Primitive Pythagorean Triplets

In article

Post by bassam king karzeddin
Why it is impossible to find a single primitive Pythagorean triplet (in
positive integers) being with all terms as powerful numbers?
Regards
Bassam King Karzeddin

What makes one positive integer more "powerful" than another?

Does badass mean numbers expressed as powers?

Note all terms of 3^2 + 4^2 = 5^2 are expressed as powers.

--
Virgil
"Mit der Dummheit kampfen Gotter selbst vergebens." (Schiller)

Peter Percival

2016-11-25 09:09:50 UTC

Post by Virgil
In article

Post by bassam king karzeddin
Why it is impossible to find a single primitive Pythagorean triplet (in
positive integers) being with all terms as powerful numbers?
Regards
Bassam King Karzeddin

What makes one positive integer more "powerful" than another?

Istr that a positive integer is called powerful if it is the product of
a square and a cube.

Post by Virgil
Does badass mean numbers expressed as powers?
Note all terms of 3^2 + 4^2 = 5^2 are expressed as powers.

--
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

Pubkeybreaker

2016-11-25 16:41:38 UTC

Post by Virgil
In article

Post by bassam king karzeddin
Why it is impossible to find a single primitive Pythagorean triplet (in
positive integers) being with all terms as powerful numbers?
Regards
Bassam King Karzeddin

What makes one positive integer more "powerful" than another?

Istr that a positive integer is called powerful if it is the product of
a square and a cube.

NO!

A powerful number is an integer N > 1 such that if p|N, then p^a exactly
divides N for some a > 1.

That is to say, that any prime dividing N must do so to a power higher than 1.

bassam king karzeddin

2016-11-26 09:22:50 UTC

Post by Virgil
In article

Post by bassam king karzeddin
Why it is impossible to find a single primitive Pythagorean triplet (in
positive integers) being with all terms as powerful numbers?
Regards
Bassam King Karzeddin

What makes one positive integer more "powerful" than another?

Istr that a positive integer is called powerful if it is the product of
a square and a cube.

NO!
A powerful number is an integer N > 1 such that if p|N, then p^a exactly
divides N for some a > 1.
That is to say, that any prime dividing N must do so to a power higher than 1.

Exactly, and this is sufficient for our defined problem, but why not for one or zero or even negative integers?

Regards
Bassam King Karzeddin

a***@gmail.com

2016-11-27 05:11:29 UTC

isn't thta just a corollary of Fermat's little theorem,

a^p = a mod p?... but,
not p-summorial!

Post by Pubkeybreaker
A powerful number is an integer N > 1 such that if p|N, then p^a exactly
divides N for some a > 1.
That is to say, that any prime dividing N must do so to a power higher than 1.

bassam king karzeddin

2016-11-27 07:06:12 UTC

Post by a***@gmail.com
isn't thta just a corollary of Fermat's little theorem,
a^p = a mod p?... but,
not p-summorial!

Your comment is not clear enough to me, nor ever you had been clear in your comments, but if you mean that my theorem is coming from Fermat (little or last theorem), I would tell you it includes it fully, but its range is much wider than that to establish new section in number theory

Regards
Bassam King Karzeddin

a***@gmail.com

2017-03-29 22:16:46 UTC

no, no, no. Fermat's theorem on bisums of one power,
is only the "last" because so-long unsolved, although
I solved the characterization of the Fermat primes,
well-after Wiles' pr00f, about a year or t00 ago.

as far as I know,
the vast majority of fermatistes do not bother
to learn what Fermat created, and
it really begins with the "little" theorem,
a^p = a modulo p, also stated as
a^(p-1) = 1 mod p

Post by bassam king karzeddin
Your comment is not clear enough to me, nor ever you had been clear in your comments, but if you mean that my theorem is coming from Fermat (little or last theorem), I would tell you it includes it fully, but its range is much wider than that to establish new section in number theory

bassam king karzeddin

2017-03-30 08:56:58 UTC

Post by a***@gmail.com
no, no, no. Fermat's theorem on bisums of one power,
is only the "last" because so-long unsolved, although
I solved the characterization of the Fermat primes,
well-after Wiles' pr00f, about a year or t00 ago.
as far as I know,
the vast majority of fermatistes do not bother
to learn what Fermat created, and
it really begins with the "little" theorem,
a^p = a modulo p, also stated as
a^(p-1) = 1 mod p

Why are you mixing things always, here we have more general conjecture than FLT, also, you had never provided anything to what had you got, or maybe you mean that your results are in your mind only, wonder!

BK

a***@gmail.com

2017-04-05 16:42:35 UTC

ah, I didn't really l00k at your result as such, but
it is clearly a consequence of the little theorem, and
you soulhd easily show such module

Post by a***@gmail.com
a^p = a mod p?... but,

N > 1 such that if p|N, then p^a exactly
divides N for some a > 1.
That is to say, that any prime dividing N must do so to a power higher than 1.

bassam king karzeddin

2017-04-05 18:14:24 UTC

Post by a***@gmail.com
ah, I didn't really l00k at your result as such, but
it is clearly a consequence of the little theorem, and
you soulhd easily show such module

Post by a***@gmail.com
a^p = a mod p?... but,

N > 1 such that if p|N, then p^a exactly
divides N for some a > 1.
That is to say, that any prime dividing N must do so to a power higher than 1.

You are so fascinated by that little adic result, but it does not help much, for sure
Really helpless and clueless as always as usual, wonder!

Good luck

BK

a***@gmail.com

2017-04-05 20:24:35 UTC

you were unable to get,
wTf I was doing ... even when I, myself,
have not really used the little theorem,
it is easy to see that it encompasses yours,
sort-of like when S.Germaine t00k care
of an infinite swath of the n-ariness,
some time après Fermatttt

Post by a***@gmail.com
ah, I didn't really l00k at your result as such, but
it is clearly a consequence of the little theorem, and
you soulhd easily show such module

Post by a***@gmail.com
a^p = a mod p?... but,

N > 1 such that if p|N, then p^a exactly
divides N for some a > 1.
That is to say, that any prime dividing N must do so to a power higher than 1.

You are so fascinated by that little adic result, but it does not help much, for sure
Really helpless and clueless as always as usual, wonder!
Good luck
BK

Vinicius Claudino Ferraz

2017-04-06 14:33:11 UTC

p = 7 prime
2p + 1 = 15 composto químico

Post by a***@gmail.com
you were unable to get,
wTf I was doing ... even when I, myself,
have not really used the little theorem,
it is easy to see that it encompasses yours,
sort-of like when S.Germaine t00k care
of an infinite swath of the n-ariness,
some time après Fermatttt

Post by a***@gmail.com
ah, I didn't really l00k at your result as such, but
it is clearly a consequence of the little theorem, and
you soulhd easily show such module

Post by a***@gmail.com
a^p = a mod p?... but,

N > 1 such that if p|N, then p^a exactly
divides N for some a > 1.
That is to say, that any prime dividing N must do so to a power higher than 1.

You are so fascinated by that little adic result, but it does not help much, for sure
Really helpless and clueless as always as usual, wonder!
Good luck
BK

a***@gmail.com

2017-04-06 19:09:18 UTC

so, what was them in sequentia

Post by Vinicius Claudino Ferraz
p = 7 prime
2p + 1 = 15 composto químico

Post by a***@gmail.com
it is easy to see that it encompasses yours,
sort-of like when S.Germaine t00k care
of an infinite swath of the n-ariness,
some time après Fermatttt

Post by a***@gmail.com
it is clearly a consequence of the little theorem, and
you should easily show such a module

Post by a***@gmail.com
a^p = a mod p?... but,

N > 1 such that if p|N, then p^a exactly
divides N for some a > 1.
That is to say, that any prime dividing N must do so to a power higher than 1.

You are so fascinated by that little adic result, but

what is it with this guy\clown --
perfectly g00d profession, of course

Arturo Magidin

2016-11-27 08:01:05 UTC

Post by Peter Percival
Istr that a positive integer is called powerful if it is the product of
a square and a cube.

NO!

True, provided we require N>1.

The two statements are equivalent.

Suppose that N is powerful; its prime factorization given by

N = p_1^{a_1} * .... * p_r^{a_r}.

Let I = { i | 1<=i<=r, a_i is even}
and J = { i | 1<=i<=r, a_i is odd}.

Note that if a_i is odd, then a_i>=3, since N is powerful.

Let x = (\prod_{k in I} p_k^{a_k}) * (\prod_{k in J} p_k^{a_k-3}). Note that this makes sense, since a_k>=3 for all k in J; and moreover that x is a square.

Let y = prod_{k in I} p_k^3. Note that y is a cube.

Finally, note that N = x*y, the product of a square and a cube.

Conversely, if N = a^2*b^3>1, and p is a prime such that p divides N, then p|a or p|b, and hence p^2|N.

--
Arturo Magidin

Peter Percival

2016-11-28 15:07:47 UTC

Post by Arturo Magidin

Post by Peter Percival
Istr that a positive integer is called powerful if it is the product of
a square and a cube.

NO!

True, provided we require N>1.

I had overlooked that.

Post by Arturo Magidin

The two statements are equivalent.
Suppose that N is powerful; its prime factorization given by
N = p_1^{a_1} * .... * p_r^{a_r}.
Let I = { i | 1<=i<=r, a_i is even}
and J = { i | 1<=i<=r, a_i is odd}.
Note that if a_i is odd, then a_i>=3, since N is powerful.
Let x = (\prod_{k in I} p_k^{a_k}) * (\prod_{k in J} p_k^{a_k-3}). Note that this makes sense, since a_k>=3 for all k in J; and moreover that x is a square.
Let y = prod_{k in I} p_k^3. Note that y is a cube.
Finally, note that N = x*y, the product of a square and a cube.
Conversely, if N = a^2*b^3>1, and p is a prime such that p divides N, then p|a or p|b, and hence p^2|N.

Peter Percival

2016-11-28 15:05:57 UTC

Post by Virgil
In article

Post by bassam king karzeddin
Why it is impossible to find a single primitive Pythagorean triplet (in
positive integers) being with all terms as powerful numbers?
Regards
Bassam King Karzeddin

What makes one positive integer more "powerful" than another?

