It's becomes hopeless right at the point where you make
those nonsense claims about the variable v.

quasi
Jerry Kraus
2016-08-24 18:14:44 UTC
Post by bassam king karzeddin
x^{2n} + y^{2m} = z^2
Doesn't have any integer solution
Actually, given the following general proof regarding x^n + y^n = z^n:

http://mathrefresher.blogspot.com/2005/05/coprime-numbers-xn-yn-zn.html

I am beginning to wonder a bit if the conjecture of the original post is in fact accurate, or provable.
Arturo Magidin
2016-08-24 18:44:46 UTC
Post by Jerry Kraus
Post by bassam king karzeddin
x^{2n} + y^{2m} = z^2
Doesn't have any integer solution
http://mathrefresher.blogspot.com/2005/05/coprime-numbers-xn-yn-zn.html
I am beginning to wonder a bit if the conjecture of the original post is in fact accurate, or provable.
I do not understand how what is in that post has anything to do with anything. The argument just shows that *IF* there is a solution with x, y, and z nonzero integers, THEN there is a solution with x, y, and z coprime. The argument to establish that can be done much more straightforwardly than the way given there:

Suppose p is a prime that divides at least two of x, y, and z. We claim that p divides the third.

If p divides x and y, then it divides both x^n and y^n, hence it divides x^n+y^n = z^n. Since p divides z^n, and p is prime, then it divides z. Thus, p divides all three.

If p divides x and z, then p divides x^n and z^n, hence it divides z^n - (x^n) = y^n. Since p is prime, then it divides y. Thus, p divides all three.

If p divides y and z, then p divides y^n and z^n, hence it divides z^n - (y^n) = x^n. Since p is prime, it then divides x. Thus, p divides all three.

Hence, if a prime p divides at least two of x, y, and z, then we can write x=pa, y=pb, z=pc, and we have

p^na^n + p^nb^n = p^nc^n; hence p^n(a^n+b^n) = p^nc^n.

Cancelling p^n, we get a^n + b^n = c^n, so (a,b,c) is also a solution.

Repeating this process as often as necessary, it follows that we may replace any solution with a solution (a,b,c) in which for every prime p, at MOST one of a, b, and c is divisible by p. Therefore, IF a solution exists, then there is a solution in which a, b, and c are pairwise coprime.

How does this relate to the situation at hand?

It is not clear that IF a solution exists then there must be a solution in which all three are coprime, because of the difference in exponents. If you try to argue as before, you would get:

Suppose (x,y,z) is a solution to x^{2n} + y^{2m} = z^2; if p is a prime, and p divides at least two of x, y, and z, then it divides at least two of (x^n, y^n, z), which is a solution to a^2 + b^2 = c^2, and hence it must divide all three. We can write x= pa, y=pb, z=pc. Then we get plugging in we get

p^{2n}a^{2n} + p^{2m}b^{2m} = p^2c^2.

Cancelling p^2 (which we can do because we are assuming m,n are at least positive) we get

p^{2(n-1)}a^{2n} + p^(2(m-1))b^{2m} = c^2.

If n,m>1, then This implies that p divides c^2, hence c; so we can write c=pd, and get

p^{2(n-1)}a^{2n} + p^{2(m-1)}b^{2m} = p^2d^2

and cancelling we can reduce by a further 2 to get

p^{2(n-2)}a^{2n} + p^{2(m-2)}b^{2m} = d^2.

We can continue doing this for min(n,m) steps. If n=m, then we will obtain

a^{2n} + b^{2m} = r^2

for some r, and we have obtained a smaller solution that has one fewer prime factors for a and b.

If n=/=m, then we will exhaust one of the powers but not the other, and the resulting equation will not necessarily yield a solution to the original equation.

Thus, I think that the best we can say in this situation is that in an arbitrary solution and for a given prime p, either all of x, y, and z are divisible by p, or else at most one is; that is, it cannot be that exactly two of them are divisible by p.

That does not preclude asking about the case in which for every prime we fall in the second case: that is, a solution with x, y, and z coprime (which would make x^n, y^m, and z a primitive pythagorean triple).
--
Arturo Magidin
bassam king karzeddin
2016-08-25 10:56:48 UTC
Post by bassam king karzeddin
x^{2n} + y^{2m} = z^2
Doesn't have any integer solution
If we abandon the restriction of the triplets being with common factor, then there are infinitely many solutions, and one can easily formulate it , here is the smallest counter example being with common factor

15^4 + 10^6 = 1025^2
bassam king karzeddin
2016-08-25 16:33:00 UTC
Post by bassam king karzeddin
x^{2n} + y^{2m} = z^2
Doesn't have any integer solution
I would like to add an additional hint for this puzzle to be solved!

"It is impossible for a primitive Pythagoras triplets in integers to be with all terms as a powerful numbers"

Powerful number is an integer which has all of it's prime factors exponents greater than one

Regards
Bassam King karzeddin
25th, Aug, 2016
quasi
2016-08-25 21:57:29 UTC
Post by bassam king karzeddin
Post by bassam king karzeddin
Conjecture: If (x, y, z) are nonzero co prime integers, and
(n, m) are positive integers > 1, then this Diophantine
x^{2n} + y^{2m} = z^2
doesn't have any integer solution
I would like to add an additional hint for this puzzle to be
solved!
"It is impossible for a primitive Pythagoras triplets in integers
to be with all terms as a powerful numbers"
Powerful number is an integer which has all of it's prime
factors exponents greater than one
I don't see how to prove the hint you provided.

quasi
quasi
2016-08-25 21:59:34 UTC
Post by bassam king karzeddin
Conjecture: If (x, y, z) are nonzero co prime integers, and
(n, m) are positive integers > 1, then this Diophantine
x^{2n} + y^{2m} = z^2
doesn't have any integer solution
As a partial result, I can prove the the above claim for the
case where m,n are both even.

quasi
Jack Fearnley
2016-08-26 03:47:48 UTC
Post by bassam king karzeddin
Conjecture: If (x, y, z) are nonzero co prime integers, and (n, m) are
x^{2n} + y^{2m} = z^2
doesn't have any integer solution
As a partial result, I can prove the the above claim for the case where
m,n are both even.
quasi
The even case is classical. The standard proof for fourth powers proves
that no sum of two fourth powers can be a square. See, for example, Hardy
& Wright Thm 226.

For the more general case, Darmon and Granville have proved that if the
sum of the reciprocals of the exponents in the general case is less than
one, then there are at most a finite number of solutions.

That is, given r,s,t such that 1/r + 1/s +1/t < 1 and coprime x,y,z,
then
x^r + y^s = z^t
has only a finite number of solutions.

I realize there may still be a heck of a lot of them though :-)

Best Regards,
Jack Fearnley
quasi
2016-08-26 07:39:39 UTC
Post by Jack Fearnley
Post by quasi
Post by bassam king karzeddin
Conjecture: If (x, y, z) are nonzero co prime integers, and
(n, m) are positive integers > 1, then this Diophantine
x^{2n} + y^{2m} = z^2
doesn't have any integer solution
As a partial result, I can prove the the above claim for the
case where m,n are both even.
The even case is classical. The standard proof for fourth
powers proves that no sum of two fourth powers can be a square.
See, for example, Hardy & Wright Thm 226.
I have a few texts (including the one you reference) that deal
with the n = 4 case of FLT, but from memory (without bothering
to check), I thought that what they showed was that no
_difference_ of positive integer 4th powers can be a positive
integer square. I see now that they prove it for sums, not
differences, and it thus appears that my proof is essentially
a duplicate of the standard proof.
Post by Jack Fearnley
For the more general case, Darmon and Granville have proved
that if the sum of the reciprocals of the exponents in the
general case is less than one, then there are at most a finite
number of solutions.
That is, given r,s,t such that 1/r + 1/s +1/t < 1 and coprime
x,y,z,then
x^r + y^s = z^t
has only a finite number of solutions.
I realize there may still be a heck of a lot of them though :-)
The OP's claim is that if m,n > 1, there are _no_ solutions.
He also indicated (although perhaps he was joking) that the
level of difficulty for the proof is no higher than that of
a challenge problem for talented high school students.

Does the OP actually have a proof? I'm skeptical.

quasi
Arturo Magidin
2016-08-26 03:48:42 UTC
Post by quasi
Post by bassam king karzeddin
Conjecture: If (x, y, z) are nonzero co prime integers, and
(n, m) are positive integers > 1, then this Diophantine
x^{2n} + y^{2m} = z^2
doesn't have any integer solution
As a partial result, I can prove the the above claim for the
case where m,n are both even.
Yes, well, you'd be ripping off Fermat there, wouldn't you? ;-)

I believe the point of the other hint ("it is impossible for a primitive pythagorean triple to consist only of integers in which every prime factor occurs more than once", or words to that effect), is that if x, y, and z were a solution in coprime integers, then x^n, y^m, and z would be a primittive pythagorean triple. If we can prove that for every prime p, if p|z then p^2|z, then the result would follow by contradiction. Though I don't see how to do that off the top of my head. I also don't know offhand of a proof that no such primitive pythagorean triple exists...
--
Arturo Magidin
quasi
2016-08-26 07:48:04 UTC
Post by Arturo Magidin
Post by quasi
Post by bassam king karzeddin
Conjecture: If (x, y, z) are nonzero co prime integers, and
(n, m) are positive integers > 1, then this Diophantine
x^{2n} + y^{2m} = z^2
doesn't have any integer solution
As a partial result, I can prove the the above claim for the
case where m,n are both even.
Yes, well, you'd be ripping off Fermat there, wouldn't you? ;-)
Unwittingly, yes.
Post by Arturo Magidin
I believe the point of the other hint ("it is impossible for
a primitive pythagorean triple to consist only of integers in
which every prime factor occurs more than once", or words to
that effect), is that if x, y, and z were a solution in coprime
integers, then x^n, y^m, and z would be a primittive pythagorean
triple. If we can prove that for every prime p, if p|z then
Yes, I understood that aspect of the hint.
Post by Arturo Magidin
Though I don't see how to do that off the top of my head.
There's no obvious way to do that.
Post by Arturo Magidin
I also don't know offhand of a proof that no such
primitive pythagorean triple exists...
I don't believe that it's a known result.

I challenge the OP to either prove it or give a reference.

quasi
bassam king karzeddin
2016-08-27 11:34:35 UTC
Post by quasi
Post by Arturo Magidin
Post by quasi
Post by bassam king karzeddin
Conjecture: If (x, y, z) are nonzero co prime integers, and
(n, m) are positive integers > 1, then this Diophantine
x^{2n} + y^{2m} = z^2
doesn't have any integer solution
As a partial result, I can prove the the above claim for the
case where m,n are both even.
Yes, well, you'd be ripping off Fermat there, wouldn't you? ;-)
Unwittingly, yes.
Post by Arturo Magidin
I believe the point of the other hint ("it is impossible for
a primitive pythagorean triple to consist only of integers in
which every prime factor occurs more than once", or words to
that effect), is that if x, y, and z were a solution in coprime
integers, then x^n, y^m, and z would be a primittive pythagorean
triple. If we can prove that for every prime p, if p|z then
Yes, I understood that aspect of the hint.
Post by Arturo Magidin
Though I don't see how to do that off the top of my head.
There's no obvious way to do that.
Post by Arturo Magidin
I also don't know offhand of a proof that no such
primitive pythagorean triple exists...
I don't believe that it's a known result.
I challenge the OP to either prove it or give a reference.
quasi
Welcome come back Quasi

I tried hard to respond from my original account at sci.math despite accessing my account, but they (moderators) never allowed me despite contacting them several times in this regard, any way this is irrelevant issue.

You may know that I love the challenge, but I don't want to spoil the joy of it right now,

About references, I really didn't borrow the idea from any where, as you see all references provided here show the almost the same elementary common knowledge that were known since many centuries in so many different forms

Note also, I posted those challenges to the world mathematicians at other forums, as Quora and MSE, but the tendency there to suppress issues by non-professional mathematicians, more especially at MSE, where they deleted many of my mere discoveries without any convenient reasons, but some are still surviving

This is not to address the issue, but I confess also that, there is only one step in my proof (most likely correct), but under heavy peer review (by myself only)

More over, I'm not so afraid of your counter examples this time!

Regards,
quasi
2016-08-27 12:20:18 UTC
Welcome back Quasi
Thanks, but I'm the one who's been here -- you're the one who's
been away, so welcome back to you!

When you first initialed this thread, I didn't recognize your
username, but now I remember you!
I tried hard to respond from my original account at sci.math
despite accessing my account, but they (moderators) never
allowed me despite contacting them several times in this
regard, any way this is irrelevant issue.
sci.math has no moderators.
You may know that I love the challenge, but I don't want to
spoil the joy of it right now,
Fair enough.
About references, I really didn't borrow the idea from any where,
as you see all references provided here show the almost the same
elementary common knowledge that were known since many centuries
in so many different forms
OK.
Note also, I posted those challenges to the world
mathematicians at other forums, as Quora and MSE, but the
tendency there to suppress issues by non-professional
mathematicians, more especially at MSE, where they deleted
many of my mere discoveries without any convenient reasons,
but some are still surviving
This is not to address the issue, but I confess also that,
there is only one step in my proof (most likely correct),
but under heavy peer review (by myself only)
Moreover, I'm not so afraid of your counterexamples
this time!
Ah yes -- as I recall (going back 10 years), you proposed a
number of very interesting conjectures, but for some of them,
I found counterxamples.

quasi
bassam king karzeddin
2016-08-27 13:18:56 UTC
bassam king karzeddin wrote:> >Welcome back Quasi
Thanks, but I'm the one who's been here -- you're the one who's
been away, so welcome back to you!
sci.math has no moderators.
Thanks Quasi

It is written in green color, moderator has to to approve the reply at sci.math, even you reply they never show the reply or any response to questions

Wonder why is this!

