Discussion:
Fermat's Last theorem short proof
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bassam king karzeddin
2007-02-21 14:51:51 UTC
Raw Message
Fermat's Last theorem short proof

We have the following general equation (using the general binomial theorem)

(x+y+z)^p =p* (x+y)*(x+z)*(y+z)*N(x,y,z) + x^p+y^p+z^p

Where
N (x, y, z) is integer function in terms of (x, y, z)
P is odd prime number
(x, y, z) are three (none zero) co prime integers?

Assuming a counter example (x, y, z) exists such that (x^p+y^p+z^p=0)

(x+y+z)^p =p* (x+y)*(x+z)*(y+z)*N (x, y, z)

CASE-1
If (p=3) implies N (x, y, z) = 1, so we have

(x+y+z)^3 =3* (x+y)*(x+z)*(y+z)

Assuming (3) does not divide (x*y*z), then it does not divide (x+y)*(x+z)*(y+z),
So the above equation does not have solution
(That is by dividing both sides by 3, you get 9 times an integer equal to an integer which is not divisible by 3, which of course is impossible
I think proof is completed for (p=3, and 3 is not a factor of (x*y*z)

My question to the specialist, is my proof a new one, more over I will not feel strange if this was known few centuries back

Thanking you a lot

Bassam King Karzeddin
Al-Hussein Bin Talal University
JORDAN
Randy Poe
2007-02-21 15:52:09 UTC
Raw Message
Post by bassam king karzeddin
Fermat's Last theorem short proof
We have the following general equation (using the general binomial theorem)
(x+y+z)^p =p* (x+y)*(x+z)*(y+z)*N(x,y,z) + x^p+y^p+z^p
Where
N (x, y, z) is integer function in terms of (x, y, z)
Are you claiming this is true in general?

Counterexample:
x=3, y=4, z=5, p=5.
(x+y+z)^p = 248832
(x^p + y^p + z^p) = 4392
(x+y)(x+z)(y+z) = 560

=> p*N(x,y,z) = (248832 - 4392)/560 = 436.5
Post by bassam king karzeddin
P is odd prime number
(x, y, z) are three (none zero) co prime integers?
Assuming a counter example (x, y, z) exists such that (x^p+y^p+z^p=0)
How can this be if x, y, z, p > 0?

- Randy
bassam king karzeddin
2007-02-23 11:56:41 UTC
Raw Message
Post by Randy Poe
How can this be if x, y, z, p > 0?
- Randy
Hi Randy
I may have forgotten to answer your second question

For a purpose of FLT I have defined (x, y, z) as integer numbers (I mean positive and negative numbers), so if

x^p+y^p+z^p=0, then obviously not all of them (x,y,z)are positive,
but in general one can see and prove (using the general binomial theorem) that the following identity holds true always

(x+y+z)^n = n*(x+y)*(x+z)*(y+z)*f(x,y,z) + x^n +y^n +z^n

where n is odd positive integer
(x,y,z) belong to C, complex numbers
f(x,y,z) is function in terms of (x,y,z)

My Regards

B.Karzeddin
Randy Poe
2007-02-23 14:55:42 UTC
Raw Message
Post by bassam king karzeddin
Post by Randy Poe
How can this be if x, y, z, p > 0?
- Randy
Hi Randy
I may have forgotten to answer your second question
What about my first question?
Post by bassam king karzeddin
For a purpose of FLT I have defined (x, y, z) as integer numbers (I mean positive and negative numbers), so if
x^p+y^p+z^p=0, then obviously not all of them (x,y,z)are positive,
OK. And after I posted I realized that perhaps you were
just writing FLT in a different form.

If x^p + y^p = z^p for p odd, then x^p + y^p + (-z)^p = 0.
Post by bassam king karzeddin
but in general one can see and prove (using the general binomial theorem) that the following identity holds true always
(x+y+z)^n = n*(x+y)*(x+z)*(y+z)*f(x,y,z) + x^n +y^n +z^n
where n is odd positive integer
(x,y,z) belong to C, complex numbers
f(x,y,z) is function in terms of (x,y,z)
Then what do you make of my counterexample? What are the
properties of f(x,y,z)? I thought in your
first post you said it was an integer. But I gave
a counterexample where n*f(x,y,z) is not an integer.

Can you sketch out how you get this identity from the
binomial theorem?