Istr that a positive integer is called powerful if it is the product of
a square and a cube.

NO!
A powerful number is an integer N > 1 such that if p|N, then p^a exactly
divides N for some a > 1.

Same thing!

Post by Pubkeybreaker
That is to say, that any prime dividing N must do so to a power higher than 1.

Virgil

2016-11-25 18:25:26 UTC

Post by Virgil
What makes one positive integer more "powerful" than another?

Istr that a positive integer is called powerful if it is the product of
a square and a cube.

Then would every x^5 be powerful?

--
Virgil
"Mit der Dummheit kampfen Gotter selbst vergebens." (Schiller)

konyberg

2016-11-25 18:59:29 UTC

Post by Virgil
What makes one positive integer more "powerful" than another?

Istr that a positive integer is called powerful if it is the product of
a square and a cube.

Then would every x^5 be powerful?
--
Virgil
"Mit der Dummheit kampfen Gotter selbst vergebens." (Schiller)

I have looked for the def of powerful number. And as I can find out; it is a number of the kind (a^2)(b^3) where a and b are positive integers.
I may have not searched enough, but that's it.

KON

konyberg

2016-11-25 19:07:07 UTC

Post by Virgil
What makes one positive integer more "powerful" than another?

Istr that a positive integer is called powerful if it is the product of
a square and a cube.

Then would every x^5 be powerful?
--
Virgil
"Mit der Dummheit kampfen Gotter selbst vergebens." (Schiller)

I have looked for the def of powerful number. And as I can find out; it is a number of the kind (a^2)(b^3) where a and b are positive integers.
I may have not searched enough, but that's it.
KON

I should add that Pubkeybraker's def is correct.

So powerful numbers are

1, 4, 8, 9, 16, 25, 27, 32, 36, 49, 64, 72, 81, 100, 108, 121, 125, 128, 144, 169, 196, 200, 216, 225, 243, 256, 288, 289, 324, 343, 361, 392, 400, 432, 441, 484, 500, 512, 529, 576, 625, 648, 675, 676, 729, 784, 800, 841, 864, 900, 961, 968, 972, 1000, ... (sequence A001694 in the OEIS).

KON

konyberg

2016-11-25 19:13:24 UTC

Post by Virgil
What makes one positive integer more "powerful" than another?

Istr that a positive integer is called powerful if it is the product of
a square and a cube.

Then would every x^5 be powerful?
--
Virgil
"Mit der Dummheit kampfen Gotter selbst vergebens." (Schiller)

I have looked for the def of powerful number. And as I can find out; it is a number of the kind (a^2)(b^3) where a and b are positive integers.
I may have not searched enough, but that's it.
KON

I should add that Pubkeybraker's def is correct.
So powerful numbers are
1, 4, 8, 9, 16, 25, 27, 32, 36, 49, 64, 72, 81, 100, 108, 121, 125, 128, 144, 169, 196, 200, 216, 225, 243, 256, 288, 289, 324, 343, 361, 392, 400, 432, 441, 484, 500, 512, 529, 576, 625, 648, 675, 676, 729, 784, 800, 841, 864, 900, 961, 968, 972, 1000, ... (sequence A001694 in the OEIS).
KON

I forgot x^5 is then a powerful number.
All numbers of the form a^2, b^3 and (a^2)(b^3) are powerful numbers.

KON

konyberg

2016-11-25 19:34:52 UTC

Post by Virgil
What makes one positive integer more "powerful" than another?

Istr that a positive integer is called powerful if it is the product of
a square and a cube.

Then would every x^5 be powerful?
--
Virgil
"Mit der Dummheit kampfen Gotter selbst vergebens." (Schiller)

I have looked for the def of powerful number. And as I can find out; it is a number of the kind (a^2)(b^3) where a and b are positive integers.
I may have not searched enough, but that's it.
KON

I should add that Pubkeybraker's def is correct.
So powerful numbers are
1, 4, 8, 9, 16, 25, 27, 32, 36, 49, 64, 72, 81, 100, 108, 121, 125, 128, 144, 169, 196, 200, 216, 225, 243, 256, 288, 289, 324, 343, 361, 392, 400, 432, 441, 484, 500, 512, 529, 576, 625, 648, 675, 676, 729, 784, 800, 841, 864, 900, 961, 968, 972, 1000, ... (sequence A001694 in the OEIS).
KON

I forgot x^5 is then a powerful number.
All numbers of the form a^2, b^3 and (a^2)(b^3) are powerful numbers.
KON

So what I want to know is (and B hasn't answered) is this.
In his opening of his tread, what does he mean by term? Is it
x^2 + y^2 = z^2, where f.inst. x = (a^2)(b^3), or x^2 = (a^2)(b^3)?

KON

bassam king karzeddin

2016-11-26 12:18:09 UTC

Post by Virgil
What makes one positive integer more "powerful" than another?

Istr that a positive integer is called powerful if it is the product of
a square and a cube.

Then would every x^5 be powerful?
--
Virgil
"Mit der Dummheit kampfen Gotter selbst vergebens." (Schiller)

Ops., that is so great conclusion of yours only, wonder if anybody helped you genious,

For additional learning, x^n is also powerful number when (n > 1), where (x & n) are integers

Keep reading, you would indeed improve in number theory (which is the crown of mathematical sciences), no juggling here is allowed as in many other branches (as the fake set theory) you always do enjoy

Regards
Bassam King Karzeddin

Post by Virgil
What makes one positive integer more "powerful" than another?

Istr that a positive integer is called powerful if it is the product of
a square and a cube.

Then would every x^5 be powerful?
--
Virgil
"Mit der Dummheit kampfen Gotter selbst vergebens." (Schiller)

Peter Percival

2016-11-28 15:09:46 UTC

Post by Virgil
What makes one positive integer more "powerful" than another?

Istr that a positive integer is called powerful if it is the product of
a square and a cube.

Then would every x^5 be powerful?

Yes, except that I overlooked a >1 requirement that Professor Magidin
pointed out.

bassam king karzeddin

2016-11-26 08:35:39 UTC

Post by Virgil
In article

Post by bassam king karzeddin
Why it is impossible to find a single primitive Pythagorean triplet (in
positive integers) being with all terms as powerful numbers?
Regards
Bassam King Karzeddin

What makes one positive integer more "powerful" than another?

Istr that a positive integer is called powerful if it is the product of
a square and a cube.

Yes, this is absolutely true, (a^2)*(b^3) is a powerful number provided that (a, b) are real integers

but for our suggested problem, we may consider (a, b) are real positive integers

Regards

Bassam King karzeddin

Post by Virgil
Does badass mean numbers expressed as powers?
Note all terms of 3^2 + 4^2 = 5^2 are expressed as powers.

Virgil

2016-11-26 22:01:59 UTC

Post by Peter Percival
Istr that a positive integer is called powerful if it is the product of
a square and a cube.

Yes, this is absolutely true, (a^2)*(b^3) is a powerful number provided that
(a, b) are real integers

Does Badass know of any unreal integers?

--
Virgil
"Mit der Dummheit kampfen Gotter selbst vergebens." (Schiller)

bassam king karzeddin

2016-11-27 07:28:17 UTC

Post by Peter Percival
Istr that a positive integer is called powerful if it is the product of
a square and a cube.

Yes, this is absolutely true, (a^2)*(b^3) is a powerful number provided that
(a, b) are real integers

Does Badass know of any unreal integers?
--

Oh, you are not shameful to come back here, where here is real mathematics that is too difficult to be spoiled by your alikes, here is (number theory) where no near solution is accepted as the case of real numbers, where no job for your alikes, but still arguing in the definitions of integers, whether they are real or complex,

Did I destroy completely your alleged complex numbers also, tell me please where in the actual existing physical space the coordination frames (XYZ-axis) are divided into positive and negative, instead of two opposite directions in physical existing sense,

So to say, your linguistic interference is only void and baseless

And I do not believe in the well established fake mathematics the way your alikes do, even though I can play well enough with them

So, the badassest (bASSam) is demolishing your alikes alleged fake intellect completely, unless you get something useful

Bassam King Karzeddin

Post by Virgil
Virgil
"Mit der Dummheit kampfen Gotter selbst vergebens." (Schiller)

Post by Peter Percival
Istr that a positive integer is called powerful if it is the product of
a square and a cube.

Yes, this is absolutely true, (a^2)*(b^3) is a powerful number provided that
(a, b) are real integers

Does Badass know of any unreal integers?
--
Virgil
"Mit der Dummheit kampfen Gotter selbst vergebens." (Schiller)

Virgil

2016-11-27 21:30:20 UTC

Post by Peter Percival
Istr that a positive integer is called powerful if it is the product of
a square and a cube.

Yes, this is absolutely true, (a^2)*(b^3) is a powerful number provided that
(a, b) are real integers

Does Badass know of any unreal integers?
--

Oh, you are not shameful

But you are! Unless you can come up with an unreal integer.

--
Virgil
"Mit der Dummheit kampfen Gotter selbst vergebens." (Schiller)

bassam king karzeddin

2016-11-28 15:01:30 UTC

Post by Peter Percival
Istr that a positive integer is called powerful if it is the product of
a square and a cube.