Can any one from moderators or sci.math explain this

Thanks again
Peter Percival
2016-08-27 14:33:29 UTC
Post by bassam king karzeddin
bassam king karzeddin wrote:> >Welcome back Quasi
Thanks, but I'm the one who's been here -- you're the one who's
been away, so welcome back to you!
sci.math has no moderators.
Thanks Quasi
The best way to post to, and read posts at, sci.math is by using a
newsreader that subscribes to a news server. Your ISP may have a news
server (though many of them have given up) or you may use, e.g.,
albasani that is free and text only.
Post by bassam king karzeddin
It is written in green color, moderator has to to approve the reply at sci.math, even you reply they never show the reply or any response to questions
Wonder why is this!
Can any one from moderators or sci.math explain this
Thanks again
--
Do, as a concession to my poor wits, Lord Darlington, just explain
to me what you really mean.
I think I had better not, Duchess. Nowadays to be intelligible is
to be found out. -- Oscar Wilde, Lady Windermere's Fan
quasi
2016-08-27 22:38:26 UTC
Post by bassam king karzeddin
Post by quasi
Welcome back Quasi
Thanks, but I'm the one who's been here -- you're the one who's
been away, so welcome back to you!
I tried hard to respond from my original account at sci.math
despite accessing my account, but they (moderators) never
allowed me despite contacting them several times in this
regard, any way this is irrelevant issue.
sci.math has no moderators.
It is written in green color, moderator has to to approve the
or any response to questions
Wonder why is this!
Math Forum is not sci.math. They enable limited read-write access
(filtered in both directions) to sci.math for users having a
Math Forum account. Thus, if you access sci.math via Math Forum,
your view of sci.math is a filtered view, moderated by staff at
Math Forum.

sci.math is part of usenet -- a worldwide network which hosts
various text and binary forums (called newsgroups). Some of the
newsgroups are moderated, some are not. The newsgroup "sci.math"
is an unmoderated group -- no one is in charge.

To access sci.math directly (not via Math Forum), you need two
things:

(1) a user account on some usenet server.

(2) an NNTP client program (i.e., a usenet newsreader program).

There are many usenet servers -- some require paid subscriptions
(typically a monthly charge), some are free. The one I use is
giganews which (for their least expensive package) costs about
five dollars a month.

There are also many NNTP client programs -- some require purchase,
some are free. The one I use is called Forte Agent. It's not free
(I don't remember the cost -- I have an old version) but it's a
one-time purchase, and it's not very expensive.
Post by bassam king karzeddin
Can any one from moderators or sci.math explain this
As I previously said, sci.math itself has no moderators, but
apparently Math Forum moderates its view of sci.math. To find
out more about that, you would have to ask the staff at
Math Forum.

quasi
Simon Roberts
2016-08-26 02:54:40 UTC
"
bassam king karzeddin

Aug 22

Conjecture: If (x, y, z) are nonzero co prime integers, and (n, m) are positive integers > 1, then this Diophantine equation :

x^{2n} + y^{2m} = z^2

Doesn't have any integer solution

"
Hint reiterated [paraphrased].

"Formulas for generating Pythagorean triples"

https://en.wikipedia.org/wiki/Formulas_for_generating_Pythagorean_triples.
quasi
2016-08-26 07:56:37 UTC
Post by bassam king karzeddin
Conjecture: If (x, y, z) are nonzero co prime integers, and
(n, m) are positive integers > 1, then this Diophantine
x^{2n} + y^{2m} = z^2
doesn't have any integer solution
Hint reiterated [paraphrased].
"Formulas for generating Pythagorean triples"
https://en.wikipedia.org/wiki/Formulas_for_generating_Pythagorean_triples.
Those formulas are well known.

The OP's hint was that if (x,y,z) is a primitive pythagorean
triple, then it's not possible for x,y,z to all be "powerful
numbers" (i.e., each of their prime factors occurs with
exponent at least 2).

I don't see how to prove that claim.

I also don't see how to resolve the original problem even
if we accept that the claim of the hint is a true claim.

quasi
konyberg
2016-08-26 21:40:18 UTC
Post by quasi
Post by bassam king karzeddin
Conjecture: If (x, y, z) are nonzero co prime integers, and
(n, m) are positive integers > 1, then this Diophantine
x^{2n} + y^{2m} = z^2
doesn't have any integer solution
Hint reiterated [paraphrased].
"Formulas for generating Pythagorean triples"
https://en.wikipedia.org/wiki/Formulas_for_generating_Pythagorean_triples.
Those formulas are well known.
The OP's hint was that if (x,y,z) is a primitive pythagorean
triple, then it's not possible for x,y,z to all be "powerful
numbers" (i.e., each of their prime factors occurs with
exponent at least 2).
I don't see how to prove that claim.
I also don't see how to resolve the original problem even
if we accept that the claim of the hint is a true claim.
quasi
I think that OP hs looked at this and think this is for all

x^4 + y^4 = z^2.

This equation has no solution in the naturals.

And this can easelly be proved.

KON
quasi
2016-08-26 22:43:18 UTC
Post by konyberg
Post by quasi
Post by bassam king karzeddin
Conjecture: If (x, y, z) are nonzero co prime integers, and
(n, m) are positive integers > 1, then this Diophantine
x^{2n} + y^{2m} = z^2
doesn't have any integer solution
Hint reiterated [paraphrased].
"Formulas for generating Pythagorean triples"
https://en.wikipedia.org/wiki/Formulas_for_generating_Pythagorean_triples.
Those formulas are well known.
The OP's hint was that if (x,y,z) is a primitive pythagorean
triple, then it's not possible for x,y,z to all be "powerful
numbers" (i.e., each of their prime factors occurs with
exponent at least 2).
I don't see how to prove that claim.
I also don't see how to resolve the original problem even
if we accept that the claim of the hint is a true claim.
I think that OP hs looked at this and think this is for all
x^4 + y^4 = z^2.
No, that's not the OP's interpretation, given that the OP
(bassam king karzeddin) also posted the example
15^4 + 10^6 = 1025^2.

quasi
quasi
2016-08-28 09:39:42 UTC
Conjecture: If (x, y, z) are nonzero coprime integers,
and (n, m) are positive integers > 1, then this Diophantine
x^{2n} + y^{2m} = z^2
doesn't have any integer solution.
I can prove this now.

It's actually quite easy (but fun, nevertheless).

Thanks bassam -- sorry for doubting you on this one. But don't
worry -- based on history, I'm sure I'll have good reason to
doubt some of your future claims.

In fact, although it's now clear that your claim with regard to
the above diophantine equation is valid, I still doubt that you
have a proof of your other claim -- the one where you asserted
that there does not exist a primitive pythagorean triple for
which all three components are powerful numbers.

quasi
quasi
2016-08-28 12:14:12 UTC
Post by quasi
Conjecture: If (x, y, z) are nonzero coprime integers,
and (n, m) are positive integers > 1, then this Diophantine
x^{2n} + y^{2m} = z^2
doesn't have any integer solution.
I can prove this now.
Oops -- I take it back.

My proof has a silly (essentially unfixable) error.

But my previously announced proof (that there are no solutions
for the case where gcd(m,n) > 1) is still good.
Post by quasi
It's actually quite easy (but fun, nevertheless).
Everything's easy to prove once an undetected logical error is
Post by quasi
Thanks bassam -- sorry for doubting you on this one.
I'm doubting once again.
Post by quasi
But don't worry -- based on history, I'm sure I'll have good
reason to doubt some of your future claims.
In fact, although it's now clear that your claim with regard to
the above diophantine equation is valid,
It seemed clear, but it was an illusion.
Post by quasi
I still doubt that you have a proof of your other claim -- the
one where you asserted that there does not exist a primitive
pythagorean triple for which all three components are powerful
numbers.
That doubt is still there -- even stronger now.

quasi
Peter Percival
2016-08-28 12:21:29 UTC
Post by quasi
Post by quasi
Conjecture: If (x, y, z) are nonzero coprime integers,
and (n, m) are positive integers > 1, then this Diophantine
x^{2n} + y^{2m} = z^2
doesn't have any integer solution.
I can prove this now.
Oops -- I take it back.
My proof has a silly (essentially unfixable) error.
But my previously announced proof (that there are no solutions
for the case where gcd(m,n) > 1) is still good.
Beggin' yer pardon Mr q, and just for my edification, the original

* x,y,z non-zero coprime integers

* n,m integers > 1.

Does your gcd(m,n) > 1 replace the second of those?
--
Do, as a concession to my poor wits, Lord Darlington, just explain
to me what you really mean.
I think I had better not, Duchess. Nowadays to be intelligible is
to be found out. -- Oscar Wilde, Lady Windermere's Fan
quasi
2016-08-28 20:43:45 UTC
Post by Peter Percival
Post by quasi
Post by quasi
Conjecture: If (x, y, z) are nonzero coprime integers,
and (n, m) are positive integers > 1, then this Diophantine
x^{2n} + y^{2m} = z^2
doesn't have any integer solution.
I can prove this now.
Oops -- I take it back.
My proof has a silly (essentially unfixable) error.
But my previously announced proof (that there are no solutions
for the case where gcd(m,n) > 1) is still good.
Beggin' yer pardon Mr q, and just for my edification, the original
* x,y,z non-zero coprime integers
* n,m integers > 1.
Does your gcd(m,n) > 1 replace the second of those?

Of course, for positive integers m,n

gcd(m,n) > 1 => n,m > 1

so in that sense, it could be viewed as a replacement.

quasi
Virgil
2016-08-28 21:48:40 UTC
Post by quasi
Post by Peter Percival
Post by quasi
Post by quasi
Conjecture: If (x, y, z) are nonzero coprime integers,
and (n, m) are positive integers > 1, then this Diophantine
x^{2n} + y^{2m} = z^2
doesn't have any integer solution.
I can prove this now.
Oops -- I take it back.
My proof has a silly (essentially unfixable) error.
But my previously announced proof (that there are no solutions
for the case where gcd(m,n) > 1) is still good.
Beggin' yer pardon Mr q, and just for my edification, the original
* x,y,z non-zero coprime integers
* n,m integers > 1.
Does your gcd(m,n) > 1 replace the second of those?
Of course, for positive integers m,n
gcd(m,n) > 1 => n,m > 1
so in that sense, it could be viewed as a replacement.
quasi
Does your "n,m > 1 "mean "m > 1 AND n > 1"?
--
Virgil
"Mit der Dummheit kampfen Gotter selbst vergebens." (Schiller)
bassam king karzeddin
2016-08-29 15:13:24 UTC
Post by Virgil
Post by quasi
Post by Peter Percival
Post by quasi
Post by quasi
Conjecture: If (x, y, z) are nonzero coprime integers,
and (n, m) are positive integers > 1, then this Diophantine
x^{2n} + y^{2m} = z^2
doesn't have any integer solution.
I can prove this now.
Oops -- I take it back.
My proof has a silly (essentially unfixable) error.
But my previously announced proof (that there are no solutions
for the case where gcd(m,n) > 1) is still good.
Beggin' yer pardon Mr q, and just for my edification, the original
* x,y,z non-zero coprime integers
* n,m integers > 1.
Does your gcd(m,n) > 1 replace the second of those?
Of course, for positive integers m,n
gcd(m,n) > 1 => n,m > 1
so in that sense, it could be viewed as a replacement.
quasi
Does your "n,m > 1 "mean "m > 1 AND n > 1"?
--
Virgil
"Mit der Dummheit kampfen Gotter selbst vergebens." (Schiller)
Good note of you, in the article given in reference at MSE, I defined (n, m) being two distinct positive integers, which includes generally the root of (quadratic, cubic, 4th , quintic, 6th, 7th, ..., nth) degree equations but not the first degree equation, may be I was hinting other formula that I will provide a link to it

Thanks for the note Virgil
bassam king karzeddin
2016-08-28 18:05:22 UTC
Post by quasi
Post by quasi
Conjecture: If (x, y, z) are nonzero coprime integers,
and (n, m) are positive integers > 1, then this Diophantine
x^{2n} + y^{2m} = z^2
doesn't have any integer solution.
I can prove this now.
Oops -- I take it back.
My proof has a silly (essentially unfixable) error.
But my previously announced proof (that there are no solutions
for the case where gcd(m,n) > 1) is still good.
Post by quasi
It's actually quite easy (but fun, nevertheless).
Everything's easy to prove once an undetected logical error is
Post by quasi
Thanks bassam -- sorry for doubting you on this one.
I'm doubting once again.
Post by quasi
But don't worry -- based on history, I'm sure I'll have good
reason to doubt some of your future claims.
In fact, although it's now clear that your claim with regard to
the above diophantine equation is valid,
It seemed clear, but it was an illusion.
Post by quasi
I still doubt that you have a proof of your other claim -- the
one where you asserted that there does not exist a primitive
pythagorean triple for which all three components are powerful
numbers.
That doubt is still there -- even stronger now.
quasi
You have the full right to be skeptical Quasi, even if I post an assumed correct proof for this puzzle!

