bassam king karzeddin

2016-11-23 19:17:53 UTC

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PermalinkRegards

Bassam King Karzeddin

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bassam king karzeddin

2016-11-23 19:17:53 UTC

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PermalinkRegards

Bassam King Karzeddin

Virgil

2016-11-25 02:13:25 UTC

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PermalinkWhy it is impossible to find a single primitive Pythagorean triplet (in

positive integers) being with all terms as powerful numbers?

Regards

Bassam King Karzeddin

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

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

Peter Percival

2016-11-25 09:09:50 UTC

Reply

PermalinkIn article

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 square and a cube.

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

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

Reply

PermalinkIn article

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 square and a cube.

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

Reply

PermalinkIn article

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 square and a cube.

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.

Regards

Bassam King Karzeddin

a***@gmail.com

2016-11-27 05:11:29 UTC

Reply

Permalinka^p = a mod p?... but,

not p-summorial!

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

Reply

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

a^p = a mod p?... but,

not p-summorial!

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.

Regards

Bassam King Karzeddin

a***@gmail.com

2017-03-29 22:16:46 UTC

Reply

Permalinkis 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

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

Reply

Permalinkno, 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

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

BK

a***@gmail.com

2017-04-05 16:42:35 UTC

Reply

Permalinkit is clearly a consequence of the little theorem, and

you soulhd easily show such module

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.

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

Reply

Permalinkah, 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

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.

divides N for some a > 1.

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

Really helpless and clueless as always as usual, wonder!

Good luck

BK

a***@gmail.com

2017-04-05 20:24:35 UTC

Reply

PermalinkwTf 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

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

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.

divides N for some a > 1.

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

Really helpless and clueless as always as usual, wonder!

Good luck

BK

Vinicius Claudino Ferraz

2017-04-06 14:33:11 UTC

Reply

Permalink2p + 1 = 15 composto químico

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

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

a^p = a mod p?... but,

divides N for some a > 1.

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

Really helpless and clueless as always as usual, wonder!

Good luck

BK

a***@gmail.com

2017-04-06 19:09:18 UTC

Reply

Permalinkp = 7 prime

2p + 1 = 15 composto químico

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

it is clearly a consequence of the little theorem, and

you should easily show such a module

a^p = a mod p?... but,

divides N for some a > 1.

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

perfectly g00d profession, of course

Arturo Magidin

2016-11-27 08:01:05 UTC

Reply

PermalinkIstr that a positive integer is called powerful if it is the product of

a square and a cube.

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.

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

Arturo Magidin

Peter Percival

2016-11-28 15:07:47 UTC

Reply

PermalinkIstr that a positive integer is called powerful if it is the product of

a square and a cube.

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.

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.

--

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

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

Peter Percival

2016-11-28 15:05:57 UTC

Reply

PermalinkIn article

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 square and a cube.

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

divides N for some a > 1.

--

Do, as a concession to my poor wits, Lord Darlington, just explain

to me what you really mean.

I think I had better not, Duchess. Nowadays to be intelligible is

to be found out. -- Oscar Wilde, Lady Windermere's Fan

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

Virgil

2016-11-25 18:25:26 UTC

Reply

PermalinkIstr that a positive integer is called powerful if it is the product of

a square and a cube.

--

Virgil

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

Virgil

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

konyberg

2016-11-25 18:59:29 UTC

Reply

PermalinkIstr that a positive integer is called powerful if it is the product of

a square and a cube.

--

Virgil

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

I may have not searched enough, but that's it.

KON

konyberg

2016-11-25 19:07:07 UTC

Reply

PermalinkIstr that a positive integer is called powerful if it is the product of

a square and a cube.

--

Virgil

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

I may have not searched enough, but that's it.

KON

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

Reply

PermalinkIstr that a positive integer is called powerful if it is the product of

a square and a cube.

--

Virgil

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

I may have not searched enough, but that's it.

KON

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

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

Reply

PermalinkIstr that a positive integer is called powerful if it is the product of

a square and a cube.

--

Virgil

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

I may have not searched enough, but that's it.

KON

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

All numbers of the form a^2, b^3 and (a^2)(b^3) are powerful numbers.