Yes, this is absolutely true, (a^2)*(b^3) is a powerful number provided that
(a, b) are real integers

Does Badass know of any unreal integers?
--

Oh, you are not shameful

But you are! Unless you can come up with an unreal integer.
--
Virgil
"Mit der Dummheit kampfen Gotter selbst vergebens." (Schiller)

So easy question, I have infinitely infinite number of those unreal integers, also I have the (magical tool) in mathematics that turns them into real by those living in the paradise of fools

Consider the infinite sequence of 9 digits as this:

(999... ), this is unreal integer since you can not fix it on the real line number, but by giving you the magical tool, you would turn it immediately to real integer, guess what is this tool, no need to guess, take it freely, it is the (dot) which is the (decimal notation), place it anywhere between the digits where you leave a finite number of digits to the left of your decimal notation, then you would certainly turn it not only to one integer but also to infinitely many, see here for explanation

(0.999... ), your unreal number is turned magically to one, see here also

(9.999... ), your your unreal number is turned magically to 10, see here also

And so on, so the powerful tool of decimal notation can magically turn every unreal number to infinitely many infinite real numbers or integers whichever you like, and now it is too good chance for secretive researchers to make towers and volumes of those new created numbers, so easy and wonderful

Consider the infinite distinct digits of (what they call it - pi) which may be considered as a distinct infinity, since infinities are also infinite! and play with it the same way I explained above with your magical tool, and have more exciting fun for many centuries studying those only few trillions of digits of it, to come up with more exciting discoveries.... etc,

Consider Zero, which is really unreal integer (think about it)

Consider that mirror image of one which is called negative one, and try to touch it the same way you touch the real integer one you marked on a number line, and see how unreal they indeed are, not only that but take its root also to cross not only the mirror but also to the unreal world of beautiful
imagination, that might take you much faster than light from one galaxy to another far galaxy in no time almost, wonderland and fairy world of pure imagination, you can see your old age or your past lives or childhood also,

So fantastic is mathematics when it operates well enough into physics, is not it?

If you need more of unreal integers I do have many

Regards
Bassam King Karzeddin
28th, Nov., 2016

bassam king karzeddin

2016-11-26 07:46:46 UTC

Post by Virgil
In article

Post by bassam king karzeddin
Why it is impossible to find a single primitive Pythagorean triplet (in
positive integers) being with all terms as powerful numbers?
Regards
Bassam King Karzeddin

What makes one positive integer more "powerful" than another?
Does badass mean numbers expressed as powers?
Note all terms of 3^2 + 4^2 = 5^2 are expressed as powers.
--
Virgil
"Mit der Dummheit kampfen Gotter selbst vergebens." (Schiller)

Virgil wrote:(25, Nov., 2016)

What makes one positive integer more "powerful" than another?

Does badass mean numbers expressed as powers?

Note all terms of 3^2 + 4^2 = 5^2 are expressed as powers.
--
Virgil
"Mit der Dummheit kampfen Gotter selbst vergebens." (Schiller)
****************************************************************************
Bassam Answer

Please get a minimum understanding of number theory concepts before you get into my topics and make fun of yourself (with online documentation).

This question is certainly not for your a like’s level of comprehension

You (Virgil) didn’t even bother to waste few seconds to ask Google what is a powerful number (that was well defined long ago) before your blindness hearted towards me and few others

A powerful number is a real positive integer with its entire prime factors exponents are greater than one

So, your had provided a triplet with only one term being as powerful number ( 3, 2^2, 5), which is too poor example to my question, so you are advised not to provide any more fun

Let the badass bASSam (you always describe) teaches you the unforgettable lesson about the meaning of primitive Pythagorean triplet (that was defend thousands of years back) during Babylon’s era,

The Primitive Pythagorean Triplet (PTP), is a right angle triangle with integer sides that are co prime to each other

In symbolic notations in real positive integers, (2ab, a^2 – b^2, a^2 + b^2), where (a > b) are two distinct positive co prime integers with different polarity,

Please (Virgil), do not ask me about the meaning of polarity before you consult Google, but take a hint it is (odd - even) for (a & b) or vice versa

As I told you before, your number theory is too poor, and I really wonder if any one who proved himself too poor in number theory can still argue day and night in a subject as mathematics

Those are called “Mythematickers” and beyond any doubt!

Regards
Bassam King Karzeddin
26th, Nov., 2016

Virgil

2016-11-26 21:57:45 UTC

Post by Virgil
In article

Post by bassam king karzeddin
Why it is impossible to find a single primitive Pythagorean triplet (in
positive integers) being with all terms as powerful numbers?
Regards
Bassam King Karzeddin

--
Virgil
"Mit der Dummheit kampfen Gotter selbst vergebens." (Schiller)

konyberg

2016-11-25 12:03:01 UTC

Post by bassam king karzeddin
Why it is impossible to find a single primitive Pythagorean triplet (in positive integers) being with all terms as powerful numbers?
Regards
Bassam King Karzeddin

This is not compleatly true. If a number of the form a^2b^3 (powerful number) and b = 1, then what?

KON

Peter Percival

2016-11-25 15:07:01 UTC

Post by bassam king karzeddin
Why it is impossible to find a single primitive Pythagorean triplet (in positive integers) being with all terms as powerful numbers?
Regards
Bassam King Karzeddin

This is not compleatly true. If a number of the form a^2b^3 (powerful number) and b = 1, then what?
KON

If, in x^2 + y^2 = z^2, it is x, y and z that are the terms to which
Bassam makes reference, then (u^2)^2 + (v^2)^2 = (w^2)^2 has no postive
integral solutions.

konyberg

2016-11-25 16:16:43 UTC

Post by bassam king karzeddin
Why it is impossible to find a single primitive Pythagorean triplet (in positive integers) being with all terms as powerful numbers?
Regards
Bassam King Karzeddin

This is not compleatly true. If a number of the form a^2b^3 (powerful number) and b = 1, then what?
KON

If, in x^2 + y^2 = z^2, it is x, y and z that are the terms to which
Bassam makes reference, then (u^2)^2 + (v^2)^2 = (w^2)^2 has no postive
integral solutions.
--
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

I should perhaps written a^2b^3 a bit more clear: (a^2)(b^3). This is the definition of a powerful number when a and b are positive integers.
And that is why I say that Bassam's claim is wrong. If b = 1 then you get a equation like x^2 + y^2 = z^2. And this is the common pythagorean equation.

KON

bassam king karzeddin

2016-11-26 08:47:38 UTC

Post by bassam king karzeddin
Why it is impossible to find a single primitive Pythagorean triplet (in positive integers) being with all terms as powerful numbers?
Regards
Bassam King Karzeddin

This is not compleatly true. If a number of the form a^2b^3 (powerful number) and b = 1, then what?
KON

After all explanations that were provided, unfortunately it is you who had been wrong

Bassam King Karzeddin

bassam king karzeddin

2016-11-26 08:44:29 UTC

Post by bassam king karzeddin
Why it is impossible to find a single primitive Pythagorean triplet (in positive integers) being with all terms as powerful numbers?
Regards
Bassam King Karzeddin

This is not compleatly true. If a number of the form a^2b^3 (powerful number) and b = 1, then what?
KON

If, in x^2 + y^2 = z^2, it is x, y and z that are the terms to which
Bassam makes reference, then (u^2)^2 + (v^2)^2 = (w^2)^2 has no postive
integral solutions.

Yes, (x, y, z) are the terms of the pythagorean triplet, where this is the simplest case triplet (x^2, y^2, z^2), that Fermat proved by infinite descent not to possess any solution in real integers

Bassam King Karzeddin

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

bassam king karzeddin

2016-11-26 08:38:28 UTC

Post by bassam king karzeddin
Why it is impossible to find a single primitive Pythagorean triplet (in positive integers) being with all terms as powerful numbers?
Regards
Bassam King Karzeddin

This is not compleatly true. If a number of the form a^2b^3 (powerful number) and b = 1, then what?
KON

So, according to the definition of powerful number (a^2) is a powerful number

Bassam King Karzeddin

bert

2016-11-25 16:28:50 UTC

Post by bassam king karzeddin
Why it is impossible to find a single primitive
Pythagorean triplet (in positive integers) being
with all terms as powerful numbers?

I suppose you mean (a^i)^2 + (b^j)^2 = (c^k)^2,
where all of (i, j, k) are greater than unity.

We would then have 1/2i + 1/2j + 1/2k < 1, so
you are asking if any of the solutions for the
Fermat-Catalan conjecture meet your condition.
--

bassam king karzeddin

2016-11-26 11:53:27 UTC

Post by bert

Post by bassam king karzeddin
Why it is impossible to find a single primitive
Pythagorean triplet (in positive integers) being
with all terms as powerful numbers?

I suppose you mean (a^i)^2 + (b^j)^2 = (c^k)^2,
where all of (i, j, k) are greater than unity.

Yes, but my little conjecture covers a much wider range than you spot

Post by bert
We would then have 1/2i + 1/2j + 1/2k < 1, so
you are asking if any of the solutions for the
Fermat-Catalan conjecture meet your condition.
--

As explained above, those only a special cases of my little conjecture

Regards
Bassam King Karzeddin

konyberg

2016-11-25 16:42:01 UTC

Post by bassam king karzeddin
Why it is impossible to find a single primitive Pythagorean triplet (in positive integers) being with all terms as powerful numbers?
Regards
Bassam King Karzeddin

Or consider this:

(x^2)(b^3) + (y^2)(b^3) = (z^2)(b^3)

Now all terms are powerful numbers, and there are an infinite number of solutions in integers.

If you mean, by term, that we have types like ((a^2)(b^3))^2, then this I think you should discuss with Fermat or Wiles :)

KON

bassam king karzeddin

2016-11-27 08:22:58 UTC

Post by bassam king karzeddin
Why it is impossible to find a single primitive Pythagorean triplet (in positive integers) being with all terms as powerful numbers?
Regards
Bassam King Karzeddin

(x^2)(b^3) + (y^2)(b^3) = (z^2)(b^3)
Now all terms are powerful numbers, and there are an infinite number of solutions in integers.

Yes, but this was not my original problem
This is called addition of two powerful numbers

Post by konyberg
If you mean, by term, that we have types like ((a^2)(b^3))^2, then this I think you should discuss with Fermat or Wiles :)
KON

Believe me, both of them can not help in this case

Regards
Bassam King Karzeddin
27th, Nov., 2016

Virgil

2016-11-27 21:33:12 UTC

Post by konyberg
If you mean, by term, that we have types like ((a^2)(b^3))^2, then this I
think you should discuss with Fermat or Wiles :)
KON

Believe me, both of them can not help in this case

Fermat, being dead, may not be able to, but Wiles, being alive,
could easily straighten Badass out, if he chose to.