But, to save time, I posted a formula at SME, which is quite relevant in this regard

http://mathoverflow.net/questions/208169/quintic-equation?sgp=2

Bassam Karzeddin
quasi
2016-08-28 21:36:16 UTC
Post by bassam king karzeddin
Post by quasi
Post by quasi
Conjecture: If (x, y, z) are nonzero coprime integers,
and (n, m) are positive integers > 1, then this Diophantine
x^{2n} + y^{2m} = z^2
doesn't have any integer solution.
I can prove this now.
Oops -- I take it back.
My proof has a silly (essentially unfixable) error.
But my previously announced proof (that there are no solutions
for the case where gcd(m,n) > 1) is still good.
Post by quasi
It's actually quite easy (but fun, nevertheless).
Everything's easy to prove once an undetected logical error is
Post by quasi
Thanks bassam -- sorry for doubting you on this one.
I'm doubting once again.
Post by quasi
But don't worry -- based on history, I'm sure I'll have good
reason to doubt some of your future claims.
In fact, although it's now clear that your claim with regard to
the above diophantine equation is valid,
It seemed clear, but it was an illusion.
Post by quasi
I still doubt that you have a proof of your other claim -- the
one where you asserted that there does not exist a primitive
pythagorean triple for which all three components are powerful
numbers.
That doubt is still there -- even stronger now.
You have the full right to be skeptical Quasi, even if I post
an assumed correct proof for this puzzle!
Will you eventually post your proof in sci.math?
Post by bassam king karzeddin
But, to save time, I posted a formula at SME, which is quite
relevant in this regard
http://mathoverflow.net/questions/208169/quintic-equation?sgp=2
I don't see how it relates to the problem of this thread.

quasi
bassam king karzeddin
2016-08-29 16:19:44 UTC
Post by quasi
Post by bassam king karzeddin
Post by quasi
Post by quasi
Conjecture: If (x, y, z) are nonzero coprime integers,
and (n, m) are positive integers > 1, then this Diophantine
x^{2n} + y^{2m} = z^2
doesn't have any integer solution.
I can prove this now.
Oops -- I take it back.
My proof has a silly (essentially unfixable) error.
But my previously announced proof (that there are no solutions
for the case where gcd(m,n) > 1) is still good.
Post by quasi
It's actually quite easy (but fun, nevertheless).
Everything's easy to prove once an undetected logical error is
Post by quasi
Thanks bassam -- sorry for doubting you on this one.
I'm doubting once again.
Post by quasi
But don't worry -- based on history, I'm sure I'll have good
reason to doubt some of your future claims.
In fact, although it's now clear that your claim with regard to
the above diophantine equation is valid,
It seemed clear, but it was an illusion.
Post by quasi
I still doubt that you have a proof of your other claim -- the
one where you asserted that there does not exist a primitive
pythagorean triple for which all three components are powerful
numbers.
That doubt is still there -- even stronger now.
You have the full right to be skeptical Quasi, even if I post
an assumed correct proof for this puzzle!
Will you eventually post your proof in sci.math?
Post by bassam king karzeddin
But, to save time, I posted a formula at SME, which is quite
relevant in this regard
http://mathoverflow.net/questions/208169/quintic-equation?sgp=2
I don't see how it relates to the problem of this thread.
quasi
Actually, sci.math is the best place to post good results, because no one is allowed and capable of removing the content of someone else for his own innocent purpose or anything else , nor he is allowed to modify or re edit it, and this is applicable to the original poster (OP), once the post had been replied!

So distinguished from other Forums as MSE or Quora or else, where they keep everything under the control of big cats

Please, someone correct me if I'm wrong in my understanding to the policy of sci.math

At least the link provided above is relevant ethically, how?

Note that, the question contains a proved discovery with old documented references, but unpublished and (before Wiles proof of FLT) - (Date is so irrelevant to this matter), are faced with so many downvotes in order to be deleted completely as usual (with me),

whereas an ordinary answer (nothing new) by well known professional gets many upvotes, which resulted in preventing them from deleting the content of my question till now!!

So, where shall I post the proof then!?

I'm not professional and can't make the simple process that many can do easily, as introduction, theme, history, references, typing, ...etc, ... etc

But certainly, I can solve the problem (and by elementary means only!)

Also, I offer it as challenge that anyone interested can score it, and hence take it completely, because the victory isn't for the conjecture (in most cases) but for the (standard or official prover: a well known political mathematician, who knows and obey the rules of documentation)

I also realised it is the heart of few unsolved problems presently as beal's conjecture for example!, where at least it can tell you something very relevant, as this:

"If an assumed solution in integers exists to beal's conjecture with no common factor for the three bases, so must the three integer exponent be with no common factors, which finally results in a contradiction!"

This is only one aspect of the problem, but truly speaking there may be many more to it, as determining which Diophantine equations are solvable,

OR: I'm afraid

"The code breaking", I hope not!

Regards

Bassam Karzeddin
Peter Percival
2016-08-29 17:01:53 UTC
Post by bassam king karzeddin
So, where shall I post the proof then!?
Right here. Stop yacking and get on with it.
--
Do, as a concession to my poor wits, Lord Darlington, just explain
to me what you really mean.
I think I had better not, Duchess. Nowadays to be intelligible is
to be found out. -- Oscar Wilde, Lady Windermere's Fan
bassam king karzeddin
2016-08-29 17:31:01 UTC
Post by Peter Percival
Post by bassam king karzeddin
So, where shall I post the proof then!?
Right here. Stop yacking and get on with it.
--
Do, as a concession to my poor wits, Lord Darlington, just explain
to me what you really mean.
I think I had better not, Duchess. Nowadays to be intelligible is
to be found out. -- Oscar Wilde, Lady Windermere's Fan
This one link here is an introduction:

http://hsm.stackexchange.com/questions/3257/sum-of-like-powers-in-real-numbers

And if you are a member there, please at least don't downvote it

Thanks
Arturo Magidin
2016-08-29 18:07:28 UTC
Post by bassam king karzeddin
Post by Peter Percival
Post by bassam king karzeddin
So, where shall I post the proof then!?
Right here. Stop yacking and get on with it.
--
Do, as a concession to my poor wits, Lord Darlington, just explain
to me what you really mean.
I think I had better not, Duchess. Nowadays to be intelligible is
to be found out. -- Oscar Wilde, Lady Windermere's Fan
http://hsm.stackexchange.com/questions/3257/sum-of-like-powers-in-real-numbers
That's not an "introduction" to anything, most especially not an introduction to a proof of your claims.

So, are you going to continue to dance or are you going to actually put up?

Either shit, or get off the pot.
--
Arturo Magidin
bassam king karzeddin
2016-08-29 19:39:29 UTC
Post by Arturo Magidin
Post by bassam king karzeddin
Post by Peter Percival
Post by bassam king karzeddin
So, where shall I post the proof then!?
Right here. Stop yacking and get on with it.
--
Do, as a concession to my poor wits, Lord Darlington, just explain
to me what you really mean.
I think I had better not, Duchess. Nowadays to be intelligible is
to be found out. -- Oscar Wilde, Lady Windermere's Fan
http://hsm.stackexchange.com/questions/3257/sum-of-like-powers-in-real-numbers
That's not an "introduction" to anything, most especially not an introduction to a proof of your claims.
So, are you going to continue to dance or are you going to actually put up?
Either shit, or get off the pot.
--
Arturo Magidin
Oh, ya, that is actually NOTHING better than shit, but I got a better idea, and you hopefully would come to know about it!
Arturo Magidin
2016-08-29 21:07:30 UTC
Post by bassam king karzeddin
Post by Arturo Magidin
Post by bassam king karzeddin
Post by Peter Percival
Post by bassam king karzeddin
So, where shall I post the proof then!?
Right here. Stop yacking and get on with it.
--
Do, as a concession to my poor wits, Lord Darlington, just explain
to me what you really mean.
I think I had better not, Duchess. Nowadays to be intelligible is
to be found out. -- Oscar Wilde, Lady Windermere's Fan
http://hsm.stackexchange.com/questions/3257/sum-of-like-powers-in-real-numbers
That's not an "introduction" to anything, most especially not an introduction to a proof of your claims.
So, are you going to continue to dance or are you going to actually put up?
Either shit, or get off the pot.
Oh, ya, that is actually NOTHING better than shit, but I got a better idea, and you hopefully would come to know about it!
And Karzeddin misses the point. AGAIN.

https://en.wikipedia.org/wiki/Shit_or_get_off_the_pot

You'll note that, as opposed to *your* link, this one is directly relevant to the issue.

Pull quote: ""Shit or get off the pot" is a common English language colloquial expression, used to imply a person should follow up their stated intentions with action. "
--
Arturo Magidin
bassam king karzeddin
2016-08-30 11:30:35 UTC
Post by Arturo Magidin
Post by bassam king karzeddin
Post by Arturo Magidin
Post by bassam king karzeddin
Post by Peter Percival
Post by bassam king karzeddin
So, where shall I post the proof then!?
Right here. Stop yacking and get on with it.
--
Do, as a concession to my poor wits, Lord Darlington, just explain
to me what you really mean.
I think I had better not, Duchess. Nowadays to be intelligible is
to be found out. -- Oscar Wilde, Lady Windermere's Fan
http://hsm.stackexchange.com/questions/3257/sum-of-like-powers-in-real-numbers
That's not an "introduction" to anything, most especially not an introduction to a proof of your claims.
So, are you going to continue to dance or are you going to actually put up?
Either shit, or get off the pot.
Oh, ya, that is actually NOTHING better than shit, but I got a better idea, and you hopefully would come to know about it!
And Karzeddin misses the point. AGAIN.
https://en.wikipedia.org/wiki/Shit_or_get_off_the_pot
You'll note that, as opposed to *your* link, this one is directly relevant to the issue.
Pull quote: ""Shit or get off the pot" is a common English language colloquial expression, used to imply a person should follow up their stated intentions with action. "
--
Arturo Magidin
Sorry for the misunderstanding Dr. Arturo Magidin, forgive my poor English as it is my last language, that I had learnt from non native english speakers

Also, I will not run away for long from my obligation here

Regards
Arturo Magidin
2016-08-30 14:32:11 UTC
Post by bassam king karzeddin
Post by Arturo Magidin
Post by bassam king karzeddin
Post by Arturo Magidin
Post by bassam king karzeddin
Post by Peter Percival
Post by bassam king karzeddin
So, where shall I post the proof then!?
Right here. Stop yacking and get on with it.
--
Do, as a concession to my poor wits, Lord Darlington, just explain
to me what you really mean.
I think I had better not, Duchess. Nowadays to be intelligible is
to be found out. -- Oscar Wilde, Lady Windermere's Fan
http://hsm.stackexchange.com/questions/3257/sum-of-like-powers-in-real-numbers
That's not an "introduction" to anything, most especially not an introduction to a proof of your claims.
So, are you going to continue to dance or are you going to actually put up?
Either shit, or get off the pot.
Oh, ya, that is actually NOTHING better than shit, but I got a better idea, and you hopefully would come to know about it!
And Karzeddin misses the point. AGAIN.
https://en.wikipedia.org/wiki/Shit_or_get_off_the_pot
You'll note that, as opposed to *your* link, this one is directly relevant to the issue.
Pull quote: ""Shit or get off the pot" is a common English language colloquial expression, used to imply a person should follow up their stated intentions with action. "
--
Arturo Magidin
Sorry for the misunderstanding Dr. Arturo Magidin, forgive my poor English as it is my last language,
English is not my first language either. For that matter it's not my second language, or my third language. And yet, I manage.
--
Arturo Magidin
Simon C. Roberts
2016-08-30 15:05:06 UTC
Post by Arturo Magidin
Post by bassam king karzeddin
Post by Arturo Magidin
Post by bassam king karzeddin
Post by Arturo Magidin
Post by bassam king karzeddin
Post by Peter Percival
Post by bassam king karzeddin
So, where shall I post the proof then!?
Right here. Stop yacking and get on with it.
--
Do, as a concession to my poor wits, Lord Darlington, just explain
to me what you really mean.
I think I had better not, Duchess. Nowadays to be intelligible is
to be found out. -- Oscar Wilde, Lady Windermere's Fan
http://hsm.stackexchange.com/questions/3257/sum-of-like-powers-in-real-numbers
That's not an "introduction" to anything, most especially not an introduction to a proof of your claims.
So, are you going to continue to dance or are you going to actually put up?
Either shit, or get off the pot.
Oh, ya, that is actually NOTHING better than shit, but I got a better idea, and you hopefully would come to know about it!
And Karzeddin misses the point. AGAIN.
https://en.wikipedia.org/wiki/Shit_or_get_off_the_pot
You'll note that, as opposed to *your* link, this one is directly relevant to the issue.
Pull quote: ""Shit or get off the pot" is a common English language colloquial expression, used to imply a person should follow up their stated intentions with action. "
--
Arturo Magidin
Sorry for the misunderstanding Dr. Arturo Magidin, forgive my poor English as it is my last language,
English is not my first language either. For that matter it's not my second language, or my third language. And yet, I manage.
WOW.
Post by Arturo Magidin
--
Arturo Magidin
bassam king karzeddin
2016-08-30 15:05:41 UTC
Post by Arturo Magidin
Post by bassam king karzeddin
Post by Arturo Magidin
Post by bassam king karzeddin
Post by Arturo Magidin
Post by bassam king karzeddin
Post by Peter Percival
Post by bassam king karzeddin
So, where shall I post the proof then!?
Right here. Stop yacking and get on with it.
--
Do, as a concession to my poor wits, Lord Darlington, just explain
to me what you really mean.
I think I had better not, Duchess. Nowadays to be intelligible is
to be found out. -- Oscar Wilde, Lady Windermere's Fan
http://hsm.stackexchange.com/questions/3257/sum-of-like-powers-in-real-numbers
That's not an "introduction" to anything, most especially not an introduction to a proof of your claims.
So, are you going to continue to dance or are you going to actually put up?
Either shit, or get off the pot.
Oh, ya, that is actually NOTHING better than shit, but I got a better idea, and you hopefully would come to know about it!
And Karzeddin misses the point. AGAIN.
https://en.wikipedia.org/wiki/Shit_or_get_off_the_pot
You'll note that, as opposed to *your* link, this one is directly relevant to the issue.
Pull quote: ""Shit or get off the pot" is a common English language colloquial expression, used to imply a person should follow up their stated intentions with action. "
--
Arturo Magidin
Sorry for the misunderstanding Dr. Arturo Magidin, forgive my poor English as it is my last language,
English is not my first language either. For that matter it's not my second language, or my third language. And yet, I manage.
--
Arturo Magidin
**********************************************************
Post by Arturo Magidin
English is not my first language either. For that matter it's not my second language, or my third language. And yet, I manage.
--
Arturo Magidin
Interesting, you must be super linguistic then.