KON

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

Reply

PermalinkIstr that a positive integer is called powerful if it is the product of

a square and a cube.

--

Virgil

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

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

Istr that a positive integer is called powerful if it is the product of

a square and a cube.

--

Virgil

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

Peter Percival

2016-11-28 15:09:46 UTC

Reply

PermalinkIstr that a positive integer is called powerful if it is the product of

a square and a cube.

pointed out.

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 08:35:39 UTC

Reply

PermalinkIn article

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 square and a cube.

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

Regards

Bassam King karzeddin

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

Virgil

2016-11-26 22:01:59 UTC

Reply

PermalinkIstr that a positive integer is called powerful if it is the product of

a square and a cube.

(a, b) are real integers

--

Virgil

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

Virgil

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

bassam king karzeddin

2016-11-27 07:28:17 UTC

Reply

PermalinkIstr that a positive integer is called powerful if it is the product of

a square and a cube.

(a, b) are real integers

--

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

Virgil

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

Istr that a positive integer is called powerful if it is the product of

a square and a cube.

(a, b) are real integers

--

Virgil

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

Virgil

2016-11-27 21:30:20 UTC

Reply

PermalinkIstr that a positive integer is called powerful if it is the product of

a square and a cube.

(a, b) are real integers

--

--

Virgil

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

Virgil

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

bassam king karzeddin

2016-11-28 15:01:30 UTC

Reply

PermalinkIstr that a positive integer is called powerful if it is the product of

a square and a cube.

(a, b) are real integers

--

--

Virgil

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

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

Reply

PermalinkIn article

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

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)

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

Reply

PermalinkIn article

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

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)

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

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

Virgil

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

konyberg

2016-11-25 12:03:01 UTC

Reply

PermalinkWhy it is impossible to find a single primitive Pythagorean triplet (in positive integers) being with all terms as powerful numbers?

Regards

Bassam King Karzeddin

KON

Peter Percival

2016-11-25 15:07:01 UTC

Reply

PermalinkWhy it is impossible to find a single primitive Pythagorean triplet (in positive integers) being with all terms as powerful numbers?

Regards

Bassam King Karzeddin

KON

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

konyberg

2016-11-25 16:16:43 UTC

Reply

PermalinkWhy it is impossible to find a single primitive Pythagorean triplet (in positive integers) being with all terms as powerful numbers?

Regards

Bassam King Karzeddin

KON

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

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

Reply

PermalinkWhy it is impossible to find a single primitive Pythagorean triplet (in positive integers) being with all terms as powerful numbers?

Regards

Bassam King Karzeddin

KON

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

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

bassam king karzeddin

2016-11-26 08:44:29 UTC

Reply

PermalinkWhy it is impossible to find a single primitive Pythagorean triplet (in positive integers) being with all terms as powerful numbers?

Regards

Bassam King Karzeddin

KON

Bassam makes reference, then (u^2)^2 + (v^2)^2 = (w^2)^2 has no postive

integral solutions.

Bassam King Karzeddin

--

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

Reply

PermalinkWhy it is impossible to find a single primitive Pythagorean triplet (in positive integers) being with all terms as powerful numbers?

Regards

Bassam King Karzeddin

KON

Bassam King Karzeddin

bert

2016-11-25 16:28:50 UTC

Reply

PermalinkWhy it is impossible to find a single primitive

Pythagorean triplet (in positive integers) being

with all terms as powerful numbers?

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

Reply

PermalinkWhy it is impossible to find a single primitive

Pythagorean triplet (in positive integers) being

with all terms as powerful numbers?

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.

--

Regards

Bassam King Karzeddin

konyberg

2016-11-25 16:42:01 UTC

Reply

PermalinkWhy 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.

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

Reply

PermalinkWhy it is impossible to find a single primitive Pythagorean triplet (in positive integers) being with all terms as powerful numbers?

Regards

Bassam King Karzeddin

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

This is called addition of two powerful numbers

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

Regards

Bassam King Karzeddin

27th, Nov., 2016

Virgil

2016-11-27 21:33:12 UTC

Reply

PermalinkIf 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

could easily straighten Badass out, if he chose to.