--
Virgil
"Mit der Dummheit kampfen Gotter selbst vergebens." (Schiller)

bassam king karzeddin

2016-11-28 14:16:41 UTC

Post by konyberg
If you mean, by term, that we have types like ((a^2)(b^3))^2, then this I
think you should discuss with Fermat or Wiles :)
KON

Believe me, both of them can not help in this case

Fermat, being dead, may not be able to, but Wiles, being alive,
could easily straighten Badass out, if he chose to.
--
Virgil
"Mit der Dummheit kampfen Gotter selbst vergebens." (Schiller)

I think I had already answered you, but never mind again

Of course I can not ask Fermat himself so far , and I really asked Wiles by mail twice through his link page provided by Adam berg at Quora, where I never received any reply, so anyone with known contacts may asks him instead of me

Regards
Bassam King Karzeddin

Post by konyberg
If you mean, by term, that we have types like ((a^2)(b^3))^2, then this I
think you should discuss with Fermat or Wiles :)
KON

Believe me, both of them can not help in this case

Fermat, being dead, may not be able to, but Wiles, being alive,
could easily straighten Badass out, if he chose to.
--
Virgil
"Mit der Dummheit kampfen Gotter selbst vergebens." (Schiller)

bassam king karzeddin

2016-11-27 11:21:51 UTC

Post by bassam king karzeddin
Why it is impossible to find a single primitive Pythagorean triplet (in positive integers) being with all terms as powerful numbers?
Regards
Bassam King Karzeddin

(x^2)(b^3) + (y^2)(b^3) = (z^2)(b^3)

Note here this is too simple to generate infinitely many solution to

(x^2 + y^2 = z^2), and then multiply all with common factor (b^3), which yields in summing of two distinct powerful numbers equals to another distinct powerful number,

But can you do it with all terms being coprime powerful integers?

Post by konyberg
Now all terms are powerful numbers, and there are an infinite number of solutions in integers.
If you mean, by term, that we have types like ((a^2)(b^3))^2, then this I think you should discuss with Fermat or Wiles :)
KON

here is wikipedia definition of powerful numbers: https://en.wikipedia.org/wiki/Powerful_number

Regards
Bassam King Karzeddin

bassam king karzeddin

2016-11-26 12:06:28 UTC

Post by bassam king karzeddin
Why it is impossible to find a single primitive Pythagorean triplet (in positive integers) being with all terms as powerful numbers?
Regards
Bassam King Karzeddin

In fact it is not a conjecture, but a theorem (at least for me), where once stated, then many ready proofs should spring out by genius number theorist immediately.

But the margin of my time is too narrow to write it down,
And I really do not want to spoil the joy of it, so enjoy it forever mathematicians

Regards
Bassam King Karzeddin
26th, Nov., 2016

Post by bassam king karzeddin
Why it is impossible to find a single primitive Pythagorean triplet (in positive integers) being with all terms as powerful numbers?
Regards
Bassam King Karzeddin

a***@gmail.com

2017-03-31 05:32:26 UTC

a-hem, what?

Post by bassam king karzeddin
Why it is impossible to find a single primitive Pythagorean triplet (in positive integers) being with all terms as powerful numbers?

a***@gmail.com

2017-04-02 01:27:32 UTC

there are also pythagorean quadruplets,
at least for the right & left tetrahedra inscribed in the sphere, but
I don't know that there are integral solutions
for the ky00b-corner one, at all ... but,
there may be old results, three

bassam king karzeddin

2017-04-02 14:02:20 UTC

Post by a***@gmail.com
there are also pythagorean quadruplets,
at least for the right & left tetrahedra inscribed in the sphere, but
I don't know that there are integral solutions
for the ky00b-corner one, at all ... but,
there may be old results, three

With your four posts into my topic, had added nothing but too incomprehensible
nonsense words for sure

So, write in your language, since Google can translate it better, for sure

BK

a***@gmail.com

2017-04-03 00:45:24 UTC

they are just of the form,
a^2 + b^2 + c^2 = d^2, being the three face-diagonals
of rectangular boxes;
the other p.t is A^2 + B^2 + C^2 = D^2, but
I don;t know id there are any integer solutions

Post by a***@gmail.com
there are also pythagorean quadruplets,
for the ky00b-corner one, at all ... but,
there may be old results, three

a***@gmail.com

2017-04-03 06:30:21 UTC

d is the hypotneuse of the right tetrahedron, or
of the left tetrahedron. really, of ocurse,
there are no right trigona, because
you just flip the paper over et voila

Post by a***@gmail.com
a^2 + b^2 + c^2 = d^2, being the three face-diagonals
of rectangular boxes;
the other p.t is A^2 + B^2 + C^2 = D^2, but
I don;t know id there are any integer solutions

Post by a***@gmail.com
there are also pythagorean quadruplets,
for the ky00b-corner one, at all ... but,
there may be old results, three

a***@gmail.com

2017-04-04 02:52:29 UTC

it is inscribed in the sphere on a diameter, but
I don't really have a compleat method thereto; really,
the main exercise is to show the relationship
of this pair of forms, within the same x,y,z box & so forth,
using small integers as til pr00f obtained

Post by a***@gmail.com
d is the hypotneuse of the right tetrahedron, or
of the left tetrahedron. really, of course,
there are no right trigona, because
you just flip the paper over et voila

Post by a***@gmail.com
a^2 + b^2 + c^2 = d^2, being the three face-diagonals
of rectangular boxes;
the other p.t is A^2 + B^2 + C^2 = D^2, but
I don't know ifthere are any integer solutions

Post by a***@gmail.com
there are also pythagorean quadruplets,
for the ky00b-corner one, at all ... but,
there may be old results, three

a***@gmail.com

2017-04-04 08:23:36 UTC

it is really quite simple,
as given by Sierpinski for all of the rational cuboids,
or pythagorean tetruplets

Post by a***@gmail.com
a^2 + b^2 + c^2 = d^2, being the three face-diagonals
of rectangular boxes -- i.e that is that
there are also pythagorean quadruplets;
the other p.t is A^2 + B^2 + C^2 = D^2, but
I don't know if there are any integer solutions

bassam king karzeddin

2017-04-04 08:33:47 UTC

Post by a***@gmail.com
it is really quite simple,
as given by Sierpinski for all of the rational cuboids,
or pythagorean tetruplets

there are many symbolic forms to get a square is a sum of three squares, FOR SURE

Simple numeric example is (13,12, 4, 3), so how can we read your mind?

BK

a***@gmail.com

2017-04-05 04:04:52 UTC

yes, 3^2 + 4^2 + 4^2*3^2 = 13^2, but
making three regular tetragona on the edges,
that's just a lot of unnecessary work;
so, the problem is to provide spatial l00ns pr00f

Post by a***@gmail.com
as given by Sierpinski for all of the rational cuboids,
or pythagorean tetruplets

Simple numeric example is (13,12, 4, 3), so how can we read your mind?

f'surely, but there are t00 forms;
if you want to conjecture that the second form has whole number solvency,
that's your new conjectural:
are there any, or are they just the same?

bassam king karzeddin

2016-11-27 08:33:39 UTC

torsdag 24. november 2016 15.50.05 UTC+1 skrev bassam

Post by bassam king karzeddin
Why it is impossible to find a single primitive

Pythagorean triplet (in positive integers) being with
all terms as powerful numbers?

Post by bassam king karzeddin
Regards
Bassam King Karzeddin

(x^2)(b^3) + (y^2)(b^3) = (z^2)(b^3)
Now all terms are powerful numbers, and there are an
infinite number of solutions in integers.
If you mean, by term, that we have types like
((a^2)(b^3))^2, then this I think you should discuss
with Fermat or Wiles :)
KON
torsdag 24. november 2016 15.50.05 UTC+1 skrev bassam

Post by bassam king karzeddin
Why it is impossible to find a single primitive

Pythagorean triplet (in positive integers) being with
all terms as powerful numbers?

Post by bassam king karzeddin
Regards
Bassam King Karzeddin

(x^2)(b^3) + (y^2)(b^3) = (z^2)(b^3)
Now all terms are powerful numbers, and there are an
infinite number of solutions in integers.

Yes, but this was not my original stated problem
This is called addition of two powerful numbers equals to another powerful number

If you mean, by term, that we have types like
((a^2)(b^3))^2, then this I think you should discuss
with Fermat or Wiles :)

Believe me, both of Fermat and Wiles can not help in this case, you may verify yourself

Regards
Bassam King Karzeddin
27th, Nov., 2016

KON

bassam king karzeddin

2016-11-28 08:25:21 UTC

Post by Virgil
In article

Post by bassam king karzeddin
,

Post by konyberg
If you mean, by term, that we have types like

((a^2)(b^3))^2, then this I

Post by konyberg
think you should discuss with Fermat or Wiles :)
KON

Believe me, both of them can not help in this case

Fermat, being dead, may not be able to, but Wiles,
being alive,
could easily straighten Badass out, if he chose to.
--
Virgil
"Mit der Dummheit kampfen Gotter selbst vergebens."
(Schiller)

I had already contacted Wiles twice in the past, but never heard of any reply

About Fermat, true I could not contact him so far

Bassam King Karzeddin

b***@gmail.com

2016-11-29 12:59:17 UTC

Post by bassam king karzeddin
I had already contacted Wiles twice in the past, but never heard of any reply
About Fermat, true I could not contact him so far
Bassam King Karzeddin

Don't you think the delusions are on your side. For
example when I read the following:

"Note that, the assumption holds true always for x as any arbitrary length from one to three & the trisection of the angle is exact (without any approximation), also this is not a claim of a solution to the angle trisection problem, but a very little insight to classify the trisectible angle which may be further developed by interested people."
http://math.stackexchange.com/q/1419425/4414

What further developments do you have in mind particularly?

bassam king karzeddin

2016-11-30 06:51:43 UTC

Am Montag, 28. November 2016 17:10:06 UTC+1 schrieb

Post by bassam king karzeddin
I had already contacted Wiles twice in the past,

but never heard of any reply

Post by bassam king karzeddin
About Fermat, true I could not contact him so far
Bassam King Karzeddin

Don't you think the delusions are on your side. For
"Note that, the assumption holds true always for x as
any arbitrary length from one to three & the
trisection of the angle is exact (without any
approximation), also this is not a claim of a
solution to the angle trisection problem, but a very
little insight to classify the trisectible angle
which may be further developed by interested people."
http://math.stackexchange.com/q/1419425/4414
What further developments do you have in mind
particularly?