- Randy
Raymond Burhoe
2007-02-23 15:11:04 UTC
Raw Message
On Feb 23, 6:56 am, bassam king karzeddin
Post by bassam king karzeddin
Post by Randy Poe
How can this be if x, y, z, p > 0?
- Randy
Hi Randy
I may have forgotten to answer your second question
What about my first question?
Post by bassam king karzeddin
For a purpose of FLT I have defined (x, y, z) as
integer numbers (I mean positive and negative
numbers), so if
Post by bassam king karzeddin
x^p+y^p+z^p=0, then obviously not all of them
(x,y,z)are positive,
OK. And after I posted I realized that perhaps you
were
just writing FLT in a different form.
If x^p + y^p = z^p for p odd, then x^p + y^p + (-z)^p
= 0.
Yes, I think that's what he's doing. That means there will be integers a, b, c, and d, where

x + y = a^p

z + y = b^p

z + x = c^p

X + y + z = pabcd, where d is a multiple of p for p > 3.

Thus, p*N(x,y,z) = (pd)^p

I think his assumption that d is coprime to p is based on wrongly extending what is true for p = 3 into higher values of p.
Post by bassam king karzeddin
but in general one can see and prove (using the
general binomial theorem) that the following identity
holds true always
Post by bassam king karzeddin
(x+y+z)^n = n*(x+y)*(x+z)*(y+z)*f(x,y,z) + x^n +y^n
+z^n
Post by bassam king karzeddin
where n is odd positive integer
(x,y,z) belong to C, complex numbers
f(x,y,z) is function in terms of (x,y,z)
Then what do you make of my counterexample? What are
the
properties of f(x,y,z)? I thought in your
first post you said it was an integer. But I gave
a counterexample where n*f(x,y,z) is not an integer.
Can you sketch out how you get this identity from the
binomial theorem?
- Randy
bassam king karzeddin
2007-02-25 14:43:52 UTC
Raw Message
Post by bassam king karzeddin
On Feb 23, 6:56 am, bassam king karzeddin
Post by bassam king karzeddin
Post by Randy Poe
How can this be if x, y, z, p > 0?
- Randy
Hi Randy
I may have forgotten to answer your second
question
What about my first question?
Post by bassam king karzeddin
For a purpose of FLT I have defined (x, y, z) as
integer numbers (I mean positive and negative
numbers), so if
Post by bassam king karzeddin
x^p+y^p+z^p=0, then obviously not all of them
(x,y,z)are positive,
OK. And after I posted I realized that perhaps you
were
just writing FLT in a different form.
If x^p + y^p = z^p for p odd, then x^p + y^p +
(-z)^p
= 0.
Yes, I think that's what he's doing. That means
there will be integers a, b, c, and d, where
x + y = a^p
z + y = b^p
z + x = c^p
X + y + z = pabcd, where d is a multiple of p for p >
3.
Thus, p*N(x,y,z) = (pd)^p
I think his assumption that d is coprime to p is
based on wrongly extending what is true for p = 3
into higher values of p.
Can you support your claim by anexample???
Post by bassam king karzeddin
Post by bassam king karzeddin
but in general one can see and prove (using the
general binomial theorem) that the following
identity
holds true always
Post by bassam king karzeddin
(x+y+z)^n = n*(x+y)*(x+z)*(y+z)*f(x,y,z) + x^n
+y^n
+z^n
Post by bassam king karzeddin
where n is odd positive integer
(x,y,z) belong to C, complex numbers
f(x,y,z) is function in terms of (x,y,z)
Then what do you make of my counterexample? What
are
the
properties of f(x,y,z)? I thought in your
first post you said it was an integer. But I gave
a counterexample where n*f(x,y,z) is not an
integer.
Can you sketch out how you get this identity from
the
binomial theorem?
- Randy
Dan Cass
2007-02-23 19:13:51 UTC
Raw Message
On Feb 21, 9:51 am, bassam king karzeddin
Post by bassam king karzeddin
Fermat's Last theorem short proof
We have the following general equation (using the
general binomial theorem)
Post by bassam king karzeddin
(x+y+z)^p =p* (x+y)*(x+z)*(y+z)*N(x,y,z) +
x^p+y^p+z^p
Post by bassam king karzeddin
Where
N (x, y, z) is integer function in terms of (x, y,
z)
Are you claiming this is true in general?
x=3, y=4, z=5, p=5.
(x+y+z)^p = 248832
(x^p + y^p + z^p) = 4392
(x+y)(x+z)(y+z) = 560
Correction: (x+y)(x+z)(y+z) = 504,
and then (248832 - 4392)/504 is 485 = 5*97

Based on a few Maple calcs, I think the identity
(x+y+z)^p
=p*(x+y)*(x+z)*(y+z)*N(x,y,z) + x^p+y^p+z^p
gives an integer coefficient polynomial for N(x,y,z)
whenever p is odd and at least 3.