--

Virgil

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

Virgil

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

bassam king karzeddin

2016-11-28 14:16:41 UTC

Reply

PermalinkIf 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

could easily straighten Badass out, if he chose to.

--

Virgil

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

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

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

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

Reply

PermalinkWhy 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 + 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?

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

Regards

Bassam King Karzeddin

bassam king karzeddin

2016-11-26 12:06:28 UTC

Reply

PermalinkWhy it is impossible to find a single primitive Pythagorean triplet (in positive integers) being with all terms as powerful numbers?

Regards

Bassam King Karzeddin

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

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

Reply

PermalinkWhy 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

Reply

Permalinkat 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

Reply

Permalinkthere 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

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

Reply

Permalinka^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

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

Reply

Permalinkof the left tetrahedron. really, of ocurse,

there are no right trigona, because

you just flip the paper over et voila

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

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

Reply

PermalinkI 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

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

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

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

Reply

Permalinkas given by Sierpinski for all of the rational cuboids,

or pythagorean tetruplets

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

Reply

Permalinkit is really quite simple,

as given by Sierpinski for all of the rational cuboids,

or pythagorean tetruplets

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

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

Reply

Permalinkmaking three regular tetragona on the edges,

that's just a lot of unnecessary work;

so, the problem is to provide spatial l00ns pr00f

as given by Sierpinski for all of the rational cuboids,

or pythagorean tetruplets

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

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

Reply

Permalinktorsdag 24. november 2016 15.50.05 UTC+1 skrev bassam

Pythagorean triplet (in positive integers) being with

all terms as powerful numbers?

(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

Pythagorean triplet (in positive integers) being with

all terms as powerful numbers?

(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 problemPythagorean triplet (in positive integers) being with

all terms as powerful numbers?

(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

Pythagorean triplet (in positive integers) being with

all terms as powerful numbers?

(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.

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((a^2)(b^3))^2, then this I think you should discuss

with Fermat or Wiles :)

Regards

Bassam King Karzeddin

27th, Nov., 2016

KON

bassam king karzeddin

2016-11-28 08:25:21 UTC

Reply

PermalinkIn article

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

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)

About Fermat, true I could not contact him so far

Bassam King Karzeddin

b***@gmail.com

2016-11-29 12:59:17 UTC

Reply

PermalinkI 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

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

Reply

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

but never heard of any reply

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 regardbut never heard of any reply

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?

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

Reply

Permalinkonly 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

Reply

PermalinkAgain not my problem, its only your prolobololoblem.

bassam king karzeddin

2016-12-01 07:50:04 UTC

Reply

PermalinkI 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.

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

bassam king karzeddin

2017-03-29 16:26:27 UTC

Reply

PermalinkWhy 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, 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

Reply

PermalinkBut, 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

Reply

PermalinkWhy it is impossible to find a single primitive Pythagorean triplet (in positive integers) being with all terms as powerful numbers?

Regards

Bassam King Karzeddin

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

Reply

PermalinkI 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

Reply

PermalinkWhy 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 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

bassam king karzeddin

2017-04-06 07:04:27 UTC

Reply

Permalinkyou talking to yourself like a busted record,

maybe with a nice accent of st00pid

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

Reply

PermalinkI 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

where's your right clownshoe?

a***@gmail.com

2017-04-08 22:38:36 UTC

Reply

Permalink
a***@gmail.com

2017-04-10 16:27:07 UTC

Reply

Permalinkit 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

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

Reply

Permalinkthis 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

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?

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

Reply

Permalinkwith mathematics, doesn't even know the result of

-1 * -1 = ?

bassam king karzeddin

2017-04-12 08:49:49 UTC

Reply

PermalinkI guess it is in his ass, since BKK is a little stiff

with mathematics, doesn't even know the result of

-1 * -1 = ?

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

Reply

PermalinkI guess it is in his ass, since BKK is a little stiff

with mathematics, doesn't even know the result of

-1 * -1 = ?

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

bassam king karzeddin

2017-04-12 13:56:21 UTC

Reply

PermalinkI guess it is in his ass, since BKK is a little stiff

with mathematics, doesn't even know the result of

-1 * -1 = ?