I do not know your real intention in this regard

Do you want to relate my conjecture to somebody else (most likely who never heard about it from history), as you did before without answering my straight question (Did Euler or Lambert found a solution to
(ax^n + bx^m + c = 0)?

How is this related to my question here?

Go and prove it if you can

Bassam King Karzeddin

b***@gmail.com

2016-11-30 16:04:09 UTC

I answered your question about a*x^n+b*x^n+c=0,
only you seem too stupid to follow the answer
and fill in the blanks and see that its an
instance of Euler/Lambert.

Not my problem, its only your prolobololoblem.

b***@gmail.com

2016-11-30 16:15:28 UTC

Also I don't see that this question is

Post by b***@gmail.com
What further developments do you have in mind
particularly?

Again not my problem, its only your prolobololoblem.

bassam king karzeddin

2016-12-01 07:50:04 UTC

Post by b***@gmail.com
I answered your question about a*x^n+b*x^n+c=0,
only you seem too stupid to follow the answer
and fill in the blanks and see that its an
instance of Euler/Lambert.
Not my problem, its only your prolobololoblem.
I answered your question about a*x^n+b*x^n+c=0,
only you seem too stupid to follow the answer
and fill in the blanks and see that its an
instance of Euler/Lambert.
Not my problem, its only your prolobololoblem.

Yes, but you provided a cheating answer as your character always do,

You had imposed a special condition (a^m = (b^n)*(c^(m-n))), to make it work the way you desire, whereas I never assumed this imposed condition, so here you fail beyond doubt, and Euler and Lambert did solve only this particular case that you assumed,

So, Go and check and comeback telling the truth Big Layer

But how can a Liar say the truth? wonder!

Bassam King Karzeddin

b***@gmail.com

2016-12-01 07:56:26 UTC

Hey Genius, be more Genius.
Laughing my ass off.

bassam king karzeddin

2017-03-29 16:26:27 UTC

Post by bassam king karzeddin
Why it is impossible to find a single primitive Pythagorean triplet (in positive integers) being with all terms as powerful numbers?
Regards
Bassam King Karzeddin

So, this must be hidden by the Professional mathematicians Fools, instead of being front page news in mathematics

Why, because they can not have an answer to this one line statement not containing any mathematical symbols also, wonder!

Because also not stated by any of those alleged genius classified by an authority, nor stated by any of those well known famous that are sleeping peacefully in the history

Because this could had been stated even from the Pythagorean ages before BC

And definitely because this is super mathematics that would uncover all your grand mathematicians dirty tricks that spoiled mathematics for ever

Because the complete inability to face a mature even by all top mathematicians on earth

Because it is a proof of your inferior and permanent inherited mental and so sever egoistic incurable diseases

Because many other reasons that may not be mentioned to leave you some very little and so negligible and unnoticeable dignity

Yes of course, this must be hidden for ever, but get tortured for ever with it as I would not provide you with any answer to it, (for sure)

And once the future artificial intelligence does it, your own grandchildren would laugh at you by the huge ignorance you already documented (sure)

So, hide it morons, it is not even worth to mention

Regards
Bassam king karzeddin
29th, March, 2017

bassam king karzeddin

2017-03-30 13:38:08 UTC

Why do not you come out of your worm holes, and say a word of truth?

But, this would keep chasing you to your last digits on earth, it carries so many secrets within, it would always reminds you that you are certainly inferiors, not even with any alleged talent or intellectuality, nor any knowledge seekers, but certainly a business and a fiction self seekers

And watch out those previously debating about a number (one) that does not even fit to any of my given triangle here

As if playing with words and definitions would give them a clue, wonder?

And make sure that your current Big masters can not help you in this regard

Also, nothing you would certainly find as helpful from your own old ancestors shameful history too!

And you would stand helpless as usual to save your grand children from this designed puzzle for you!

Here, no one can cheat, here are only integers, and existing numbers are not like atoms with epsilon radius and delta distance where easily you can play and cheat

Here is a challenge to your fake intellectuality, for sure

But I have many more, not even worth to post them to foolish people for sure

But, it may be worth to post them to the artificial new born not of your kind

And the shameful, would never learn any new lesson, either from the present nor from the history, for sure

Regards
Bassam King Karzeddin
30th, March, 2017

bassam king karzeddin

2017-03-30 13:52:51 UTC

Post by bassam king karzeddin
Why it is impossible to find a single primitive Pythagorean triplet (in positive integers) being with all terms as powerful numbers?
Regards
Bassam King Karzeddin

Why do not you come out of your worm holes, and say a word of truth?

But, this would keep chasing you to your last digits on earth, it carries so many secrets within, it would always reminds you that you are certainly inferiors, not even with any alleged talent or intellectuality, nor any knowledge seekers, but certainly a business and a fiction self seekers

And watch out carefully those previously debating about a number (one) that does not even fit to any of my given triangle here

As if playing with words and definitions would give them a clue, wonder?

And make sure that your current Big masters can not help you in this regard, for sure

Also, nothing you would certainly find as helpful from your own old ancestors shameful history too!

And you would stand helpless as usual to save your grand children from this designed puzzle for you!

Here, no one can cheat, here are only integers, and existing numbers are not like atoms with epsilon radius and delta distance where easily you can play and cheat, and making so much dance for the nonsense daily products of yours

Here is a challenge to your fake intellectuality, for sure

But I have many more, not even worth to post them to foolish people any more, and you would hope this site is moderated, since it becomes so easy work for the thieves to hide it forever, as the case with that unnamed site SE, or Quora, or Wikipedia, and many alikes moron sites for kids mainly

But, it may be worth to post them to the artificial new born not of your kind, nor of your dirty inherited traits

And the shameful, would never learn any new lesson, either from the present nor from the history, for sure

And I shall make you hate the destiny that make you as a mathematicians, for sure

And still, you would never realize why a skilled football player boy is much more worth than all of you on earth, including your all great grand ancestors

Regards
Bassam King Karzeddin
30th, March, 2017

a***@gmail.com

2017-03-31 05:20:54 UTC

do you have to be an a-hole at all times when online?

I mean, let us suppose that you are just a nice guy, in person,
when your facility with English would be accented
in some interesting way

bassam king karzeddin

2017-04-05 09:30:20 UTC

Post by bassam king karzeddin
Why it is impossible to find a single primitive Pythagorean triplet (in positive integers) being with all terms as powerful numbers?
Regards
Bassam King Karzeddin

I am laughing so loudly about your little ability to do any real mathematics

This is of course impossible, is not it? wonder!
but you do not know why, for sure

And you may need the proof, but frankly you do not deserve to see them, since then you would hate them, because it would show you how too tiny you are

How ignorant you had been?, how innocent you were acting? how helpless and pointless you are, how...?, for sure

So, go and hide it again and again, but so unfortunately you can not hide it completely as you like, but definitely you do not like it, but so unfortunately, it is not up to your silly choice for sure

So, enjoy it for the rest of your meaningless life

BK

a***@gmail.com

2017-04-05 20:50:40 UTC

you talking to yourself like a busted record,
maybe with a nice accent of st00pid

bassam king karzeddin

2017-04-06 07:04:27 UTC

Post by a***@gmail.com
you talking to yourself like a busted record,
maybe with a nice accent of st00pid

Stupidity is well defined, when someone jump randomly into issues that are too difficult for him, providing only a very negligible hint that is slightly better than nothing, and thinking that he has solved the problem,

If you can solve it, then do it, but if you cannot keep silent and learn, and do note throw arbitrary words about your tetrahedron, and you cannot do it for sure

BK

a***@gmail.com

2017-04-07 05:49:39 UTC

by far the best b00k of synthetic geometry,
I call, "the tetrahedron & ho wto use it,"
because a g00d chink of it is about a)
spheres, and b)
tetrahedra. or, you could say,
the continuous tetrahedron & the dyscrete sphere

Post by bassam king karzeddin
about your tetrahedron, and you cannot do it for sure

where's your right clownshoe?

a***@gmail.com

2017-04-08 22:38:36 UTC

yeah, well you're not getting it, back! <tosses leftshoe that away>

Post by a***@gmail.com
where's your right clownshoe?

a***@gmail.com

2017-04-10 16:27:07 UTC

this b00k was published in '958, and
it is perfectly elementary -- you don't need any planar stuff
to do it, although one has to use isometry
to get the very few figures -- just using a T-skware and
a 30-60-90 trigon e.g. however,
he also has a b00k of planar geometry,
with a few things in it that are not implicit to the spatial treatise

Post by a***@gmail.com
by far the best b00k of synthetic geometry,
I call, "the tetrahedron & how to use it,"
because a g00d chink of it is about a)
spheres, and b)
tetrahedra. or, you could say,
the continuous tetrahedron & the dyscrete sphere
where's your right clownshoe?

bassam king karzeddin

2017-04-10 18:01:57 UTC

Post by a***@gmail.com
this b00k was published in '958, and
it is perfectly elementary -- you don't need any planar stuff
to do it, although one has to use isometry
to get the very few figures -- just using a T-skware and
a 30-60-90 trigon e.g. however,
he also has a b00k of planar geometry,
with a few things in it that are not implicit to the spatial treatise

Moron: Why don't you understand the hidden message?

And what was this only message here?

It is like many other messages of mine before, wonder!

Where I state facts without stating the proofs, knowing that you have no means to it, for sure

And here, in this regard, nobody, no books, nothing else can help you, for sure

But may be in the future, the highest form of new born of artificial intelligence can help in this regard, but so unfortunately, you would not be there to listen to the perpetual facts

And what facts?