Bassam Karzeddin
bassam king karzeddin
2016-08-29 17:41:36 UTC
Post by Peter Percival
Post by bassam king karzeddin
So, where shall I post the proof then!?
Right here. Stop yacking and get on with it.
This one link here is an introduction:

http://hsm.stackexchange.com/questions/3257/sum-of-like-powers-in-real-numbers

And if you are a member there, please at least don't downvote it

Thanks
Post by Peter Percival
--
Do, as a concession to my poor wits, Lord Darlington, just explain
to me what you really mean.
I think I had better not, Duchess. Nowadays to be intelligible is
to be found out. -- Oscar Wilde, Lady Windermere's Fan
a***@gmail.com
2016-08-30 20:22:17 UTC
didn't Fermat prove x^4 + y^4 <> z^2
bassam king karzeddin
2016-08-31 07:24:45 UTC
Post by a***@gmail.com
didn't Fermat prove x^4 + y^4 <> z^2
didn't Fermat prove x^4 + y^4 <> z^2
Of course he did, for the sum and difference, which was called the infinite descent, his method extended to another form as (x^4 (+/-) y^4 = 2z^2), in integers has no solution!
bassam king karzeddin
2016-09-01 15:52:45 UTC
Post by quasi
Post by bassam king karzeddin
Post by quasi
Post by quasi
Conjecture: If (x, y, z) are nonzero coprime integers,
and (n, m) are positive integers > 1, then this Diophantine
x^{2n} + y^{2m} = z^2
doesn't have any integer solution.
I can prove this now.
Oops -- I take it back.
My proof has a silly (essentially unfixable) error.
But my previously announced proof (that there are no solutions
for the case where gcd(m,n) > 1) is still good.
Post by quasi
It's actually quite easy (but fun, nevertheless).
Everything's easy to prove once an undetected logical error is
Post by quasi
Thanks bassam -- sorry for doubting you on this one.
I'm doubting once again.
Post by quasi
But don't worry -- based on history, I'm sure I'll have good
reason to doubt some of your future claims.
In fact, although it's now clear that your claim with regard to
the above diophantine equation is valid,
It seemed clear, but it was an illusion.
Post by quasi
I still doubt that you have a proof of your other claim -- the
one where you asserted that there does not exist a primitive
pythagorean triple for which all three components are powerful
numbers.
That doubt is still there -- even stronger now.
You have the full right to be skeptical Quasi, even if I post
an assumed correct proof for this puzzle!
Will you eventually post your proof in sci.math?
Post by bassam king karzeddin
But, to save time, I posted a formula at SME, which is quite
relevant in this regard
http://mathoverflow.net/questions/208169/quintic-equation?sgp=2
I don't see how it relates to the problem of this thread.
quasi
The proof:
Why not starting demonstrating with the simplest case here, that is when (n = m =p), where (p) is any prime number, so the conjecture we have in magazine can be reduced to the following case:

(x^{2p} + y^{2p} = z^2) , where (x, y) are co prim positive integers, then the conjecture says no integer solution (generally)

The simplest proof:

Clearly, the case (p = 2), was Fermatâ€™s solution, so we are left with odd primes only for (p), where this only represent a Pythagorean triplets identity due to even exponents

Assume, (x = (x_1) (x_2), where (x) is odd integer and, gcd(x_1, x_2) = 1, where x_1, x_2 are positive co prime integers

By substitution we get this form of equation

(2x^p)2+ ((x_1)^{2p} â€“ (x_2)^{2p} )^2 = ((x_1)^{2p} + (x_2)^{2p} )^2

Then we have: ( (x_1) ^ {2p} + (x_2) ^ {2p} = 2z), which is possible, (z is odd integer)

And therefore, the following must be equal:

(x_1)^ {2p} - (x_2) ^ {2p} = 2y^p, (y is even integer), factoring the term, and taking into consideration that
Gcd((x_1) â€“ (x_2), x_1) + (x_2)) = 2, so we have:

(x_1)^ p + (x_2) p = 2(y_1) ^p, where (y_1) is odd integer â€¦ (Eqn. 1)

And, y = (y_1)(y_2), and gcd((y_1), (y_2)) = 1, and ((y_1), (y_2)) positive integers with different polarity, thus we have

(x_1)^ p - (x_2) p = (y_2) ^p, where (y_2) is even integer here,

Note that if you reverse the order of polarity, you would arrive at the same equations, and if you consider non coprimeness also will be validated for this case, since it is Fermat's only

So, according to the proof of Fermatâ€™s last theorem by Sir Andrew wiles & Taylor (1995), we have at least one of the above equations is impossible, for all odd primes >3, and for (p = 3), was proved a century after Fermatâ€™s famous conjecture

Conclusion: no integer solution exists for our proposed conjecture, and the proof is completed for this particular case, and there is another one!

The rest cases would continue
quasi
2016-09-01 22:23:06 UTC
...>>
...>>Conjecture: If (x, y, z) are nonzero coprime integers,
...>>and (n, m) are positive integers > 1, then this
...>>
...>> x^{2n} + y^{2m} = z^2
...>>
...>>doesn't have any integer solution.
Why not starting demonstrating with the simplest case here,
that is when (n = m =p), where (p) is any prime number,
But note, as I previously indicated, I was able to prove the
conjecture for the case where gcd(m,n) > 1.

I was unable to prove the conjecture for any _other_ case.
so the conjecture we have in magazine
In what magazine?
(x^{2p} + y^{2p} = z^2) , where (x, y) are co prime positive
integers, then the conjecture says no integer solution
(generally)
Clearly, the case (p = 2), was Fermatâ€™s solution, so we are
left with odd primes only for (p), where this only represent
a Pythagorean triplets identity due to even exponents
Assume, (x = (x_1) (x_2), where (x) is odd integer and,
gcd(x_1, x_2) = 1, where x_1, x_2 are positive co prime
integers
What are you doing?

Are you trying to exploit the fact that (x^p,y^p,z) is
a primitive pythagorean triplet?

If so, you can get expressions for x^p,y^p,z in terms of two
new variables.

You don't get an expression for x.

Moreover, writing x = (x_1)(x_2) seems useless since it allows
the trivial factorization x = (x)(1) (i.e., x_1 = x and x_2 = 1).
By substitution we get this form of equation
(2x^p)2+ ((x_1)^{2p} â€“ (x_2)^{2p} )^2
= ((x_1)^{2p} + (x_2)^{2p} )^2
Then we have: ( (x_1) ^ {2p} + (x_2) ^ {2p} = 2z),
I have no idea how you got the above equation.

How does the factorization of x give you anything about y and z?

quasi
bassam king karzeddin
2016-09-03 09:37:19 UTC
Post by quasi
...>>
...>>Conjecture: If (x, y, z) are nonzero coprime integers,
...>>and (n, m) are positive integers > 1, then this
...>>
...>> x^{2n} + y^{2m} = z^2
...>>
...>>doesn't have any integer solution.
Why not starting demonstrating with the simplest case here,
that is when (n = m =p), where (p) is any prime number,
But note, as I previously indicated, I was able to prove the
conjecture for the case where gcd(m,n) > 1.
Yes of course, let gcd(m,n) =k, where k is odd integer, and impossible to be even integer, since in this even case you would get the same form that Fermat had proved (a^4 + b^4 = c^2), impossible in integers,

So, let (p) be an odd prime number that divides k, so eventually, you arrive at this proposed case here , another case is gcd (m, n) =1, is even much easier!, I will come to it
Post by quasi
I was unable to prove the conjecture for any _other_ case.
so the conjecture we have in magazine
In what magazine?
I mean world magazine, (google groups that is open to the world freely)!
Post by quasi
(x^{2p} + y^{2p} = z^2) , where (x, y) are co prime positive
integers, then the conjecture says no integer solution
(generally)
Clearly, the case (p = 2), was Fermatâ€™s solution, so we are
left with odd primes only for (p), where this only represent
a Pythagorean triplets identity due to even exponents
Assume, (x = (x_1) (x_2), where (x) is odd integer and,
gcd(x_1, x_2) = 1, where x_1, x_2 are positive co prime
integers
What are you doing?
Are you trying to exploit the fact that (x^p,y^p,z) is
a primitive pythagorean triplet?
Yes, our conjecture assumes only that (is a pythagorean triplet)
Post by quasi
If so, you can get expressions for x^p,y^p,z in terms of two
new variables.
And that is why all the terms are expressed in two variables only, (x_1) & (x_2)
Post by quasi
You don't get an expression for x.
X = (X_1)*(X_2), as specified above
Post by quasi
Moreover, writing x = (x_1)(x_2) seems useless since it allows
the trivial factorization x = (x)(1) (i.e., x_1 = x and x_2 = 1).
Yes of course, and I never said that one of the factors of (x) can't be one
Post by quasi
By substitution we get this form of equation
(2x^p)^2+ ((x_1)^{2p} â€“ (x_2)^{2p} )^2
= ((x_1)^{2p} + (x_2)^{2p} )^2
Then we have: ( (x_1) ^ {2p} + (x_2) ^ {2p} = 2z),
I have no idea how you got the above equation.
How does the factorization of x give you anything about y and z?
quasi
The solution of the following Diophantine equation

x^n + y^2 = z^2

in integers would be so easily and self proved identity as the following,

2y = (x_1)^n - (x_2)^n, and 2z= (x_1)^n - (x_2)^n, where x is odd integer,

x = (x_1)*(x_2), and y therefore even integer, and z is odd integer, this is the same as Pythagoras triplet when (n) is even integer
Post by quasi
...>>
...>>Conjecture: If (x, y, z) are nonzero coprime integers,
...>>and (n, m) are positive integers > 1, then this
...>>
...>> x^{2n} + y^{2m} = z^2
...>>
...>>doesn't have any integer solution.
Why not starting demonstrating with the simplest case here,
that is when (n = m =p), where (p) is any prime number,
But note, as I previously indicated, I was able to prove the
conjecture for the case where gcd(m,n) > 1.
I was unable to prove the conjecture for any _other_ case.
so the conjecture we have in magazine
In what magazine?
(x^{2p} + y^{2p} = z^2) , where (x, y) are co prime positive
integers, then the conjecture says no integer solution
(generally)
Clearly, the case (p = 2), was Fermatâ€™s solution, so we are
left with odd primes only for (p), where this only represent
a Pythagorean triplets identity due to even exponents
Assume, (x = (x_1) (x_2), where (x) is odd integer and,
gcd(x_1, x_2) = 1, where x_1, x_2 are positive co prime
integers
What are you doing?
Are you trying to exploit the fact that (x^p,y^p,z) is
a primitive pythagorean triplet?
If so, you can get expressions for x^p,y^p,z in terms of two
new variables.
You don't get an expression for x.
Moreover, writing x = (x_1)(x_2) seems useless since it allows
the trivial factorization x = (x)(1) (i.e., x_1 = x and x_2 = 1).
By substitution we get this form of equation
(2x^p)2+ ((x_1)^{2p} â€“ (x_2)^{2p} )^2
= ((x_1)^{2p} + (x_2)^{2p} )^2
Then we have: ( (x_1) ^ {2p} + (x_2) ^ {2p} = 2z),
I have no idea how you got the above equation.
It is the same Pythagorean identity Quasi!
Post by quasi
How does the factorization of x give you anything about y and z?
quasi
Given any positive integer >2, you can so simply find out all possible primitive Pythagorean triplets, and from that number only, very easy Quasi

Please check again Quasi, and would be grateful to you if you can mistake me

It is as you once said, Too....Easy !

Regards

Basaam King Karzeddin
quasi
2016-09-03 11:02:14 UTC
Post by bassam king karzeddin
Post by quasi
...>>
...>>Conjecture: If (x, y, z) are nonzero coprime integers,
...>>and (n, m) are positive integers > 1, then this
...>>
...>> x^{2n} + y^{2m} = z^2
...>>
...>>doesn't have any integer solution.
Why not starting demonstrating with the simplest case here,
that is when (n = m =p), where (p) is any prime number,
But note, as I previously indicated, I was able to prove the
conjecture for the case where gcd(m,n) > 1.
Yes of course, let gcd(m,n) =k, where k is odd integer, and
impossible to be even integer, since in this even case you
would get the same form that Fermat had proved (a^4 + b^4 = c^2),
impossible in integers,
So, let (p) be an odd prime number that divides k, so eventually,
you arrive at this proposed case here , another case is gcd
(m, n) = 1, is even much easier!,
Easier? I don't believe it.

In any case, your posted "proof" for the case m = n = p, where p
is an odd prime looks like nonsense to me.
Post by bassam king karzeddin
I will come to it
Most likely, it will be just more nonsense..
Post by bassam king karzeddin
Post by quasi
I was unable to prove the conjecture for any _other_ case.
so the conjecture we have in magazine
In what magazine?
I mean world magazine, (google groups that is open to the world
freely)!
You really have no concept of what sci.math is.

As previously explained, sci.math is part of usenet, a worldwide
network of user groups.

Google Groups and Math Forum each provide a partly filtered
two-way portal to sci.math, but they don't "own" sci.math.