For p=3 we have N(x,y,z)=1,
For p=5 I get N(x,y,z)=x^2+y^2+z^2+xy+xz+yz
Note this correctly gives N(3,4,5) = 97.

If this approach could even be pushed to a valid
proof of FLT in the case n=3, it would be interesting...
Dan Cass
2007-02-23 19:23:07 UTC
Raw Message
Post by Dan Cass
On Feb 21, 9:51 am, bassam king karzeddin
Post by bassam king karzeddin
Fermat's Last theorem short proof
We have the following general equation (using
the
general binomial theorem)
Post by bassam king karzeddin
(x+y+z)^p =p* (x+y)*(x+z)*(y+z)*N(x,y,z) +
x^p+y^p+z^p
Post by bassam king karzeddin
Where
N (x, y, z) is integer function in terms of (x,
y,
z)
Are you claiming this is true in general?
x=3, y=4, z=5, p=5.
(x+y+z)^p = 248832
(x^p + y^p + z^p) = 4392
(x+y)(x+z)(y+z) = 560
Correction: (x+y)(x+z)(y+z) = 504,
and then (248832 - 4392)/504 is 485 = 5*97
Based on a few Maple calcs, I think the identity
(x+y+z)^p
=p*(x+y)*(x+z)*(y+z)*N(x,y,z) + x^p+y^p+z^p
gives an integer coefficient polynomial for N(x,y,z)
whenever p is odd and at least 3.
NO I only think it works for p an odd PRIME...
Post by Dan Cass
For p=3 we have N(x,y,z)=1,
For p=5 I get N(x,y,z)=x^2+y^2+z^2+xy+xz+yz
Note this correctly gives N(3,4,5) = 97.
If this approach could even be pushed to a valid
proof of FLT in the case n=3, it would be
interesting...
bassam king karzeddin
2007-02-24 18:35:15 UTC
Raw Message
Post by bassam king karzeddin
Post by Dan Cass
On Feb 21, 9:51 am, bassam king karzeddin
Post by bassam king karzeddin
Fermat's Last theorem short proof
We have the following general equation (using
the
general binomial theorem)
Post by bassam king karzeddin
(x+y+z)^p =p* (x+y)*(x+z)*(y+z)*N(x,y,z) +
x^p+y^p+z^p
Post by bassam king karzeddin
Where
N (x, y, z) is integer function in terms of
(x,
Post by Dan Cass
y,
z)
Are you claiming this is true in general?
x=3, y=4, z=5, p=5.
(x+y+z)^p = 248832
(x^p + y^p + z^p) = 4392
(x+y)(x+z)(y+z) = 560
Correction: (x+y)(x+z)(y+z) = 504,
and then (248832 - 4392)/504 is 485 = 5*97
Based on a few Maple calcs, I think the identity
(x+y+z)^p
=p*(x+y)*(x+z)*(y+z)*N(x,y,z) + x^p+y^p+z^p
gives an integer coefficient polynomial for
N(x,y,z)
Post by Dan Cass
whenever p is odd and at least 3.
NO I only think it works for p an odd PRIME...
So, you got it, and my proof (sorry Fermat's wounderful proof) is clear now when p is not a factor of x*y*z
Post by bassam king karzeddin
Post by Dan Cass
For p=3 we have N(x,y,z)=1,
For p=5 I get N(x,y,z)=x^2+y^2+z^2+xy+xz+yz
Note this correctly gives N(3,4,5) = 97.
If this approach could even be pushed to a valid
proof of FLT in the case n=3, it would be
interesting...
But, is it a new one???
And If so, why not the best Journals come down here to learn???
Doesn't Mathematics feeds them???
Shoudn't they be loyal to the science they eat???
Is not here a permanent written pages to the planet for free???
Is not here a place for science mathematics???

I WILL KEEP WONDERING!!!

My Regards

Bassam Karzeddin
Al Hussein bin Talal University
JORDAN
Randy Poe
2007-02-26 15:51:20 UTC
Raw Message
Post by Dan Cass
On Feb 21, 9:51 am, bassam king karzeddin
Post by bassam king karzeddin
Fermat's Last theorem short proof
We have the following general equation (using the
general binomial theorem)
Post by bassam king karzeddin
(x+y+z)^p =p* (x+y)*(x+z)*(y+z)*N(x,y,z) +
x^p+y^p+z^p
Post by bassam king karzeddin
Where
N (x, y, z) is integer function in terms of (x, y,
z)
Are you claiming this is true in general?
x=3, y=4, z=5, p=5.
(x+y+z)^p = 248832
(x^p + y^p + z^p) = 4392
(x+y)(x+z)(y+z) = 560
Correction: (x+y)(x+z)(y+z) = 504,
and then (248832 - 4392)/504 is 485 = 5*97
Based on a few Maple calcs, I think the identity
(x+y+z)^p
=p*(x+y)*(x+z)*(y+z)*N(x,y,z) + x^p+y^p+z^p
gives an integer coefficient polynomial for N(x,y,z)
whenever p is odd and at least 3.
Ah. OK. Glad somebody answered. OP refused to.
(Not sure how I got that arithmetic wrong.)