The fact that I keep always documenting and printing the biggest shame upon your alikes foreheads for ever

Because those future artificial newborns need to have fun and laugh loudly too

And those shameless tiny insects achievements would be analysed more carefully to be placed in proper more appropriate fun sections

And, I shall not bother to solve it too, just to save you from this dirty inevitable destiny, since you don't deserve the remedy to be healed completely from its lating curse upon your tiny intelligence, for sure

So, enjoy it and get tortured with it as long as you are alive, and once you finish you would permanently reax from its lasting curse, for sure

However, I have many more written in my posts, and many others not bothered to write yet, wonder!

Regards
Bassam King Karzeddin
10 th, April, 2017

b***@gmail.com

2017-04-10 19:42:48 UTC

I guess it is in his ass, since BKK is a little stiff
with mathematics, doesn't even know the result of

-1 * -1 = ?

Post by a***@gmail.com
where's your right clownshoe?

bassam king karzeddin

2017-04-12 08:49:49 UTC

Post by b***@gmail.com
I guess it is in his ass, since BKK is a little stiff
with mathematics, doesn't even know the result of
-1 * -1 = ?

Post by a***@gmail.com
where's your right clownshoe?

You would never understand this super logic, just because you were programmed to follow the illogic

IF (+)*(+) = (+), It implies directly that
(-)*(-) = (-), but if you make it otherwise, then you are braking the natural symmetry and replacing it deliberately by artificial non existing symmetry except in the FOOLS minds, for a very dirty reason to make unnecessary fake and dirty extra business, for sure

Where (+) or (-) are nothing more than a direction and its natural opposite direction from a start point on a real number line, but the natural fact would not create any extra business for sure

And remember the simplest fact that a straight line has no actual beginning nor actual end by its mere geometrical definition

And this simplest fact would take few more centuries to be comprehended fully by the mainstream professional morons, wonder!

Bassam King Karzeddin
12th, April, 2017

s***@googlemail.com

2017-04-12 10:31:07 UTC

Post by b***@gmail.com
I guess it is in his ass, since BKK is a little stiff
with mathematics, doesn't even know the result of
-1 * -1 = ?

Post by a***@gmail.com
where's your right clownshoe?

You would never understand this super logic, just because you were programmed to follow the illogic
IF (+)*(+) = (+), It implies directly that
(-)*(-) = (-), but if you make it otherwise, then you are braking the natural symmetry and replacing it deliberately by artificial non existing symmetry except in the FOOLS minds, for a very dirty reason to make unnecessary fake and dirty extra business, for sure
Where (+) or (-) are nothing more than a direction and its natural opposite direction from a start point on a real number line, but the natural fact would not create any extra business for sure
And remember the simplest fact that a straight line has no actual beginning nor actual end by its mere geometrical definition
And this simplest fact would take few more centuries to be comprehended fully by the mainstream professional morons, wonder!
Bassam King Karzeddin
12th, April, 2017

how does it imply that? Do you know that the + and * operators in rings can mean many different things?

bassam king karzeddin

2017-04-12 13:56:21 UTC

Post by s***@googlemail.com

Post by b***@gmail.com
I guess it is in his ass, since BKK is a little stiff
with mathematics, doesn't even know the result of
-1 * -1 = ?

Post by a***@gmail.com
where's your right clownshoe?

You would never understand this super logic, just because you were programmed to follow the illogic
IF (+)*(+) = (+), It implies directly that
(-)*(-) = (-), but if you make it otherwise, then you are braking the natural symmetry and replacing it deliberately by artificial non existing symmetry except in the FOOLS minds, for a very dirty reason to make unnecessary fake and dirty extra business, for sure
Where (+) or (-) are nothing more than a direction and its natural opposite direction from a start point on a real number line, but the natural fact would not create any extra business for sure
And remember the simplest fact that a straight line has no actual beginning nor actual end by its mere geometrical definition
And this simplest fact would take few more centuries to be comprehended fully by the mainstream professional morons, wonder!
Bassam King Karzeddin
12th, April, 2017

how does it imply that? Do you know that the + and * operators in rings can mean many different things?

Do not pretend, I had already explain that quite many times in my posts, and it is not worth to explain and prove it for each person independently

And even I do, people would still follow what was taught to them by an authority, but the rules are also normal, since the cheap sheep would always follow the shepherd

BK

bassam king karzeddin

2017-04-12 15:02:23 UTC

Post by s***@googlemail.com

Post by b***@gmail.com
I guess it is in his ass, since BKK is a little stiff
with mathematics, doesn't even know the result of
-1 * -1 = ?

Post by a***@gmail.com
where's your right clownshoe?

You would never understand this super logic, just because you were programmed to follow the illogic
IF (+)*(+) = (+), It implies directly that
(-)*(-) = (-), but if you make it otherwise, then you are braking the natural symmetry and replacing it deliberately by artificial non existing symmetry except in the FOOLS minds, for a very dirty reason to make unnecessary fake and dirty extra business, for sure
Where (+) or (-) are nothing more than a direction and its natural opposite direction from a start point on a real number line, but the natural fact would not create any extra business for sure
And remember the simplest fact that a straight line has no actual beginning nor actual end by its mere geometrical definition
And this simplest fact would take few more centuries to be comprehended fully by the mainstream professional morons, wonder!
Bassam King Karzeddin
12th, April, 2017

how does it imply that? Do you know that the + and * operators in rings can mean many different things?

Do not pretend, I had already explain that quite many times in my posts, and it is not worth to explain and prove it for each person independently
And even I do, people would still follow what was taught to them by an authority, but the rules are also normal, since the cheap sheep would always follow the shepherd
BK

For sure

BK

a***@gmail.com

2017-04-10 21:50:34 UTC

n.b, there is no essential difference
between the skew tetragon & the tetrahedron; indeed,
once can quantumfy the tetrahedron by the four dihedrals,
and one edgelength to get the size

Post by a***@gmail.com
this b00k was published in '958, and
to do it, although one has to use isometry
to get the very few figures -- just using a T-skware and
a 30-60-90 trigon e.g. however,

because a g00d chunk of it is about a)
spheres, and b)
tetrahedra. or, you could say,
the continuous tetrahedron & the dyscrete sphere

So, enjoy it and get tortured with it as long as you are alive, and once you finish you would permanently reax from its lasting curse, for sure
However, I have many more written in my posts, and many others not bothered to write yet, wonder!
10 th, April, 2017

a***@gmail.com

2017-04-11 19:30:46 UTC

and that is the fundament of tetrahedronometry, bad00mp

Post by a***@gmail.com
between the skew tetragon & the tetrahedron; indeed,
once can quantumfy the tetrahedron by the four dihedrals,

Post by a***@gmail.com
to get the very few figures -- just using a T-skware and
a 30-60-90 trigon e.g. however,

because a g00d chunk of it is about a)
spheres, and b)
tetrahedra. or, you could say,
the continuous tetrahedron & the dyscrete sphere

bassam king karzeddin

2017-05-10 07:21:43 UTC

Post by bassam king karzeddin
Why it is impossible to find a single primitive Pythagorean triplet (in positive integers) being with all terms as powerful numbers?
Regards
Bassam King Karzeddin

This wonderful conjecture (FOR YOU ONLY) is really needing a good prize to motivate mainly the number theorists amateurs, (not necessarily professional mathematicians and not necessarily all other branches of mathematics), for sure
And the so many professional betrayers of the Queen who feeds them especially those dumps historians in mathematics would be very soon cursed and painted with shame from the top of their empty heads to the bottom of their feet for ignoring deliberately this rare and peculiar challenge for not being cooked up at their smelling kitchen of mathematics for sure

BK

bassam king karzeddin

2017-05-11 14:58:11 UTC

Post by bassam king karzeddin
Why it is impossible to find a single primitive Pythagorean triplet (in positive integers) being with all terms as powerful numbers?
Regards
Bassam King Karzeddin

And at least this one could have been guessed (not necessarily proved) since Babylon times, but frankly, there was nobody to even guess it correctly, for sure

And guess how many centuries this half line would take the mathematicians to be proved correctly beyond any little doubt

BKK

bassam king karzeddin

2017-07-19 18:45:23 UTC

Post by bassam king karzeddin
Why it is impossible to find a single primitive Pythagorean triplet (in positive integers) being with all terms as powerful numbers?
Regards
Bassam King Karzeddin

See whatever you don't like to see, only one line statement without any single mathematical notation

It will chase you for sure

BKK

bassam king karzeddin

2017-07-29 15:52:35 UTC

Post by bassam king karzeddin
Why it is impossible to find a single primitive Pythagorean triplet (in positive integers) being with all terms as powerful numbers?
Regards
Bassam King Karzeddin

See whatever you don't like to see, only one line statement without any single mathematical notation
It will chase you for sure
BKK

I also try to make it especially for you only in one word, but I couldn't manage so far, so can you? wonder!

OK, LET me increase the prize for you, so how much is suitable for this half line conjecture do you think its worth?

BKK

bassam king karzeddin

2017-07-23 08:58:12 UTC

Post by bassam king karzeddin
Why it is impossible to find a single primitive Pythagorean triplet (in positive integers) being with all terms as powerful numbers?
Regards
Bassam King Karzeddin

This is really a wonderful piece of stamp of shame printed on the forehead of every number theorist and professional mathematician who ignore it so DELIBERATELY for so tiny egoistic problems that are too unnoticeable for sure since this was not yet taken to the highest levels of investigations

BKK

bassam king karzeddin

2017-07-29 16:32:10 UTC

On Wednesday, July 19, 2017 at 9:45:30 PM UTC+3,

On Thursday, November 24, 2016 at 5:50:05 PM UTC+3,

Post by bassam king karzeddin
Why it is impossible to find a single primitive

Pythagorean triplet (in positive integers) being with
all terms as powerful numbers?