Most sci.math regulars (myself included) access sci.math
directly from a usenet server via an NNTP "newsreader"
program; not from Google Groups; not from Math Forum.
Post by bassam king karzeddin
Post by quasi
(x^{2p} + y^{2p} = z^2) , where (x, y) are co prime positive
integers, then the conjecture says no integer solution
(generally)
Clearly, the case (p = 2), was Fermatâ€™s solution, so we are
left with odd primes only for (p), where this only represent
a Pythagorean triplets identity due to even exponents
Assume, (x = (x_1) (x_2), where (x) is odd integer and,
gcd(x_1, x_2) = 1, where x_1, x_2 are positive co prime
integers
What are you doing?
Are you trying to exploit the fact that (x^p,y^p,z) is
a primitive pythagorean triplet?
Yes, our conjecture assumes only that (is a pythagorean triplet)
If so, you can get expressions for x^p,y^p,z in terms of two
new variables.
And that is why all the terms are expressed in two variables
only, (x_1) & (x_2)
But the pythagorean triple formulas give you integer valued
expressions, in terms of two new parameters, for x^p,y^p,z;
not for x,y,z.
Post by bassam king karzeddin
Post by quasi
You don't get an expression for x.
X = (X_1)*(X_2), as specified above
Post by quasi
Moreover, writing x = (x_1)(x_2) seems useless since it allows
the trivial factorization x = (x)(1) (i.e., x_1 = x and x_2 = 1).
Yes of course, and I never said that one of the factors of (x)
can't be one
So if you assume x1 = x, x2 = 1, how would the rest of your
argument proceed?
Post by bassam king karzeddin
Post by quasi
By substitution we get this form of equation
(2x^p)^2+ ((x_1)^{2p} â€“ (x_2)^{2p} )^2
= ((x_1)^{2p} + (x_2)^{2p} )^2
Then we have: ( (x_1) ^ {2p} + (x_2) ^ {2p} = 2z),
I have no idea how you got the above equation.
How does the factorization of x give you anything about y and z?
The solution of the following Diophantine equation
x^n + y^2 = z^2
in integers would be so easily and self proved identity as
the following,
2y = (x_1)^n - (x_2)^n, and 2z= (x_1)^n - (x_2)^n, where
x is odd integer,
???
Post by bassam king karzeddin
x = (x_1)*(x_2), and y therefore even integer, and z is odd
integer, this is the same as Pythagoras triplet when (n) is
even integer
???
Post by bassam king karzeddin
Given any positive integer >2, you can so simply find out all
possible primitive Pythagorean triplets, and from that number
only, very easy Quasi
???
Post by bassam king karzeddin
Please check again Quasi, and would be grateful to you if you
can mistake me
It is as you once said, Too....Easy !
Sorry, bassam -- I'm now convinced you have no proof, not even
for the case gcd(m,n) > 1, which is, in fact, a readily provable
case (I have a proof).

As you've previously indicated, you are not a mathematician,
so it's reasonable to expect that your arguments would not be
expressed at the same level of logical precision as would be
expected from a mathematician. But there is a level of
imprecision below which it becomes virtually impossible for
mathematical language. Your posted arguments are definitely
below that threshold.

I don't expect much -- probably more nonsense, but let's see you
(bassam king karzeddin) prove either or both of the following
particular cases. I can prove the first one; I don't have a proof
of the second; I don't believe you can prove either one.

(1) There do not exist pairwise coprime nonzero integers x,y,z
such that x^6 + y^6 = z^2.

(2) There do not exist pairwise coprime nonzero integers x,y,z
such that x^6 + y^4 = z^2.

quasi
bassam king karzeddin
2016-09-03 12:06:28 UTC
Post by quasi
Post by bassam king karzeddin
Post by quasi
...>>
...>>Conjecture: If (x, y, z) are nonzero coprime integers,
...>>and (n, m) are positive integers > 1, then this
...>>
...>> x^{2n} + y^{2m} = z^2
...>>
...>>doesn't have any integer solution.
Why not starting demonstrating with the simplest case here,
that is when (n = m =p), where (p) is any prime number,
But note, as I previously indicated, I was able to prove the
conjecture for the case where gcd(m,n) > 1.
Yes of course, let gcd(m,n) =k, where k is odd integer, and
impossible to be even integer, since in this even case you
would get the same form that Fermat had proved (a^4 + b^4 = c^2),
impossible in integers,
So, let (p) be an odd prime number that divides k, so eventually,
you arrive at this proposed case here , another case is gcd
(m, n) = 1, is even much easier!,
Easier? I don't believe it.
In any case, your posted "proof" for the case m = n = p, where p
is an odd prime looks like nonsense to me.
Post by bassam king karzeddin
I will come to it
Most likely, it will be just more nonsense..
Post by bassam king karzeddin
Post by quasi
I was unable to prove the conjecture for any _other_ case.
so the conjecture we have in magazine
In what magazine?
I mean world magazine, (google groups that is open to the world freely)!
You really have no concept of what sci.math is.
As previously explained, sci.math is part of usenet, a worldwide
network of user groups.
Google Groups and Math Forum each provide a partly filtered
two-way portal to sci.math, but they don't "own" sci.math.
Most sci.math regulars (myself included) access sci.math
directly from a usenet server via an NNTP "newsreader"
program; not from Google Groups; not from Math Forum.
Post by bassam king karzeddin
Post by quasi
(x^{2p} + y^{2p} = z^2) , where (x, y) are co prime positive
integers, then the conjecture says no integer solution
(generally)
Clearly, the case (p = 2), was Fermatâ€™s solution, so we are
left with odd primes only for (p), where this only represent
a Pythagorean triplets identity due to even exponents
Assume, (x = (x_1) (x_2), where (x) is odd integer and,
gcd(x_1, x_2) = 1, where x_1, x_2 are positive co prime
integers
What are you doing?
Are you trying to exploit the fact that (x^p,y^p,z) is
a primitive pythagorean triplet?
Yes, our conjecture assumes only that (is a pythagorean triplet)
If so, you can get expressions for x^p,y^p,z in terms of two
new variables.
And that is why all the terms are expressed in two variables
only, (x_1) & (x_2)
But the pythagorean triple formulas give you integer valued
expressions, in terms of two new parameters, for x^p,y^p,z;
not for x,y,z.
Post by bassam king karzeddin
Post by quasi
You don't get an expression for x.
X = (X_1)*(X_2), as specified above
Post by quasi
Moreover, writing x = (x_1)(x_2) seems useless since it allows
the trivial factorization x = (x)(1) (i.e., x_1 = x and x_2 = 1).
Yes of course, and I never said that one of the factors of (x) can't be one
So if you assume x1 = x, x2 = 1, how would the rest of your
argument proceed?
Post by bassam king karzeddin
Post by quasi
By substitution we get this form of equation
(2x^p)^2+ ((x_1)^{2p} â€“ (x_2)^{2p} )^2
= ((x_1)^{2p} + (x_2)^{2p} )^2
Then we have: ( (x_1) ^ {2p} + (x_2) ^ {2p} = 2z),
I have no idea how you got the above equation.
How does the factorization of x give you anything about y and z?
The solution of the following Diophantine equation
x^n + y^2 = z^2
in integers would be so easily and self proved identity as
the following,
2y = (x_1)^n - (x_2)^n, and 2z= (x_1)^n - (x_2)^n, where
x is odd integer,
???
Post by bassam king karzeddin
x = (x_1)*(x_2), and y therefore even integer, and z is odd
integer, this is the same as Pythagoras triplet when (n) is
even integer
???
Post by bassam king karzeddin
Given any positive integer >2, you can so simply find out all
possible primitive Pythagorean triplets, and from that number
only, very easy Quasi
???
Post by bassam king karzeddin
Please check again Quasi, and would be grateful to you if you can mistake me
It is as you once said, Too....Easy !
Sorry, bassam -- I'm now convinced you have no proof, not even
for the case gcd(m,n) > 1, which is, in fact, a readily provable
case (I have a proof).
As you've previously indicated, you are not a mathematician,
so it's reasonable to expect that your arguments would not be
expressed at the same level of logical precision as would be
expected from a mathematician. But there is a level of
imprecision below which it becomes virtually impossible for
mathematical language. Your posted arguments are definitely
below that threshold.
I don't expect much -- probably more nonsense, but let's see you
(bassam king karzeddin) prove either or both of the following
particular cases. I can prove the first one; I don't have a proof
of the second; I don't believe you can prove either one.
(1) There do not exist pairwise coprime nonzero integers x,y,z
such that x^6 + y^6 = z^2.
(2) There do not exist pairwise coprime nonzero integers x,y,z
such that x^6 + y^4 = z^2.
quasi
In both examples you challenge, are assumed Pythagorean triplet (due to even exponents), so no harm if you assume (x = a*b) in both of your examples here, where (a, b) are assumed positive odd integers, (a>b), gcd(a, b) =1, so we have all the terms are expressible in two variables (a & b), (seem done in my submitted proof)

So, we have first (x^6 + y^6 = z^2), substituting in (a & b), we get then:

4(ab)^6 + (a^6 â€“ b^6)^2 = (a^6 + b^6)^2

So, (2z = a^6 + b^6), which is possible, where z is odd integer

And, (2y^3 = a^6 â€“ b^6 = (a^3 â€“ b^3)*(a^3 + b^3),

Note that gcd ((a^3 + b^3), (a^3 â€“ b^3)) = 2 ONLY, since both (a & b) are assumed odd integers, but (y) is even, factoring the terms:

(a^3 + y^3 = 2c^3), and (a^3 â€“ b^3 = d^3), where (y = c*d), and gcd (c, d) = 1, where also (c) is odd and (d )is even integer,

The contradiction here, we arrived at Fermat last theorem that must have solution in order to solve our conjecture,
a^3 â€“ b^3 = d^3

But FLT was proved impossible to have solution; therefore this conjecture is impossible to have solution â€“proved rigorously

Please, someone helps Quasi to this simple proof, (Iâ€™m sorry Quasi)

The next one would follow

Regards

Bassam Karzeddin
030916

(C)
quasi
2016-09-03 13:01:43 UTC
Post by bassam king karzeddin
Post by quasi
I don't expect much -- probably more nonsense, but let's see
you (bassam king karzeddin) prove either or both of the
following particular cases. I can prove the first one; I don't
have a proof of the second; I don't believe you can prove
either one.
(1) There do not exist pairwise coprime nonzero integers x,y,z
such that x^6 + y^6 = z^2.
(2) There do not exist pairwise coprime nonzero integers x,y,z
such that x^6 + y^4 = z^2.
In both examples you challenge, are assumed Pythagorean triplet
(due to even exponents), so no harm if you assume (x = a*b) in
both of your examples here, where (a, b) are assumed positive
odd integers, (a>b), gcd(a, b) =1, so we have all the terms
are expressible in two variables (a & b), (seem done in my
submitted proof)
So, we have first (x^6 + y^6 = z^2), substituting in (a & b),
4(ab)^6 + (a^6 â€“ b^6)^2 = (a^6 + b^6)^2
Sure, that's an algebraic identity.
Post by bassam king karzeddin
So, (2z = a^6 + b^6),
I don't follow that at all.

You have the following:

x,y,z pairwise coprime nonzero integers.

x^6 + y^6 = z^2

x odd, x = a*b, gcd(a,b) = 1, a > b

Thus you have

4(ab)^6 + 4y^6 = (2z)^2

4(ab)^6 + (a^6 â€“ b^6)^2 = (a^6 + b^6)^2

But you haven't proved 4y^6 = (a^6 - b^6)^2, so you
can't justify the claim 2z = a^6 + b^6.
Post by bassam king karzeddin
Please, someone helps Quasi to this simple proof, (Iâ€™m sorry
Quasi)
I _have_ a simple proof of (1) -- yours appears flawed, or at
best incomplete.

Is anyone else able to follow bassam's argument?
Post by bassam king karzeddin
The next one would follow.
I don't have a proof of (2), but I don't believe you have one
either -- at least not a proof that would convince anyone at the
level, say, of an undergraduate math major.

quasi
bassam king karzeddin
2016-09-03 14:09:29 UTC
Post by quasi
Post by bassam king karzeddin
Post by quasi
I don't expect much -- probably more nonsense, but let's see
you (bassam king karzeddin) prove either or both of the
following particular cases. I can prove the first one; I don't
have a proof of the second; I don't believe you can prove
either one.
(1) There do not exist pairwise coprime nonzero integers x,y,z
such that x^6 + y^6 = z^2.
(2) There do not exist pairwise coprime nonzero integers x,y,z
such that x^6 + y^4 = z^2.
In both examples you challenge, are assumed Pythagorean triplet
(due to even exponents), so no harm if you assume (x = a*b) in
both of your examples here, where (a, b) are assumed positive
odd integers, (a>b), gcd(a, b) =1, so we have all the terms
are expressible in two variables (a & b), (seem done in my
submitted proof)
So, we have first (x^6 + y^6 = z^2), substituting in (a & b),
4(ab)^6 + (a^6 â€“ b^6)^2 = (a^6 + b^6)^2
Sure, that's an algebraic identity.
Post by bassam king karzeddin
So, (2z = a^6 + b^6),
I don't follow that at all.
x,y,z pairwise coprime nonzero integers.
x^6 + y^6 = z^2
x odd, x = a*b, gcd(a,b) = 1, a > b
Thus you have
4(ab)^6 + 4y^6 = (2z)^2
4(ab)^6 + (a^6 â€“ b^6)^2 = (a^6 + b^6)^2
But you haven't proved 4y^6 = (a^6 - b^6)^2, so you
can't justify the claim 2z = a^6 + b^6.
Post by bassam king karzeddin
Please, someone helps Quasi to this simple proof, (Iâ€™m sorry Quasi)
I _have_ a simple proof of (1) -- yours appears flawed, or at
best incomplete.
Is anyone else able to follow bassam's argument?
Post by bassam king karzeddin
The next one would follow.
I don't have a proof of (2), but I don't believe you have one
either -- at least not a proof that would convince anyone at the
level, say, of an undergraduate math major.
quasi
Consider this triplet (2(ab)^3, (a^6 - b^6), (a^6 + b^6)), as integer sides of right angle triangle, This is Pythagorean original identity only as (2ab, (a^2 - b^2), (a^2 + b^2), but I considered both (a, b) as odd cubes, that is all!