OK, we'll take that as valid. Now let's look at the rest
of the argument.
Post by Dan Cass
Assuming a counter example (x, y, z) exists such that (x^p+y^p+z^p=0)
(x+y+z)^p =p* (x+y)*(x+z)*(y+z)*N (x, y, z)
OK. So if a FLT counter-example exists, then there will
be x, y >0, z<0 such that this identity holds.
Post by Dan Cass
CASE-1
If (p=3) implies N (x, y, z) = 1,
Why does that follow?

Perhaps there is something about N(x,y,z) we are not being
told. I will ask OP for the third time to provide his
proof of this identity (which I will now accept as true)
so I understand what N(x,y,z) is.

It would be nice if he could answer the question I just

If for the third time he refuses, I'll abandon this thread.

- Randy
bassam king karzeddin
2007-02-26 16:53:10 UTC
Raw Message
Post by bassam king karzeddin
Post by Dan Cass
On Feb 21, 9:51 am, bassam king karzeddin
Post by bassam king karzeddin
Fermat's Last theorem short proof
We have the following general equation (using
the
Post by Dan Cass
general binomial theorem)
Post by bassam king karzeddin
(x+y+z)^p =p* (x+y)*(x+z)*(y+z)*N(x,y,z) +
x^p+y^p+z^p
Post by bassam king karzeddin
Where
N (x, y, z) is integer function in terms of (x,
y,
Post by Dan Cass
z)
Are you claiming this is true in general?
x=3, y=4, z=5, p=5.
(x+y+z)^p = 248832
(x^p + y^p + z^p) = 4392
(x+y)(x+z)(y+z) = 560
Correction: (x+y)(x+z)(y+z) = 504,
and then (248832 - 4392)/504 is 485 = 5*97
Based on a few Maple calcs, I think the identity
(x+y+z)^p
=p*(x+y)*(x+z)*(y+z)*N(x,y,z) + x^p+y^p+z^p
gives an integer coefficient polynomial for
N(x,y,z)
Post by Dan Cass
whenever p is odd and at least 3.
Ah. OK. Glad somebody answered. OP refused to.
(Not sure how I got that arithmetic wrong.)
OK, we'll take that as valid. Now let's look at the
rest
of the argument.
Post by Dan Cass
Assuming a counter example (x, y, z) exists such
that (x^p+y^p+z^p=0)
Post by Dan Cass
(x+y+z)^p =p* (x+y)*(x+z)*(y+z)*N (x, y, z)
OK. So if a FLT counter-example exists, then there
will
be x, y >0, z<0 such that this identity holds.
Post by Dan Cass
CASE-1
If (p=3) implies N (x, y, z) = 1,
Why does that follow?
Perhaps there is something about N(x,y,z) we are not
being
told. I will ask OP for the third time to provide his
proof of this identity (which I will now accept as
true)
so I understand what N(x,y,z) is.
It would be nice if he could answer the question I
just
If for the third time he refuses, I'll abandon this
- Randy
Hi Randy

I will answer you

Given two identical sets of things and a balance,and you are asked to put one set on only one side of the balance, so the second set is on the other side of the balance, then equilibrium state is attained, no matter if you keep rearranging them in deferent manners only on each side
So, do the multiplication please, then all terms will be canceled, and you will get (0=0)