Post by bassam king karzeddin
Regards
Bassam King Karzeddin

See whatever you don't like to see, only one line

statement without any single mathematical notation

It will chase you for sure
BKK

I also try to make it especially for you only in one
e word, but I couldn't manage so far, so can you?
wonder!
OK, LET me increase the prize for you, so how much
h is suitable for this half line conjecture do you
think its worth?
BKK

Actually, your opinions on issues that are not of your business is worthless for sure

BKK

bassam king karzeddin

2017-07-31 14:34:59 UTC

Post by bassam king karzeddin
Why it is impossible to find a single primitive Pythagorean triplet (in positive integers) being with all terms as powerful numbers?
Regards
Bassam King Karzeddin

When I tell you this is impossible then you must believe that is impossible for sure, and if any student asks you why this is impossible, tell him simply that the KING said that (that is all)
BKK

bassam king karzeddin

2017-08-01 12:08:44 UTC

Post by bassam king karzeddin
Why it is impossible to find a single primitive Pythagorean triplet (in positive integers) being with all terms as powerful numbers?
Regards
Bassam King Karzeddin

And this one also must have crashed your supercomputers to ashes, for sure

But I know that you are smart enough to manage and fabricate an old suitable reference for sure

So, hurry up (Wikipedia writers and secretive researchers) and manage as always as usual since you know how to write it officially

After all who cares about what had been published in such a dirty place where intellectual property is given to almost everyone so freely for sure

BKK

bassam king karzeddin

2017-08-06 12:46:34 UTC

Post by bassam king karzeddin
Why it is impossible to find a single primitive Pythagorean triplet (in positive integers) being with all terms as powerful numbers?
Regards
Bassam King Karzeddin

Because this one actually isn't any conjecture, but only a theorem (at least for me), then why bother to teach it to those cheap sheep, they would never appreciate it unless it is imposed on their tiny skulls by an authority for sure

So, let them enjoy it the rest of their meaningless life for sure

BKK

bassam king karzeddin

2017-08-07 09:26:26 UTC

Post by bassam king karzeddin
Why it is impossible to find a single primitive Pythagorean triplet (in positive integers) being with all terms as powerful numbers?
Regards
Bassam King Karzeddin

OK, LET us think of increasing the prize
BKK

bassam king karzeddin

2017-08-09 11:39:28 UTC

Post by bassam king karzeddin
Why it is impossible to find a single primitive Pythagorean triplet (in positive integers) being with all terms as powerful numbers?
Regards
Bassam King Karzeddin

Because simply I stated that, and you have no other choice for sure

BKK

bassam king karzeddin

2017-09-12 08:45:18 UTC

Post by bassam king karzeddin
Why it is impossible to find a single primitive Pythagorean triplet (in positive integers) being with all terms as powerful numbers?
Regards
Bassam King Karzeddin

So, it is indeed impossible, and you must believe it, especially that you have no other choice, and I am who is stating

Or, do you find it hard to accept it from me, wonder!

But hey, don't you believe (proudly) in the proof of Fermat's last theorem even without getting almost any hint? wonder!

BKK

bassam king karzeddin

2017-09-13 16:37:26 UTC

Should I announce a bigger prize for you to understand this so simple theorem? wonder!

BKK

bassam king karzeddin

2017-09-18 07:55:38 UTC

Post by bassam king karzeddin
Why it is impossible to find a single primitive Pythagorean triplet (in positive integers) being with all terms as powerful numbers?
Regards
Bassam King Karzeddin

Do you understand what id a powerful number first? wonder!
BKK

bassam king karzeddin

2017-09-20 18:11:33 UTC

Post by bassam king karzeddin
Why it is impossible to find a single primitive Pythagorean triplet (in positive integers) being with all terms as powerful numbers?
Regards
Bassam King Karzeddin

And I know that is quite hard for mathematicians, but the lazy don't bother to publish it widely from its original source, as no honesty without a reason

BKK

bassam king karzeddin

2017-09-24 16:16:47 UTC

Post by bassam king karzeddin
Why it is impossible to find a single primitive Pythagorean triplet (in positive integers) being with all terms as powerful numbers?
Regards
Bassam King Karzeddin

Just imagine how little and how so tiny and how unnoticeable is the world of mathematicians for sure

A new theorem had been stated since a long time, not a single professional historian could claim older sources nor a single genius mathematician could also refute it, but still, doesn't count to be adopted by any mathematical establishment, especially that it is published in a Usenet as sci.math

As if the word publish means only in a Journals, (not learning any lesson from
the history)
But just imagine if this (half line theorem without a single mathematical notation) was announced by a top known genius mathematician or an alleged top (Journal or University), then how the mathematical press would immediately act with so much music and so many analysis with especially those many so professional talents that acquire very long tongues and deepest throats (that are good for something else)

So where are they hiding now from this simplest challenge? wonder!

Don't we live in an era of supercomputers also? wonder!

I also announced a modest prize for you genius mathematicians

Is it your so negligible and so unnoticeable dignity that make you pretend to be deaf and so blind? wonder!

Or is it the inherited dishonesty that the vast majority of the professional mathematicians acquire? no wonder!

Aren't you so shameful of your little self-being as real traitors to the science that feeds, drinks, and cloths you? wonder!

Or is it that your inferiority complex that makes you have hated a lot of real challenges for the sake of making easy baseless maths based on non-existing concepts as many as here (infinities, epsilon, delta, Approximations, famous meaningless cut, Average, endless sequences, ...etc), wonder!

So, if this is the case then you really don't need to worry at all, because you are dancing alone in the dark

And just imagine the football world where a little boy only scores a goal of a win in a competitive football match, then see how the whole world first-page coloured photos in newspapers, in the world press, in every TV station, ... etc

Just compare yourself and certainly, you would find yourself facing a mirror mathematicians, but wait and don't break the mirror...

And I was only comparing you with a class of another little category, and there are much higher categories for sure

But why all this is happening with a class of people who generally think that they were born genius and very distinct from others, but the fact is really shocking, but why?

It is because mathematicians had been deceived and upset-minded for so many centuries by now for the sake of worshipping some few devils that mislead them globally, otherwise, if true mathematicians are sufficient number, then they would never accept to be less than the top leaders of this mad world that are generally governed by mad people who can't even comprehend a simplest theorem like the Pythagorean even you assign the best teachers for them for all their lives, and they will destroy the world over your heads for sure

Just look and listen carefully around, but those helpless wouldn't change their inevitable destiny unless they learn the biggest unforgottable lesson for sure

BKK

bassam king karzeddin

2017-09-26 09:13:53 UTC

Post by bassam king karzeddin
Why it is impossible to find a single primitive Pythagorean triplet (in positive integers) being with all terms as powerful numbers?
Regards
Bassam King Karzeddin

Just imagine how little and how so tiny and how unnoticeable is the world of mathematicians for sure
A new theorem had been stated since a long time, not a single professional historian could claim older sources nor a single genius mathematician could also refute it, but still, doesn't count to be adopted by any mathematical establishment, especially that it is published in a Usenet as sci.math
As if the word publish means only in a Journals, (not learning any lesson from
the history)
But just imagine if this (half line theorem without a single mathematical notation) was announced by a top known genius mathematician or an alleged top (Journal or University), then how the mathematical press would immediately act with so much music and so many analysis with especially those many so professional talents that acquire very long tongues and deepest throats (that are good for something else)
So where are they hiding now from this simplest challenge? wonder!
Don't we live in an era of supercomputers also? wonder!
I also announced a modest prize for you genius mathematicians
Is it your so negligible and so unnoticeable dignity that make you pretend to be deaf and so blind? wonder!
Or is it the inherited dishonesty that the vast majority of the professional mathematicians acquire? no wonder!
Aren't you so shameful of your little self-being as real traitors to the science that feeds, drinks, and cloths you? wonder!
Or is it that your inferiority complex that makes you have hated a lot of real challenges for the sake of making easy baseless maths based on non-existing concepts as many as here (infinities, epsilon, delta, Approximations, famous meaningless cut, Average, endless sequences, ...etc), wonder!
So, if this is the case then you really don't need to worry at all, because you are dancing alone in the dark
And just imagine the football world where a little boy only scores a goal of a win in a competitive football match, then see how the whole world first-page coloured photos in newspapers, in the world press, in every TV station, ... etc
Just compare yourself and certainly, you would find yourself facing a mirror mathematicians, but wait and don't break the mirror...
And I was only comparing you with a class of another little category, and there are much higher categories for sure
But why all this is happening with a class of people who generally think that they were born genius and very distinct from others, but the fact is really shocking, but why?
It is because mathematicians had been deceived and upset-minded for so many centuries by now for the sake of worshipping some few devils that mislead them globally, otherwise, if true mathematicians are sufficient number, then they would never accept to be less than the top leaders of this mad world that are generally governed by mad people who can't even comprehend a simplest theorem like the Pythagorean even you assign the best teachers for them for all their lives, and they will destroy the world over your heads for sure
Just look and listen carefully around, but those helpless wouldn't change their inevitable destiny unless they learn the biggest unforgottable lesson for sure
BKK

And observe carefully here the complete absence in this thread of few well-known TROLLS (at least in sci.math – Google) as (Dan, Jan, Markus, Zelos, … and other so strange forgotten creatures), where at the same time they, they usually do appear very often in my other threads in many much less talkative topics than real challenging mathematics as number theory where generally any mathematician can say repeatedly anything that he learnt blindly from books

I really wonder about those incomprehensible characters

And I know that they would run away and hide immediately once they read the first line, and not because they don’t understand any big issue in mathematics, but certainly because they have no word to add here, and they would be so glad if anyone could prove me wrong, and they would immediately appear in this case
But here in such area, there isn’t any hope for anyone to turn it down (for sure)

And not because that I can’t prove it, of course, I can and (so easily), but my real intention is to make my Theorems (which seem only as Conjectures for you all) as a suitable and permanent curse and real punishment for such common typo mathematicians in the era of free word internet publishing, which will soon invade and throw away the official traditional publishing as peer-reviewed

Noting that, if the well-known Euler’s (sum of powers wrong conjecture) was only announced today then most likely it will be refuted in the same day mainly due to computer engineers and mathematicians programmers, but that actually took few centuries to be refuted

Ref. https://en.wikipedia.org/wiki/Euler%27s_sum_of_powers_conjecture

I also challenge all those (themselves) called world Peer-reviewed (Top Journals and Universities) to prove me wrong in any of my PUBLISHED theorems

In fact, I want to make a great fun of them mainly for school students and another non-professional-mathematicians, especially for footballers and in the best area they feel so proud (for sure)