So, do square them please, then you would have the following:

4(ab)^6 + (a^6 â€“ b^6)^2 = (a^6 + b^6)^2 , that I provided earlier?

Note that all the terms of this triple are divisible by 4, (then becomes primitive), that is why you might got confused while seeing the terms

2y^3 = a^6 - b^6, and 2z = a^6 + b^6 , so divide all by 4, then it is clear, I'm quite sure you would realize it very soon,

I know also there is another proof, and may be many more, since this is an obvious absolute fact (at least to me)
Post by quasi
Post by bassam king karzeddin
Post by quasi
I don't expect much -- probably more nonsense, but let's see
you (bassam king karzeddin) prove either or both of the
following particular cases. I can prove the first one; I don't
have a proof of the second; I don't believe you can prove
either one.
(1) There do not exist pairwise coprime nonzero integers x,y,z
such that x^6 + y^6 = z^2.
(2) There do not exist pairwise coprime nonzero integers x,y,z
such that x^6 + y^4 = z^2.
In both examples you challenge, are assumed Pythagorean triplet
(due to even exponents), so no harm if you assume (x = a*b) in
both of your examples here, where (a, b) are assumed positive
odd integers, (a>b), gcd(a, b) =1, so we have all the terms
are expressible in two variables (a & b), (seem done in my
submitted proof)
So, we have first (x^6 + y^6 = z^2), substituting in (a & b),
4(ab)^6 + (a^6 â€“ b^6)^2 = (a^6 + b^6)^2
Sure, that's an algebraic identity.
but, if we deal it in integers, it becomes diophantine equation also,
Post by quasi
Post by bassam king karzeddin
So, (2z = a^6 + b^6),
I don't follow that at all.
Post by quasi
x,y,z pairwise coprime nonzero integers.
x^6 + y^6 = z^2
x odd, x = a*b, gcd(a,b) = 1, a > b
Thus you have
4(ab)^6 + 4y^6 = (2z)^2
4(ab)^6 + (a^6 â€“ b^6)^2 = (a^6 + b^6)^2
But you haven't proved 4y^6 = (a^6 - b^6)^2, so you
Take the Sqrt of both sides in positive sense

then, 2y^3 = a^6 - b^6, as explained above!
Post by quasi
can't justify the claim 2z = a^6 + b^6.
Post by bassam king karzeddin
Please, someone helps Quasi to this simple proof, (Iâ€™m sorry Quasi)
I _have_ a simple proof of (1) -- yours appears flawed, or at
best incomplete.
Most likely you have one, that we didn't see yet,
Post by quasi
Is anyone else able to follow bassam's argument?
Post by bassam king karzeddin
The next one would follow.
I don't have a proof of (2), but I don't believe you have one
either -- at least not a proof that would convince anyone at the
level, say, of an undergraduate math major.
quasi
I would like to note that doesn't require any high level in Math to convey,

Just a little revision and every thing would be so clear,

We should remember the primitive triplet Identity (Odd + Even = Odd)

It would be nice of you if you post your proof also

Regards

Bassam Karzeddin
quasi
2016-09-03 21:46:56 UTC
Post by bassam king karzeddin
Post by quasi
Post by bassam king karzeddin
Post by quasi
I don't expect much -- probably more nonsense, but let's see
you (bassam king karzeddin) prove either or both of the
following particular cases. I can prove the first one; I don't
have a proof of the second; I don't believe you can prove
either one.
(1) There do not exist pairwise coprime nonzero integers x,y,z
such that x^6 + y^6 = z^2.
(2) There do not exist pairwise coprime nonzero integers x,y,z
such that x^6 + y^4 = z^2.
In both examples you challenge, are assumed Pythagorean triplet
(due to even exponents), so no harm if you assume (x = a*b) in
both of your examples here, where (a, b) are assumed positive
odd integers, (a>b), gcd(a, b) =1, so we have all the terms
are expressible in two variables (a & b), (seem done in my
submitted proof)
So, we have first (x^6 + y^6 = z^2), substituting in (a & b),
4(ab)^6 + (a^6 â€“ b^6)^2 = (a^6 + b^6)^2
Sure, that's an algebraic identity.
Post by bassam king karzeddin
So, (2z = a^6 + b^6),
I don't follow that at all.
x,y,z pairwise coprime nonzero integers.
x^6 + y^6 = z^2
x odd, x = a*b, gcd(a,b) = 1, a > b
Thus you have
4(ab)^6 + 4y^6 = (2z)^2
4(ab)^6 + (a^6 â€“ b^6)^2 = (a^6 + b^6)^2
But you haven't proved 4y^6 = (a^6 - b^6)^2, so you
can't justify the claim 2z = a^6 + b^6.
Post by bassam king karzeddin
Please, someone helps Quasi to this simple proof,
(Iâ€™m sorry Quasi)
I _have_ a simple proof of (1) -- yours appears flawed,
or at best incomplete.
Most likely you have one, that we didn't see yet,
I'll post it later this week, but it's a busy week, so probably
Thursday.
Post by bassam king karzeddin
Post by quasi
Is anyone else able to follow bassam's argument?
No response yet from other readers.
Post by bassam king karzeddin
Post by quasi
Post by bassam king karzeddin
The next one would follow.
I don't have a proof of (2), but I don't believe you have one
either -- at least not a proof that would convince anyone at
the level, say, of an undergraduate math major.
Consider this triplet (2(ab)^3, (a^6 - b^6), (a^6 + b^6)), as
integer sides of right angle triangle, This is Pythagorean
original identity only as (2ab, (a^2 - b^2), (a^2 + b^2), but
I considered both (a, b) as odd cubes, that is all!
4(ab)^6 + (a^6 â€“ b^6)^2 = (a^6 + b^6)^2 , that I provided
earlier.
Note that all the terms of this triple are divisible by 4,
(then becomes primitive), that is why you might got confused
while seeing the terms
2y^3 = a^6 - b^6, and 2z = a^6 + b^6 , so divide all by 4,
then it is clear, I'm quite sure you would realize it very
soon,
Without loss of generality we can assume x is a positive
integer.

You are assuming x is odd. No problem.

Then you write x = a*b, where a,b coprime positive integers
with a > b.

That implies x > 1, but still no problem.

You have two primitive pythagorean triples ...

(x^3,y^3,z) [by hypothesis]

((ab)^2,(a^6-b^6)/2,(a^6 + b^6)/2) [algebraic identity]

Since x = a*b, the first components of the two triples are equal.

But Tthat does not imply that the triples are equal.

Thus, your claim that z must equal (a^6 + b^6)/2 is an
unsupported claim.

To illustrate the flaw in your argument, here are two distinct
primitive pythagorean triples which have equal first components:

15, 8, 17

15, 112, 113
Post by bassam king karzeddin
I would like to note that doesn't require any high level in
Math to convey,
Just a little revision and every thing would be so clear,
We should remember the primitive triplet Identity
(Odd + Even = Odd)
It would be nice of you if you post your proof also.
I'll post it no later than Thursday.

quasi
bassam king karzeddin
2016-09-04 12:41:13 UTC
Post by quasi
Post by bassam king karzeddin
Post by quasi
Post by bassam king karzeddin
Post by quasi
I don't expect much -- probably more nonsense, but let's see
you (bassam king karzeddin) prove either or both of the
following particular cases. I can prove the first one; I don't
have a proof of the second; I don't believe you can prove
either one.
(1) There do not exist pairwise coprime nonzero integers x,y,z
such that x^6 + y^6 = z^2.
(2) There do not exist pairwise coprime nonzero integers x,y,z
such that x^6 + y^4 = z^2.
In both examples you challenge, are assumed Pythagorean triplet
(due to even exponents), so no harm if you assume (x = a*b) in
both of your examples here, where (a, b) are assumed positive
odd integers, (a>b), gcd(a, b) =1, so we have all the terms
are expressible in two variables (a & b), (seem done in my
submitted proof)
4(ab)^6 + (a^6 â€“ b^6)^2 = (a^6 + b^6)^2
Sure, that's an algebraic identity.
Post by bassam king karzeddin
So, (2z = a^6 + b^6),
I don't follow that at all.
x,y,z pairwise coprime nonzero integers.
x^6 + y^6 = z^2
the primitive triplet for the above example (x^3, y^2, z), where y is only even integer, where also when multiplied by two, would results on other similar triplet with double size of the original triangle, (2x^3, 2y^3, 2z), where all sides in this triplet are even integers!, thus we have

4x^6 + 4y^3 = 4z^2
Post by quasi
Post by bassam king karzeddin
Post by quasi
x odd, x = a*b, gcd(a,b) = 1, a > b
Thus you have
4(ab)^6 + 4y^6 = (2z)^2
4(ab)^6 + (a^6 â€“ b^6)^2 = (a^6 + b^6)^2
Note here, all the terms of the primitive original triplet are multiplied by 4, so divide the identity by 4, would get back the original triplet (SO...EASY)
Post by quasi
Post by bassam king karzeddin
Post by quasi
But you haven't proved 4y^6 = (a^6 - b^6)^2, so you
can't justify the claim 2z = a^6 + b^6.
I hope, this time would be convinced
Post by quasi
Post by bassam king karzeddin
Post by quasi
Post by bassam king karzeddin
Please, someone helps Quasi to this simple proof,
(Iâ€™m sorry Quasi)
I _have_ a simple proof of (1) -- yours appears flawed,
or at best incomplete.
It is indeed more than rigorous, that is why no body is capable to criticise it, otherwise you would certainly see many
Post by quasi
Post by bassam king karzeddin
Most likely you have one, that we didn't see yet,
I'll post it later this week, but it's a busy week, so probably
Thursday.
Post by bassam king karzeddin
Post by quasi
Is anyone else able to follow bassam's argument?
No response yet from other readers.
And it is generally rare that anyone would respond to an absolute fact, but many responses would be there for a silly matters
Post by quasi
Post by bassam king karzeddin
Post by quasi
Post by bassam king karzeddin
The next one would follow.
I don't have a proof of (2), but I don't believe you have one
either -- at least not a proof that would convince anyone at
the level, say, of an undergraduate math major.
Do you really think that I was conjecturing randomly!?
Post by quasi
Post by bassam king karzeddin
Consider this triplet (2(ab)^3, (a^6 - b^6), (a^6 + b^6)), as
integer sides of right angle triangle, This is Pythagorean
original identity only as (2ab, (a^2 - b^2), (a^2 + b^2), but
I considered both (a, b) as odd cubes, that is all!
4(ab)^6 + (a^6 â€“ b^6)^2 = (a^6 + b^6)^2 , that I provided
earlier.
Note that all the terms of this triple are divisible by 4,
(then becomes primitive), that is why you might got confused
while seeing the terms
2y^3 = a^6 - b^6, and 2z = a^6 + b^6 , so divide all by 4,
then it is clear, I'm quite sure you would realize it very
soon,
Without loss of generality we can assume x is a positive
integer.
You are assuming x is odd. No problem.
Then you write x = a*b, where a,b coprime positive integers
with a > b.
That implies x > 1, but still no problem.
Yes no problem at all whenever we consider primitive triplets
Post by quasi
You have two primitive pythagorean triples ...
(x^3,y^3,z) [by hypothesis]
Yes
Post by quasi
((ab)^2,(a^6-b^6)/2,(a^6 + b^6)/2) [algebraic identity]
Wait here, this is the algebraic identity, not the one you mentioned above!