A CUBE CAN'T BE TRIPLED,
AN EQUATION IS BETTER THAN A CIVILIZATION

I HOPE THAT CAN HELP

MY REGARDS

Bassam Karzeddin
AL Hussein bin Talal University
JORDAN
Randy Poe
2007-02-26 17:49:22 UTC
Raw Message
Post by bassam king karzeddin
Post by bassam king karzeddin
Post by Dan Cass
On Feb 21, 9:51 am, bassam king karzeddin
Post by bassam king karzeddin
Fermat's Last theorem short proof
We have the following general equation (using
the
Post by Dan Cass
general binomial theorem)
Post by bassam king karzeddin
(x+y+z)^p =p* (x+y)*(x+z)*(y+z)*N(x,y,z) +
x^p+y^p+z^p
Post by bassam king karzeddin
Where
N (x, y, z) is integer function in terms of (x,
y,
Post by Dan Cass
z)
Are you claiming this is true in general?
x=3, y=4, z=5, p=5.
(x+y+z)^p = 248832
(x^p + y^p + z^p) = 4392
(x+y)(x+z)(y+z) = 560
Correction: (x+y)(x+z)(y+z) = 504,
and then (248832 - 4392)/504 is 485 = 5*97
Based on a few Maple calcs, I think the identity
(x+y+z)^p
=p*(x+y)*(x+z)*(y+z)*N(x,y,z) + x^p+y^p+z^p
gives an integer coefficient polynomial for
N(x,y,z)
Post by Dan Cass
whenever p is odd and at least 3.
Ah. OK. Glad somebody answered. OP refused to.
(Not sure how I got that arithmetic wrong.)
OK, we'll take that as valid. Now let's look at the
rest
of the argument.
Post by Dan Cass
Assuming a counter example (x, y, z) exists such
that (x^p+y^p+z^p=0)
Post by Dan Cass
(x+y+z)^p =p* (x+y)*(x+z)*(y+z)*N (x, y, z)
OK. So if a FLT counter-example exists, then there
will
be x, y >0, z<0 such that this identity holds.
Post by Dan Cass
CASE-1
If (p=3) implies N (x, y, z) = 1,
Why does that follow?
Perhaps there is something about N(x,y,z) we are not
being
told. I will ask OP for the third time to provide his
proof of this identity (which I will now accept as
true)
so I understand what N(x,y,z) is.
It would be nice if he could answer the question I
just
If for the third time he refuses, I'll abandon this
- Randy
Hi Randy
I will answer you
Given two identical sets of things and a balance,and you are asked to put one set on only one side of the balance, so the second set is on the other side of the balance, then equilibrium state is attained, no matter if you keep rearranging them in deferent manners only on each side
So, do the multiplication please, then all terms will be canceled, and you will get (0=0)
A CUBE CAN'T BE TRIPLED,
AN EQUATION IS BETTER THAN A CIVILIZATION
I HOPE THAT CAN HELP
I now follow most of your argument. I'll just repeat my
last statement:

"what you've shown as far as I can tell is that
if x^3 + y^3 + z^3 = 0 has a solution, then one of
x, y or z must be divisible by 3. "

I agree that if p=3 and (x,y,z) is a FLT counter
example, then assuming xyz <> 0 mod 3 leads to
a contradiction. Hence (x,y,z) FLT counter example
implies that xyz = 0 mod 3.

- Randy

bassam king karzeddin
2007-02-26 16:55:05 UTC
Raw Message
Post by bassam king karzeddin
Post by Dan Cass
On Feb 21, 9:51 am, bassam king karzeddin
Post by bassam king karzeddin
Fermat's Last theorem short proof
We have the following general equation (using
the
Post by Dan Cass
general binomial theorem)
Post by bassam king karzeddin
(x+y+z)^p =p* (x+y)*(x+z)*(y+z)*N(x,y,z) +
x^p+y^p+z^p
Post by bassam king karzeddin
Where
N (x, y, z) is integer function in terms of (x,
y,
Post by Dan Cass
z)
Are you claiming this is true in general?
x=3, y=4, z=5, p=5.
(x+y+z)^p = 248832
(x^p + y^p + z^p) = 4392
(x+y)(x+z)(y+z) = 560
Correction: (x+y)(x+z)(y+z) = 504,
and then (248832 - 4392)/504 is 485 = 5*97
Based on a few Maple calcs, I think the identity
(x+y+z)^p
=p*(x+y)*(x+z)*(y+z)*N(x,y,z) + x^p+y^p+z^p
gives an integer coefficient polynomial for
N(x,y,z)
Post by Dan Cass
whenever p is odd and at least 3.
Ah. OK. Glad somebody answered. OP refused to.
(Not sure how I got that arithmetic wrong.)
OK, we'll take that as valid. Now let's look at the
rest
of the argument.
Post by Dan Cass
Assuming a counter example (x, y, z) exists such
that (x^p+y^p+z^p=0)
Post by Dan Cass
(x+y+z)^p =p* (x+y)*(x+z)*(y+z)*N (x, y, z)
OK. So if a FLT counter-example exists, then there
will
be x, y >0, z<0 such that this identity holds.
Post by Dan Cass
CASE-1
If (p=3) implies N (x, y, z) = 1,
Why does that follow?
Perhaps there is something about N(x,y,z) we are not
being
told. I will ask OP for the third time to provide his
proof of this identity (which I will now accept as
true)
so I understand what N(x,y,z) is.
It would be nice if he could answer the question I
just
If for the third time he refuses, I'll abandon this
- Randy
Hi Randy