Regards
Bassam King Karzeddin
Sep. 26, 2017

konyberg

2017-09-27 20:27:41 UTC

Post by bassam king karzeddin
Why it is impossible to find a single primitive Pythagorean triplet (in positive integers) being with all terms as powerful numbers?
Regards
Bassam King Karzeddin

Just imagine how little and how so tiny and how unnoticeable is the world of mathematicians for sure
A new theorem had been stated since a long time, not a single professional historian could claim older sources nor a single genius mathematician could also refute it, but still, doesn't count to be adopted by any mathematical establishment, especially that it is published in a Usenet as sci.math
As if the word publish means only in a Journals, (not learning any lesson from
the history)
But just imagine if this (half line theorem without a single mathematical notation) was announced by a top known genius mathematician or an alleged top (Journal or University), then how the mathematical press would immediately act with so much music and so many analysis with especially those many so professional talents that acquire very long tongues and deepest throats (that are good for something else)
So where are they hiding now from this simplest challenge? wonder!
Don't we live in an era of supercomputers also? wonder!
I also announced a modest prize for you genius mathematicians
Is it your so negligible and so unnoticeable dignity that make you pretend to be deaf and so blind? wonder!
Or is it the inherited dishonesty that the vast majority of the professional mathematicians acquire? no wonder!
Aren't you so shameful of your little self-being as real traitors to the science that feeds, drinks, and cloths you? wonder!
Or is it that your inferiority complex that makes you have hated a lot of real challenges for the sake of making easy baseless maths based on non-existing concepts as many as here (infinities, epsilon, delta, Approximations, famous meaningless cut, Average, endless sequences, ...etc), wonder!
So, if this is the case then you really don't need to worry at all, because you are dancing alone in the dark
And just imagine the football world where a little boy only scores a goal of a win in a competitive football match, then see how the whole world first-page coloured photos in newspapers, in the world press, in every TV station, ... etc
Just compare yourself and certainly, you would find yourself facing a mirror mathematicians, but wait and don't break the mirror...
And I was only comparing you with a class of another little category, and there are much higher categories for sure
But why all this is happening with a class of people who generally think that they were born genius and very distinct from others, but the fact is really shocking, but why?
It is because mathematicians had been deceived and upset-minded for so many centuries by now for the sake of worshipping some few devils that mislead them globally, otherwise, if true mathematicians are sufficient number, then they would never accept to be less than the top leaders of this mad world that are generally governed by mad people who can't even comprehend a simplest theorem like the Pythagorean even you assign the best teachers for them for all their lives, and they will destroy the world over your heads for sure
Just look and listen carefully around, but those helpless wouldn't change their inevitable destiny unless they learn the biggest unforgottable lesson for sure
BKK

And observe carefully here the complete absence in this thread of few well-known TROLLS (at least in sci.math – Google) as (Dan, Jan, Markus, Zelos, … and other so strange forgotten creatures), where at the same time they, they usually do appear very often in my other threads in many much less talkative topics than real challenging mathematics as number theory where generally any mathematician can say repeatedly anything that he learnt blindly from books
I really wonder about those incomprehensible characters
And I know that they would run away and hide immediately once they read the first line, and not because they don’t understand any big issue in mathematics, but certainly because they have no word to add here, and they would be so glad if anyone could prove me wrong, and they would immediately appear in this case
But here in such area, there isn’t any hope for anyone to turn it down (for sure)
And not because that I can’t prove it, of course, I can and (so easily), but my real intention is to make my Theorems (which seem only as Conjectures for you all) as a suitable and permanent curse and real punishment for such common typo mathematicians in the era of free word internet publishing, which will soon invade and throw away the official traditional publishing as peer-reviewed
Noting that, if the well-known Euler’s (sum of powers wrong conjecture) was only announced today then most likely it will be refuted in the same day mainly due to computer engineers and mathematicians programmers, but that actually took few centuries to be refuted
Ref. https://en.wikipedia.org/wiki/Euler%27s_sum_of_powers_conjecture
I also challenge all those (themselves) called world Peer-reviewed (Top Journals and Universities) to prove me wrong in any of my PUBLISHED theorems
In fact, I want to make a great fun of them mainly for school students and another non-professional-mathematicians, especially for footballers and in the best area they feel so proud (for sure)
Regards
Bassam King Karzeddin
Sep. 26, 2017

The answer is so easy that non here has thought you were serious. It is not solved. If you have a solution; give it.

KON

bassam king karzeddin

2017-09-30 08:55:59 UTC

Post by bassam king karzeddin
Why it is impossible to find a single primitive Pythagorean triplet (in positive integers) being with all terms as powerful numbers?
Regards
Bassam King Karzeddin

Just imagine how little and how so tiny and how unnoticeable is the world of mathematicians for sure
A new theorem had been stated since a long time, not a single professional historian could claim older sources nor a single genius mathematician could also refute it, but still, doesn't count to be adopted by any mathematical establishment, especially that it is published in a Usenet as sci.math
As if the word publish means only in a Journals, (not learning any lesson from
the history)
But just imagine if this (half line theorem without a single mathematical notation) was announced by a top known genius mathematician or an alleged top (Journal or University), then how the mathematical press would immediately act with so much music and so many analysis with especially those many so professional talents that acquire very long tongues and deepest throats (that are good for something else)
So where are they hiding now from this simplest challenge? wonder!
Don't we live in an era of supercomputers also? wonder!
I also announced a modest prize for you genius mathematicians
Is it your so negligible and so unnoticeable dignity that make you pretend to be deaf and so blind? wonder!
Or is it the inherited dishonesty that the vast majority of the professional mathematicians acquire? no wonder!
Aren't you so shameful of your little self-being as real traitors to the science that feeds, drinks, and cloths you? wonder!
Or is it that your inferiority complex that makes you have hated a lot of real challenges for the sake of making easy baseless maths based on non-existing concepts as many as here (infinities, epsilon, delta, Approximations, famous meaningless cut, Average, endless sequences, ...etc), wonder!
So, if this is the case then you really don't need to worry at all, because you are dancing alone in the dark
And just imagine the football world where a little boy only scores a goal of a win in a competitive football match, then see how the whole world first-page coloured photos in newspapers, in the world press, in every TV station, ... etc
Just compare yourself and certainly, you would find yourself facing a mirror mathematicians, but wait and don't break the mirror...
And I was only comparing you with a class of another little category, and there are much higher categories for sure
But why all this is happening with a class of people who generally think that they were born genius and very distinct from others, but the fact is really shocking, but why?
It is because mathematicians had been deceived and upset-minded for so many centuries by now for the sake of worshipping some few devils that mislead them globally, otherwise, if true mathematicians are sufficient number, then they would never accept to be less than the top leaders of this mad world that are generally governed by mad people who can't even comprehend a simplest theorem like the Pythagorean even you assign the best teachers for them for all their lives, and they will destroy the world over your heads for sure
Just look and listen carefully around, but those helpless wouldn't change their inevitable destiny unless they learn the biggest unforgottable lesson for sure
BKK

And observe carefully here the complete absence in this thread of few well-known TROLLS (at least in sci.math – Google) as (Dan, Jan, Markus, Zelos, … and other so strange forgotten creatures), where at the same time they, they usually do appear very often in my other threads in many much less talkative topics than real challenging mathematics as number theory where generally any mathematician can say repeatedly anything that he learnt blindly from books
I really wonder about those incomprehensible characters
And I know that they would run away and hide immediately once they read the first line, and not because they don’t understand any big issue in mathematics, but certainly because they have no word to add here, and they would be so glad if anyone could prove me wrong, and they would immediately appear in this case
But here in such area, there isn’t any hope for anyone to turn it down (for sure)
And not because that I can’t prove it, of course, I can and (so easily), but my real intention is to make my Theorems (which seem only as Conjectures for you all) as a suitable and permanent curse and real punishment for such common typo mathematicians in the era of free word internet publishing, which will soon invade and throw away the official traditional publishing as peer-reviewed
Noting that, if the well-known Euler’s (sum of powers wrong conjecture) was only announced today then most likely it will be refuted in the same day mainly due to computer engineers and mathematicians programmers, but that actually took few centuries to be refuted
Ref. https://en.wikipedia.org/wiki/Euler%27s_sum_of_powers_conjecture
I also challenge all those (themselves) called world Peer-reviewed (Top Journals and Universities) to prove me wrong in any of my PUBLISHED theorems
In fact, I want to make a great fun of them mainly for school students and another non-professional-mathematicians, especially for footballers and in the best area they feel so proud (for sure)
Regards
Bassam King Karzeddin
Sep. 26, 2017

The answer is so easy that non here has thought you were serious. It is not solved. If you have a solution; give it.
KON

So, it is unsolved problem, and I don't know truly why one shouldn't be serious here

And if it isn't a truly unsolvable problem, then why the mathematical establishments announce it so as an unsolvable problem in number theory? wonder!

Or is it that they consider sci.math as a SH*T source or a reference for mathematical problems? wonder!

Then, they should have to get ready for a stage where a reference or a source may be also considered from any SH*TY reference or any rubbish source as here for example as long as it contains challenging problems (with documented dates that no one on earth could solve it yet)

Or maybe they would think about it once they deliberately conclude it from any future fabricated old references from an alleged reputable source, as usual, especially if they truly could solve it, wonder!

Isn't so strange that people came to know about so many others who wrote only notes for themselves in much older centuries, whereas something PUBLISHED REPEATEDLY to the whole WOLRD and so freely nowadays goes unnoticeable? wonder!

Also, I wouldn't be stingy with you mathematicians by not showing you the solution, but I would prefer giving a big hint to it from my PUBLISHED and well documented and also irrefutable formula (certified by others) for the real number (r), for (x^r + y^r = z^r), where (x, y, z) are distinct positive integers,

And don't pretend that you didn't see my formula that was announced repeatedly and in other sites too

However, the formula was done (in 1990) by the formal mathematical analysis that mathematicians usually are familiar

BKK

bassam king karzeddin

2017-09-30 15:22:43 UTC