((ab)^3, (a^6-b^6)/2,(a^6 + b^6)/2) [algebraic identity]
Post by quasi
Since x = a*b, the first components of the two triples are equal.
But Tthat does not imply that the triples are equal.
???,
Meaningless, and I don't understand the intention behind that!
Post by quasi
Thus, your claim that z must equal (a^6 + b^6)/2 is an
unsupported claim.
This is very obvious from the identity itself, and doesn't require any kind of support
Post by quasi
To illustrate the flaw in your argument, here are two distinct
15, 8, 17
15, 112, 113
Of course yes, an odd integer may occur in so many distinct primitive triplets, depending on the prime factors of the odd integer, I can provide you with so many examples if you would like, so where is the flow then in my reasoning?
Post by quasi
Post by bassam king karzeddin
I would like to note that doesn't require any high level in
Math to convey,
I do repeat the same words mentioned by me
Post by quasi
Post by bassam king karzeddin
Just a little revision and every thing would be so clear,
And I know that you are not mudding, but honest and skeptical to raise the truth above, no matter who would win, the fact and nothing else must win
Post by quasi
Post by bassam king karzeddin
We should remember the primitive triplet Identity
(Odd + Even = Odd)
It would be nice of you if you post your proof also.
I'll post it no later than Thursday.
quasi
We would be ager to see another proof of this so obvious fact
Post by quasi
Post by bassam king karzeddin
Post by quasi
Post by bassam king karzeddin
Post by quasi
I don't expect much -- probably more nonsense, but let's see
you (bassam king karzeddin) prove either or both of the
following particular cases. I can prove the first one; I don't
have a proof of the second; I don't believe you can prove
either one.
(1) There do not exist pairwise coprime nonzero integers x,y,z
such that x^6 + y^6 = z^2.
(2) There do not exist pairwise coprime nonzero integers x,y,z
such that x^6 + y^4 = z^2.
In both examples you challenge, are assumed Pythagorean triplet
(due to even exponents), so no harm if you assume (x = a*b) in
both of your examples here, where (a, b) are assumed positive
odd integers, (a>b), gcd(a, b) =1, so we have all the terms
are expressible in two variables (a & b), (seem done in my
submitted proof)
4(ab)^6 + (a^6 â€“ b^6)^2 = (a^6 + b^6)^2
Sure, that's an algebraic identity.
Post by bassam king karzeddin
So, (2z = a^6 + b^6),
I don't follow that at all.
x,y,z pairwise coprime nonzero integers.
x^6 + y^6 = z^2
x odd, x = a*b, gcd(a,b) = 1, a > b
Thus you have
4(ab)^6 + 4y^6 = (2z)^2
4(ab)^6 + (a^6 â€“ b^6)^2 = (a^6 + b^6)^2
But you haven't proved 4y^6 = (a^6 - b^6)^2, so you
can't justify the claim 2z = a^6 + b^6.
Post by bassam king karzeddin
Please, someone helps Quasi to this simple proof,
(Iâ€™m sorry Quasi)
I _have_ a simple proof of (1) -- yours appears flawed,
or at best incomplete.
Most likely you have one, that we didn't see yet,
I'll post it later this week, but it's a busy week, so probably
Thursday.
Post by bassam king karzeddin
Post by quasi
Is anyone else able to follow bassam's argument?
No response yet from other readers.
Post by bassam king karzeddin
Post by quasi
Post by bassam king karzeddin
The next one would follow.
I don't have a proof of (2), but I don't believe you have one
either -- at least not a proof that would convince anyone at
the level, say, of an undergraduate math major.
Consider this triplet (2(ab)^3, (a^6 - b^6), (a^6 + b^6)), as
integer sides of right angle triangle, This is Pythagorean
original identity only as (2ab, (a^2 - b^2), (a^2 + b^2), but
I considered both (a, b) as odd cubes, that is all!
4(ab)^6 + (a^6 â€“ b^6)^2 = (a^6 + b^6)^2 , that I provided
earlier.
Note that all the terms of this triple are divisible by 4,
(then becomes primitive), that is why you might got confused
while seeing the terms
2y^3 = a^6 - b^6, and 2z = a^6 + b^6 , so divide all by 4,
then it is clear, I'm quite sure you would realize it very
soon,
Without loss of generality we can assume x is a positive
integer.
You are assuming x is odd. No problem.
Then you write x = a*b, where a,b coprime positive integers
with a > b.
That implies x > 1, but still no problem.
You have two primitive pythagorean triples ...
(x^3,y^3,z) [by hypothesis]
((ab)^2,(a^6-b^6)/2,(a^6 + b^6)/2) [algebraic identity]
Since x = a*b, the first components of the two triples are equal.
But Tthat does not imply that the triples are equal.
Thus, your claim that z must equal (a^6 + b^6)/2 is an
unsupported claim.
To illustrate the flaw in your argument, here are two distinct
15, 8, 17
15, 112, 113
Post by bassam king karzeddin
I would like to note that doesn't require any high level in
Math to convey,
Just a little revision and every thing would be so clear,
We should remember the primitive triplet Identity
(Odd + Even = Odd)
It would be nice of you if you post your proof also.
I'll post it no later than Thursday.
quasi
Regards

Bassam Karzeddin
quasi
2016-09-04 13:34:10 UTC
Post by quasi
Post by bassam king karzeddin
Post by quasi
Post by bassam king karzeddin
Post by quasi
I don't expect much -- probably more nonsense, but let's
see you (bassam king karzeddin) prove either or both of
the following particular cases. I can prove the first one;
I don't have a proof of the second; I don't believe you
can prove either one.
(1) There do not exist pairwise coprime nonzero integers
x,y,z such that x^6 + y^6 = z^2.
(2) There do not exist pairwise coprime nonzero integers
x,y,z such that x^6 + y^4 = z^2.
In both examples you challenge, are assumed Pythagorean
triplet (due to even exponents), so no harm if you assume
(x = a*b) in both of your examples here, where (a, b) are
assumed positive odd integers, (a>b), gcd(a,b)=1, so we
have all the terms are expressible in two variables
(a & b), (seem done in my submitted proof)
4(ab)^6 + (a^6 â€“ b^6)^2 = (a^6 + b^6)^2
Sure, that's an algebraic identity.
Post by bassam king karzeddin
So, (2z = a^6 + b^6),
I don't follow that at all.
x,y,z pairwise coprime nonzero integers.
x^6 + y^6 = z^2
the primitive triplet for the above example (x^3, y^3, z),
where y is only even integer, where also when multiplied
by two, would results on other similar triplet with double
size of the original triangle, (2x^3, 2y^3, 2z), where all
sides in this triplet are even integers!, thus we have
4x^6 + 4y^3 = 4z^2
Post by quasi
Post by bassam king karzeddin
Post by quasi
x odd, x = a*b, gcd(a,b) = 1, a > b
Thus you have
4(ab)^6 + 4y^6 = (2z)^2
4(ab)^6 + (a^6 â€“ b^6)^2 = (a^6 + b^6)^2
Note here, all the terms of the primitive original triplet
are multiplied by 4, so divide the identity by 4, would get
back the original triplet (SO...EASY)
Post by quasi
Post by bassam king karzeddin
Post by quasi
But you haven't proved 4y^6 = (a^6 - b^6)^2, so you
can't justify the claim 2z = a^6 + b^6.
I hope, this time would be convinced
Post by quasi
Post by bassam king karzeddin
Post by quasi
Post by bassam king karzeddin
Please, someone helps Quasi to this simple proof,
(Iâ€™m sorry Quasi)
I _have_ a simple proof of (1) -- yours appears flawed,
or at best incomplete.
It is indeed more than rigorous,
Rigor mortis.
that is why nobody is capable to criticise it,
I criticized it.
otherwise you would certainly see many
Given your lack of coherence, many have probably stopped
Post by quasi
Post by bassam king karzeddin
Most likely you have one, that we didn't see yet,
I'll post it later this week, but it's a busy week, so probably
Thursday.
Post by bassam king karzeddin
Post by quasi
Is anyone else able to follow bassam's argument?
No response yet from other readers.
And it is generally rare that anyone would respond to an
absolute fact,
Absolute fact? You're surely hallucinating.
but many responses would be there for a silly matters
But in this case, since your argument is essentially
gibberish, many won't bother to respond.
Post by quasi
Post by bassam king karzeddin
Post by quasi
Post by bassam king karzeddin
The next one would follow.
I don't have a proof of (2), but I don't believe you
have one either -- at least not a proof that would
convince anyone at the level, say, of an undergraduate
math major.
Do you really think that I was conjecturing randomly!?
It's fine as a conjecture.

I don't accept the claim that your posted proof is correct
for the case x^6 + y^6 = z^2.

And I'm absolutely certain that you don't have a correct
proof (even though you haven't yet posted it) for the case
x^6 + y^4 = z^2.
Post by quasi
Post by bassam king karzeddin
Consider this triplet (2(ab)^3, (a^6 - b^6), (a^6 + b^6)), as
integer sides of right angle triangle, This is Pythagorean
original identity only as (2ab, (a^2 - b^2), (a^2 + b^2), but
I considered both (a, b) as odd cubes, that is all!
4(ab)^6 + (a^6 â€“ b^6)^2 = (a^6 + b^6)^2 , that I provided
earlier.
Note that all the terms of this triple are divisible by 4,
(then becomes primitive), that is why you might got confused
while seeing the terms
2y^3 = a^6 - b^6, and 2z = a^6 + b^6 , so divide all by 4,
then it is clear, I'm quite sure you would realize it very
soon,
Without loss of generality we can assume x is a positive
integer.
You are assuming x is odd. No problem.
Then you write x = a*b, where a,b coprime positive integers
with a > b.
That implies x > 1, but still no problem.
Yes no problem at all whenever we consider primitive triplets
Post by quasi
You have two primitive pythagorean triples ...
(x^3,y^3,z) [by hypothesis]
Yes
Post by quasi
((ab)^2,(a^6-b^6)/2,(a^6 + b^6)/2) [algebraic identity]
Wait here, this is the algebraic identity, not the one you
mentioned above!
((ab)^3, (a^6-b^6)/2,(a^6 + b^6)/2) [algebraic identity]
Yes, typo -- I meant (ab)^3, not (ab)^2.
Post by quasi
Since x = a*b, the first components of the two triples are equal.
But that does not imply that the triples are equal.
Meaningless, and I don't understand the intention behind that!
Your claim that z = (a^6 + b^6)/2, is equivalent to claiming
that the two triples are equal, but as I pointed out, there's
no justification for that.
Post by quasi
Thus, your claim that z must equal (a^6 + b^6)/2 is an
unsupported claim.
This is very obvious from the identity itself, and doesn't
require any kind of support
Obvious to you perhaps, but as I said, it doesn't follow.
Post by quasi
To illustrate the flaw in your argument, here are two
distinct primitive pythagorean triples which have equal
15, 8, 17
15, 112, 113
Of course yes, an odd integer may occur in so many distinct >primitive triplets, depending on the prime factors of the odd >integer, I can provide you with so many examples if you would
like, so where is the flaw then in my reasoning?
I'll repeat my objection.

You have two primitive pythagorean triples:

(x^3, y^3, z)

((ab)^3, (a^6-b^6)/2, (a^6 + b^6)/2)

You also have x = a*b.

Thus, the two triples have the same first component.

But that's not enough to prove the two triples are equal.

Thus, you can't conclude z = (a^6 + b^6)/2.
Post by quasi
Post by bassam king karzeddin
I would like to note that doesn't require any high level in
Math to convey,
I do repeat the same words mentioned by me
Post by quasi
Post by bassam king karzeddin
Just a little revision and every thing would be so clear,
And I know that you are not mudding, but honest and skeptical
to raise the truth above, no matter who would win,
Yes, definitely -- truth matters.
the fact and nothing else must win
Agreed.

But there needs to be independent verification by others.

As of now, it appears that no reputable sci.math reader has
yet stated that they accept your proof -- a proof which you
claim is totally obvious and "more than rigorous".

I wonder why?

quasi
bassam king karzeddin
2016-09-04 19:32:11 UTC
Post by quasi
Post by quasi
Post by bassam king karzeddin
Post by quasi
Post by bassam king karzeddin
Post by quasi
I don't expect much -- probably more nonsense, but let's
see you (bassam king karzeddin) prove either or both of
the following particular cases. I can prove the first one;
I don't have a proof of the second; I don't believe you
can prove either one.
(1) There do not exist pairwise coprime nonzero integers
x,y,z such that x^6 + y^6 = z^2.
(2) There do not exist pairwise coprime nonzero integers
x,y,z such that x^6 + y^4 = z^2.
In both examples you challenge, are assumed Pythagorean
triplet (due to even exponents), so no harm if you assume
(x = a*b) in both of your examples here, where (a, b) are
assumed positive odd integers, (a>b), gcd(a,b)=1, so we
have all the terms are expressible in two variables
(a & b), (seem done in my submitted proof)
4(ab)^6 + (a^6 â€“ b^6)^2 = (a^6 + b^6)^2
Sure, that's an algebraic identity.
Post by bassam king karzeddin
So, (2z = a^6 + b^6),
I don't follow that at all.
x,y,z pairwise coprime nonzero integers.
x^6 + y^6 = z^2
the primitive triplet for the above example (x^3, y^3, z),
where y is only even integer, where also when multiplied
by two, would results on other similar triplet with double
size of the original triangle, (2x^3, 2y^3, 2z), where all
sides in this triplet are even integers!, thus we have
4x^6 + 4y^3 = 4z^2
Post by quasi
Post by bassam king karzeddin
Post by quasi
x odd, x = a*b, gcd(a,b) = 1, a > b
Thus you have
4(ab)^6 + 4y^6 = (2z)^2
4(ab)^6 + (a^6 â€“ b^6)^2 = (a^6 + b^6)^2
Note here, all the terms of the primitive original triplet
are multiplied by 4, so divide the identity by 4, would get
back the original triplet (SO...EASY)
Post by quasi
Post by bassam king karzeddin
Post by quasi
But you haven't proved 4y^6 = (a^6 - b^6)^2, so you
can't justify the claim 2z = a^6 + b^6.
I hope, this time would be convinced
Post by quasi
Post by bassam king karzeddin
Post by quasi
Post by bassam king karzeddin
Please, someone helps Quasi to this simple proof,
(Iâ€™m sorry Quasi)
I _have_ a simple proof of (1) -- yours appears flawed,
or at best incomplete.
It is indeed more than rigorous,
Rigor mortis.
that is why nobody is capable to criticise it,
I criticized it.
Alone!!??

Without any hint to obvious mistake!?