I will answer you

Given two identical sets of things and a balance,and you are asked to put one set on only one side of the balance, so the second set is on the other side of the balance, then equilibrium state is attained, no matter if you keep rearranging them in deferent manners only on each side
So, do the multiplication please, then all terms will be canceled, and you will get (0=0)

A CUBE CAN'T BE TRIPLED,
AN EQUATION IS BETTER THAN A CIVILIZATION

I HOPE THAT CAN HELP

MY REGARDS

Bassam Karzeddin
AL Hussein bin Talal University
JORDAN
bassam king karzeddin
2007-02-23 12:22:42 UTC
Raw Message
Hello all

Why not come out of your holes and say a word of truth?

strangely, If some one makes a mistake then many outstanding mathematicians will appear immediately correcting or harassing the OP

I will be grateful and happier if my proof is wrong, or worthless or known before

I'm also in my own way to make a much simpler and shorter proof that you may teach your kids before going to school
Here, I will not stop, and my aim is far a head of FLT.

My Regards
B.Karzeddin
Raymond Burhoe
2007-02-23 14:32:39 UTC
Raw Message
We have the following general equation which any
mathematician can prove(using the general binomial
theorem)
(x+y+z)^p =p* (x+y)*(x+z)*(y+z)*N(x,y,z) +
x^p+y^p+z^p
Where
N (x, y, z) is odd integer function in terms of
(x, y, z) and is prime to (p)
P is odd prime number
(x, y, z) are three distinct(none zero) co prime
integers?
Assuming a counter example (x, y, z) exists such that
(x^p+y^p+z^p=0)
Then we have the following equation
(x+y+z)^p =p* (x+y)*(x+z)*(y+z)*N (x, y, z)
CASE-1
If (p=3) implies N (x, y, z) = 1, so we have
(x+y+z)^3 =3* (x+y)*(x+z)*(y+z)
Assuming (3) does not divide (x*y*z), then it does
not divide (x+y)*(x+z)*(y+z),because
All prime factors of (x+y) devide z,
All prime factors of (x+z) devide y
All prime factors of (y+z) devide x
So the above equation does not have solution ,
because 3 must be a prime factor of both sides of the
above equation
(That is by dividing both sides by 3, you get
(3)^(3k-1),on the left hand and no (3) factor on the
right hand of the above same equation,( where k is
positive integer) , and that implies our wrong
assumptiono of the existance of a counter example
which of course is impossible
I think proof is completed for (p=3, and 3 is not a
factor of (x*y*z)
P = 3 is a special case, where (x + y + z) is a multiple of 3 but not a multiple of 9 whenever x*y*z is a nonmultiple of 3. It can be proved that any prime factor of x*y*z that is coprime to (x + y)(z + y)(z + x) must have the form 2pj + 1, so that it follows from the FLT equation that (x + y + z) = 0 (mod p^2). If you want anyone to believe your proof, you're going to have to prove your claim that N(x,y,z) is coprime to p when p > 4.
The same arguments applies for any odd prime power p
,
(that is only needed to prove FLT
where (p) doesn't divide (x*y*z), as the following
If x^p+y^p+z^p=0, implies
(x+y+z)^p =p* (x+y)*(x+z)*(y+z)*N(x,y,z)
Assuming (p) does not divide (x*y*z), then (p) does
not divide (x+y)*(x+z)*(y+z),because
All prime factors of (x+y) devide z,
All prime factors of (x+z) devide y
All prime factors of (y+z) devide x
And we have gcd(p,N(x,y,z))=1
So the above equation does not have solution ,
because p must be a prime factor of both sides of the
above equation
(That is by dividing both sides by p, you get
(p)^(p*k-1),on the left hand and no (p) factor on the
right hand of the above same equation,( where k is
positive integer) , and that implies our wrong
assumptiono of the existance of a counter example
which of course is impossible
I think proof is completed for (p) any odd prime
number, where (p) doesn't divide (x*y*z)
The same arguments applies for any odd prime power p
,
(that is only needed to prove FLT
My question to the specialists, is my proof a new
one, more over I will not feel strange if this was
known few centuries back
Thanking you a lot
Bassam King Karzeddin
Al-Hussein Bin Talal University
JORDAN
Message was edited by: bassam king karzeddin
bassam king karzeddin
2007-02-25 14:35:33 UTC
Raw Message
I'm also not a mathematician and this proof is ment mainly to nonmathematicians
Post by Raymond Burhoe
We have the following general equation which any
mathematician can prove(using the general binomial
theorem)
(x+y+z)^p =p* (x+y)*(x+z)*(y+z)*N(x,y,z) +
x^p+y^p+z^p
Where
N (x, y, z) is odd integer function in terms of
(x, y, z) and is prime to (p)
P is odd prime number
(x, y, z) are three distinct(none zero) co prime
integers?
Assuming a counter example (x, y, z) exists such
that
(x^p+y^p+z^p=0)
Then we have the following equation
(x+y+z)^p =p* (x+y)*(x+z)*(y+z)*N (x, y, z)
CASE-1
If (p=3) implies N (x, y, z) = 1, so we have
(x+y+z)^3 =3* (x+y)*(x+z)*(y+z)
Assuming (3) does not divide (x*y*z), then it does
not divide (x+y)*(x+z)*(y+z),because
All prime factors of (x+y) devide z,
All prime factors of (x+z) devide y
All prime factors of (y+z) devide x
So the above equation does not have solution ,
because 3 must be a prime factor of both sides of
the
above equation
(That is by dividing both sides by 3, you get
(3)^(3k-1),on the left hand and no (3) factor on
the
right hand of the above same equation,( where k is
positive integer) , and that implies our wrong
assumptiono of the existance of a counter example
which of course is impossible
I think proof is completed for (p=3, and 3 is not
a
factor of (x*y*z)
P = 3 is a special case, where (x + y + z) is a
multiple of 3 but not a multiple of 9 whenever x*y*z
is a nonmultiple of 3. It can be proved that any
prime factor of x*y*z that is coprime to (x + y)(z +
y)(z + x) must have the form 2pj + 1, so that it
follows from the FLT equation that (x + y + z) = 0
(mod p^2). If you want anyone to believe your proof,
you're going to have to prove your claim that
N(x,y,z) is coprime to p when p > 4.
You have (x+y+z)^3 = 3*(x+y)*(x+z)*(y+z) for a counter example, you assume 3 is not a factor of xyz that implies 3 is not a factor of (x+y)(x+z)(y+z), and
in your example (x+y+z) = 9*m, so aplly to the above eqn. after dividing both sides of the above eqn.by 3, then you get 3^5 on the left and no 3 factor on the right, hence your assumption is wrong

The other part of your question is not that difficult, I think many have done the symbolic calculations and rearranged them in to a disirable shaps
Post by Raymond Burhoe
The same arguments applies for any odd prime power
p
,
(that is only needed to prove FLT
where (p) doesn't divide (x*y*z), as the following
If x^p+y^p+z^p=0, implies
(x+y+z)^p =p* (x+y)*(x+z)*(y+z)*N(x,y,z)
Assuming (p) does not divide (x*y*z), then (p)
does
not divide (x+y)*(x+z)*(y+z),because
All prime factors of (x+y) devide z,
All prime factors of (x+z) devide y
All prime factors of (y+z) devide x
And we have gcd(p,N(x,y,z))=1
So the above equation does not have solution ,
because p must be a prime factor of both sides of
the
above equation
(That is by dividing both sides by p, you get
(p)^(p*k-1),on the left hand and no (p) factor on
the
right hand of the above same equation,( where k is
positive integer) , and that implies our wrong
assumptiono of the existance of a counter example
which of course is impossible
I think proof is completed for (p) any odd prime
number, where (p) doesn't divide (x*y*z)
The same arguments applies for any odd prime power
p
,
(that is only needed to prove FLT
My question to the specialists, is my proof a new
one, more over I will not feel strange if this was
known few centuries back
Thanking you a lot
Bassam King Karzeddin
Al-Hussein Bin Talal University
JORDAN
Message was edited by: bassam king karzeddin
Jeroen
2007-02-26 12:31:31 UTC
Raw Message
Post by bassam king karzeddin
I'm also not a mathematician and this proof is ment mainly to nonmathematicians
There is no proof, because the first line is wrong given the
counterexample of Randy Poe. You stated that:

(x+y+z)^p =p* (x+y)*(x+z)*(y+z)*N(x,y,z) + x^p+y^p+z^p

Randy replied:

"Are you claiming this is true in general?

Counterexample:
x=3, y=4, z=5, p=5.
(x+y+z)^p = 248832
(x^p + y^p + z^p) = 4392
(x+y)(x+z)(y+z) = 560

=> p*N(x,y,z) = (248832 - 4392)/560 = 436.5
"

You didn't reply to this counterexample yet....