I really would be happier if I'm mistaken
Post by quasi
otherwise you would certainly see many
Given your lack of coherence, many have probably stopped
But the views are increasing, I checked the number
Post by quasi
Post by quasi
Post by bassam king karzeddin
Most likely you have one, that we didn't see yet,
I'll post it later this week, but it's a busy week, so probably
Thursday.
Post by bassam king karzeddin
Post by quasi
Is anyone else able to follow bassam's argument?
No response yet from other readers.
And it is generally rare that anyone would respond to an
absolute fact,
Absolute fact? You're surely hallucinating.
Unless you have a counter example,
Post by quasi
but many responses would be there for a silly matters
But in this case, since your argument is essentially
gibberish, many won't bother to respond.
I know that personal attitudes by the vast majority of mathematicians towards a person is much more important than the mathematics itself!, and surely you are an exception
Post by quasi
Post by quasi
Post by bassam king karzeddin
Post by quasi
Post by bassam king karzeddin
The next one would follow.
I don't have a proof of (2), but I don't believe you
have one either -- at least not a proof that would
convince anyone at the level, say, of an undergraduate
math major.
Do you really think that I was conjecturing randomly!?
It's fine as a conjecture.
I don't accept the claim that your posted proof is correct
for the case x^6 + y^6 = z^2.
I shall explain it in numerical example to get it the simple idea, :

let x = 15,

4*(15^6) + (5^6 - 3^6)^2 = (5^6 + 3^6)^2,

Forget about the right hand side, since 2z = 5^6 - 3^6 can be any integer (no problem)

Also the first term is assumed power 6 (started assumption - no problem),

what is so important is the evenly middle term that can be factored as (5^3 - 3^3) = 2*(7^2), only double an odd integer which isn't a cube or even double a cube, in order to make the equation balance to power 3 (then all powered to two), that we desire, , and (5^3 + 3^3) = (2^3)(19), that isn't either a cube, nor their multiplication is four times a cube,

And this feature is forever, for any assumed x, with all possible arrangement of its factors, being in so many distinct primitive triplets,
Post by quasi
And I'm absolutely certain that you don't have a correct
proof (even though you haven't yet posted it) for the case
x^6 + y^4 = z^2.
I wonder why should I post the sensitive proof, if my simplest one is still incomprehensible to the specialist in this field

So, let them enjoy it for few more centuries!

Or, a better idea, let it be a challenge to future generations of artificial intelligence only!
Post by quasi
Post by quasi
Post by bassam king karzeddin
Consider this triplet (2(ab)^3, (a^6 - b^6), (a^6 + b^6)), as
integer sides of right angle triangle, This is Pythagorean
original identity only as (2ab, (a^2 - b^2), (a^2 + b^2), but
I considered both (a, b) as odd cubes, that is all!
4(ab)^6 + (a^6 â€“ b^6)^2 = (a^6 + b^6)^2 , that I provided
earlier.
Note that all the terms of this triple are divisible by 4,
(then becomes primitive), that is why you might got confused
while seeing the terms
2y^3 = a^6 - b^6, and 2z = a^6 + b^6 , so divide all by 4,
then it is clear, I'm quite sure you would realize it very
soon,
Without loss of generality we can assume x is a positive
integer.
You are assuming x is odd. No problem.
Then you write x = a*b, where a,b coprime positive integers
with a > b.
That implies x > 1, but still no problem.
Yes no problem at all whenever we consider primitive triplets
Post by quasi
You have two primitive pythagorean triples ...
(x^3,y^3,z) [by hypothesis]
Yes
Post by quasi
((ab)^2,(a^6-b^6)/2,(a^6 + b^6)/2) [algebraic identity]
Wait here, this is the algebraic identity, not the one you
mentioned above!
((ab)^3, (a^6-b^6)/2,(a^6 + b^6)/2) [algebraic identity]
Yes, typo -- I meant (ab)^3, not (ab)^2.
I thought that was causing the confusion!
Post by quasi
Post by quasi
Since x = a*b, the first components of the two triples are equal.
But that does not imply that the triples are equal.
Meaningless, and I don't understand the intention behind that!
Your claim that z = (a^6 + b^6)/2, is equivalent to claiming
that the two triples are equal, but as I pointed out, there's
no justification for that.
Post by quasi
Thus, your claim that z must equal (a^6 + b^6)/2 is an
unsupported claim.
This is very obvious from the identity itself, and doesn't
require any kind of support
Obvious to you perhaps, but as I said, it doesn't follow.
It must follow, to balance the simplest identity
Post by quasi
Post by quasi
To illustrate the flaw in your argument, here are two
distinct primitive pythagorean triples which have equal
15, 8, 17
15, 112, 113
Of course yes, an odd integer may occur in so many distinct >primitive triplets, depending on the prime factors of the odd >integer, I can provide you with so many examples if you would
like, so where is the flaw then in my reasoning?
I'll repeat my objection.
(x^3, y^3, z)
((ab)^3, (a^6-b^6)/2, (a^6 + b^6)/2)
You also have x = a*b.
And y^3 = (a^6-b^6)/2, and z = (a^6 + b^6)/2)
Post by quasi
Thus, the two triples have the same first component.
But that's not enough to prove the two triples are equal.
Thus, you can't conclude z = (a^6 + b^6)/2.
Post by quasi
Post by bassam king karzeddin
I would like to note that doesn't require any high level in
Math to convey,
I do repeat the same words mentioned by me
Post by quasi
Post by bassam king karzeddin
Just a little revision and every thing would be so clear,
And I know that you are not mudding, but honest and skeptical
to raise the truth above, no matter who would win,
Yes, definitely -- truth matters.
So, victory for the truth, even it is killing
Post by quasi
the fact and nothing else must win
Agreed.
But there needs to be independent verification by others.
As of now, it appears that no reputable sci.math reader has
yet stated that they accept your proof -- a proof which you
claim is totally obvious and "more than rigorous".
I wonder why?
quasi
Absolute facts, don't necessarily require anyone acceptance, this is only our social behaviours

By the way, today I posted the same proof at overflow SE, but within few minutes, and without any comments, disappeared, but it may be still visible with clear notation at History SE, before they here me and delete it completely
quasi
2016-09-04 00:32:17 UTC
...
another case is gcd(m, n) = 1, is even much easier!
If it's so easy, let's see you prove that the equation

x^6 + y^4 = z^2

has no solutions in positive integers x,y,z with gcd(x,y,z) = 1.

I'm willing to bet that you won't be able to post a valid
proof, at least not one that would be accepted as correct by
the mainstream mathematical community.

quasi
bassam king karzeddin
2016-09-03 12:54:06 UTC
Post by quasi
Post by bassam king karzeddin
Post by quasi
...>>
...>>Conjecture: If (x, y, z) are nonzero coprime integers,
...>>and (n, m) are positive integers > 1, then this
...>>
...>> x^{2n} + y^{2m} = z^2
...>>
...>>doesn't have any integer solution.
Why not starting demonstrating with the simplest case here,
that is when (n = m =p), where (p) is any prime number,
But note, as I previously indicated, I was able to prove the
conjecture for the case where gcd(m,n) > 1.
Yes of course, let gcd(m,n) =k, where k is odd integer, and
impossible to be even integer, since in this even case you
would get the same form that Fermat had proved (a^4 + b^4 = c^2),
impossible in integers,
So, let (p) be an odd prime number that divides k, so eventually,
you arrive at this proposed case here , another case is gcd
(m, n) = 1, is even much easier!,
Easier? I don't believe it.
In any case, your posted "proof" for the case m = n = p, where p
is an odd prime looks like nonsense to me.
Post by bassam king karzeddin
I will come to it
Most likely, it will be just more nonsense..
Post by bassam king karzeddin
Post by quasi
I was unable to prove the conjecture for any _other_ case.
so the conjecture we have in magazine
In what magazine?
I mean world magazine, (google groups that is open to the world freely)!
You really have no concept of what sci.math is.
As previously explained, sci.math is part of usenet, a worldwide
network of user groups.
Google Groups and Math Forum each provide a partly filtered
two-way portal to sci.math, but they don't "own" sci.math.
Most sci.math regulars (myself included) access sci.math
directly from a usenet server via an NNTP "newsreader"
program; not from Google Groups; not from Math Forum.
Post by bassam king karzeddin
Post by quasi
(x^{2p} + y^{2p} = z^2) , where (x, y) are co prime positive
integers, then the conjecture says no integer solution
(generally)
Clearly, the case (p = 2), was Fermatâ€™s solution, so we are
left with odd primes only for (p), where this only represent
a Pythagorean triplets identity due to even exponents
Assume, (x = (x_1) (x_2), where (x) is odd integer and,
gcd(x_1, x_2) = 1, where x_1, x_2 are positive co prime
integers
What are you doing?
Are you trying to exploit the fact that (x^p,y^p,z) is
a primitive pythagorean triplet?
Yes, our conjecture assumes only that (is a pythagorean triplet)
If so, you can get expressions for x^p,y^p,z in terms of two
new variables.
And that is why all the terms are expressed in two variables
only, (x_1) & (x_2)
But the pythagorean triple formulas give you integer valued
expressions, in terms of two new parameters, for x^p,y^p,z;
not for x,y,z.
Post by bassam king karzeddin
Post by quasi
You don't get an expression for x.
X = (X_1)*(X_2), as specified above
Post by quasi
Moreover, writing x = (x_1)(x_2) seems useless since it allows
the trivial factorization x = (x)(1) (i.e., x_1 = x and x_2 = 1).
Yes of course, and I never said that one of the factors of (x) can't be one
So if you assume x1 = x, x2 = 1, how would the rest of your
argument proceed?
Post by bassam king karzeddin
Post by quasi
By substitution we get this form of equation
(2x^p)^2+ ((x_1)^{2p} â€“ (x_2)^{2p} )^2
= ((x_1)^{2p} + (x_2)^{2p} )^2
Then we have: ( (x_1) ^ {2p} + (x_2) ^ {2p} = 2z),
I have no idea how you got the above equation.
How does the factorization of x give you anything about y and z?
The solution of the following Diophantine equation
x^n + y^2 = z^2
in integers would be so easily and self proved identity as
the following,
2y = (x_1)^n - (x_2)^n, and 2z= (x_1)^n - (x_2)^n, where
x is odd integer,
???
Post by bassam king karzeddin
x = (x_1)*(x_2), and y therefore even integer, and z is odd
integer, this is the same as Pythagoras triplet when (n) is
even integer
???
Post by bassam king karzeddin
Given any positive integer >2, you can so simply find out all
possible primitive Pythagorean triplets, and from that number
only, very easy Quasi
???
Post by bassam king karzeddin
Please check again Quasi, and would be grateful to you if you can mistake me
It is as you once said, Too....Easy !
Sorry, bassam -- I'm now convinced you have no proof, not even
for the case gcd(m,n) > 1, which is, in fact, a readily provable
case (I have a proof).
As you've previously indicated, you are not a mathematician,
so it's reasonable to expect that your arguments would not be
expressed at the same level of logical precision as would be
expected from a mathematician. But there is a level of
imprecision below which it becomes virtually impossible for
mathematical language. Your posted arguments are definitely
below that threshold.
I don't expect much -- probably more nonsense, but let's see you
(bassam king karzeddin) prove either or both of the following
particular cases. I can prove the first one; I don't have a proof
of the second; I don't believe you can prove either one.
(1) There do not exist pairwise coprime nonzero integers x,y,z
such that x^6 + y^6 = z^2.
(2) There do not exist pairwise coprime nonzero integers x,y,z
such that x^6 + y^4 = z^2.
quasi
In both examples you challenge, are assumed Pythagorean triplet (due to even exponents), so no harm if you assume (x = a*b) in both of your examples here, where (a, b) are assumed positive odd integers, (a>b), gcd(a, b) =1, so we have all the terms are expressible in two variables (a & b), (seem done in my submitted proof)

So, we have first (x^6 + y^6 = z^2), substituting in (a & b), we get then:

4(ab)^6 + (a^6 â€“ b^6)^2 = (a^6 + b^6)^2

So, (2z = a^6 + b^6), which is possible, where z is odd integer

And, (2y^3 = a^6 â€“ b^6 = (a^3 â€“ b^3)*(a^3 + b^3),

Note that gcd ((a^3 + b^3), (a^3 â€“ b^3)) = 2, since both (a & b) are assumed odd integers, but (y) is even, factoring the terms:

(a^3 + y^3 = 2c^3), and (a^3 â€“ b^3 = d^3), where (y = c*d), and gcd (c, d) = 1, where also (c) is odd and (d )is even integer,

The contradiction here, we arrived at Fermat last theorem that must have solution in order to solve our conjecture,
a^3 â€“ b^3 = d^3

But FLT was proved impossible to have solution; therefore this conjecture is impossible to have solution â€“proved rigorously

Please, someone helps Quasi to this simple proof, (Iâ€™m sorry Quasi)

The next one would follow!

Note to readers at sci.math drexel: when I type from Google group, the SIGN of question mark appears (?), instead of negative sign (-), which makes the proof little unobvious, but strangely it doesn't happen at Google group?

So, replace the (?) by negative (-), in order to see the simple proof more clearly
Regards

Bassam Karzeddin
030916
Arturo Magidin
2016-09-03 20:20:53 UTC
On Saturday, September 3, 2016 at 6:02:24 AM UTC-5, quasi wrote:

[.aside.]
Post by quasi
You really have no concept of what sci.math is.
As previously explained, sci.math is part of usenet, a worldwide
network of user groups.
Google Groups and Math Forum each provide a partly filtered
two-way portal to sci.math, but they don't "own" sci.math.
Most sci.math regulars (myself included) access sci.math
directly from a usenet server via an NNTP "newsreader"
program; not from Google Groups; not from Math Forum.
Not that it matters, but:

I don't know if that is true; it certainly *used* to be true: I used to access it from a server at my graduate school. But a lot of universities have stopped hosting NNTP newsreaders. I for one had to migrate to google.groups to continue reading. A random sampling of messages I did just now revealed a bit more than half of them had google.groups IDs.

Usenet is not what it was; its ecological niche seems to have migrated to Tumblr and other media.

That does not detract from the fact that Bassam Karzeddin is indeed utterly and completely clueless and his statements betray a depth of ignorance that usually takes a lot of work to either achieve or maintain.
--
Arturo Magidin