Jeroen
bassam king karzeddin
2007-02-26 15:38:36 UTC
Raw Message
bassam king karzeddin
I'm also not a mathematician and this proof is meant
mainly to non mathematicians and was invented by a non mathematician but great genius few centuries back
There is no proof, because the first line is wrong
given the
(x+y+z)^p =p* (x+y)*(x+z)*(y+z)*N(x,y,z) +
) + x^p+y^p+z^p
"Are you claiming this is true in general?
x=3, y=4, z=5, p=5.
(x+y+z)^p = 248832
(x^p + y^p + z^p) = 4392
(x+y)(x+z)(y+z) = 560
=> p*N(x,y,z) = (248832 - 4392)/560 = 436.5
"
You didn't reply to this counterexample yet....
Jeroen
You may ask Randy himself or check replies or calculations

Regards
B.Karzeddin
Hisanobu Shinya
2007-02-26 13:49:32 UTC
Raw Message
Post by bassam king karzeddin
Fermat's Last theorem short proof
We have the following general equation (using the
general binomial theorem)
(x+y+z)^p =p* (x+y)*(x+z)*(y+z)*N(x,y,z) +
x^p+y^p+z^p
Where
N (x, y, z) is integer function in terms of (x, y, z)
P is odd prime number
(x, y, z) are three (none zero) co prime integers?
Maybe you are using the division algorithm:

Given integers a and b, there exist q and r < |a| such that

b = aq + r.

However, you have no information on q and r.
Post by bassam king karzeddin
Assuming a counter example (x, y, z) exists such that
(x^p+y^p+z^p=0)
(x+y+z)^p =p* (x+y)*(x+z)*(y+z)*N (x, y, z)
CASE-1
If (p=3) implies N (x, y, z) = 1, so we have
(x+y+z)^3 =3* (x+y)*(x+z)*(y+z)
Assuming (3) does not divide (x*y*z), then it does
not divide (x+y)*(x+z)*(y+z),
So the above equation does not have solution
(That is by dividing both sides by 3, you get 9 times
an integer equal to an integer which is not divisible
by 3, which of course is impossible
I think proof is completed for (p=3, and 3 is not a
factor of (x*y*z)
My question to the specialist, is my proof a new one,
more over I will not feel strange if this was known
few centuries back
Thanking you a lot
Bassam King Karzeddin
Al-Hussein Bin Talal University
JORDAN
Randy Poe
2007-02-26 16:04:30 UTC
Raw Message
Post by bassam king karzeddin
Fermat's Last theorem short proof
We have the following general equation (using the general binomial theorem)
(x+y+z)^p =p* (x+y)*(x+z)*(y+z)*N(x,y,z) + x^p+y^p+z^p
Where
N (x, y, z) is integer function in terms of (x, y, z)
P is odd prime number
(x, y, z) are three (none zero) co prime integers?
Assuming a counter example (x, y, z) exists such that (x^p+y^p+z^p=0)
(x+y+z)^p =p* (x+y)*(x+z)*(y+z)*N (x, y, z)
CASE-1
If (p=3) implies N (x, y, z) = 1, so we have
(x+y+z)^3 =3* (x+y)*(x+z)*(y+z)
All right, you didn't answer but somebody else did. I have
now seen a validation of this identity including the
statement that N(x,y,z) = 1 for p = 3.

In that case, yes this is true with x,y>0, z<0 if there
exists a FLT counterexample with p=3,
Post by bassam king karzeddin
Assuming (3) does not divide (x*y*z), then it does not divide (x+y)*(x+z)*(y+z),
So the above equation does not have solution
(That is by dividing both sides by 3, you get 9 times an integer equal to an integer which is not divisible by 3,
You are still leaving a lot out. Dividing both sides by 3,
I get (x+y+z)^3/3 = (x+y)*(x+z)*(y+z)

OK, if xyz is not divisible by 3, then none of x, y or z
are divisible by 3. But (x+y+z) could still be divisible
by 3.

If it is, then (x+y+z)^3 must be divisible by 27 so the
left hand side, as you say, is divisible by 9.

Why isn't the right hand side divisible by 3?

Suppose (x+y+z) = 3k for some integer k.
Then (x+y) = 3k-z, and 3 does not divide z, therefore
3 does not divide (x+y). Similarly for (x+z), (y+z).

All right, your contradiction argument seems OK to me
unless I'm missing something. I wish you'd actually
*provided* your argument instead of leaving it for