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

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

Reply

PermalinkI guess it is in his ass, since BKK is a little stiff

with mathematics, doesn't even know the result of

-1 * -1 = ?

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

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

BK

a***@gmail.com

2017-04-10 21:50:34 UTC

Reply

Permalinkbetween the skew tetragon & the tetrahedron; indeed,

once can quantumfy the tetrahedron by the four dihedrals,

and one edgelength to get the size

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

spheres, and b)

tetrahedra. or, you could say,

the continuous tetrahedron & the dyscrete sphere

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

Reply

Permalinkbetween the skew tetragon & the tetrahedron; indeed,

once can quantumfy the tetrahedron by the four dihedrals,

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

spheres, and b)

tetrahedra. or, you could say,

the continuous tetrahedron & the dyscrete sphere

bassam king karzeddin

2017-05-10 07:21:43 UTC

Reply

PermalinkWhy 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 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

Reply

PermalinkWhy 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 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

Reply

PermalinkWhy it is impossible to find a single primitive Pythagorean triplet (in positive integers) being with all terms as powerful numbers?

Regards

Bassam King Karzeddin

It will chase you for sure

BKK

bassam king karzeddin

2017-07-29 15:52:35 UTC

Reply

PermalinkWhy it is impossible to find a single primitive Pythagorean triplet (in positive integers) being with all terms as powerful numbers?

Regards

Bassam King Karzeddin

It will chase you for sure

BKK

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

Reply

PermalinkWhy it is impossible to find a single primitive Pythagorean triplet (in positive integers) being with all terms as powerful numbers?

Regards

Bassam King Karzeddin

BKK

bassam king karzeddin

2017-07-29 16:32:10 UTC

Reply

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

all terms as powerful numbers?

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

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

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

statement without any single mathematical notationIt will chase you for sure

BKK

I also try to make it especially for you only in oneBKK

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

BKK

bassam king karzeddin

2017-07-31 14:34:59 UTC

Reply

PermalinkWhy it is impossible to find a single primitive Pythagorean triplet (in positive integers) being with all terms as powerful numbers?

Regards

Bassam King Karzeddin

BKK

bassam king karzeddin

2017-08-01 12:08:44 UTC

Reply

PermalinkWhy it is impossible to find a single primitive Pythagorean triplet (in positive integers) being with all terms as powerful numbers?

Regards

Bassam King Karzeddin

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

Reply

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

BKK

bassam king karzeddin

2017-08-07 09:26:26 UTC

Reply

PermalinkWhy it is impossible to find a single primitive Pythagorean triplet (in positive integers) being with all terms as powerful numbers?

Regards

Bassam King Karzeddin

BKK

bassam king karzeddin

2017-08-09 11:39:28 UTC

Reply

PermalinkWhy it is impossible to find a single primitive Pythagorean triplet (in positive integers) being with all terms as powerful numbers?

Regards

Bassam King Karzeddin

BKK

bassam king karzeddin

2017-09-12 08:45:18 UTC

Reply

PermalinkWhy 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, 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

Reply

PermalinkBKK

bassam king karzeddin

2017-09-18 07:55:38 UTC

Reply

PermalinkWhy it is impossible to find a single primitive Pythagorean triplet (in positive integers) being with all terms as powerful numbers?

Regards

Bassam King Karzeddin

BKK

bassam king karzeddin

2017-09-20 18:11:33 UTC

Reply

PermalinkWhy it is impossible to find a single primitive Pythagorean triplet (in positive integers) being with all terms as powerful numbers?

Regards

Bassam King Karzeddin

BKK

bassam king karzeddin

2017-09-24 16:16:47 UTC

Reply

PermalinkWhy 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 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

Reply

PermalinkWhy 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 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

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

Reply

PermalinkWhy 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 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

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

KON

bassam king karzeddin

2017-09-30 08:55:59 UTC

Reply

PermalinkWhy 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 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

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

KON

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

Reply

PermalinkOn Wednesday, September 27, 2017 at 11:27:53 PM

being with all terms as powerful numbers?

how unnoticeable is the world of mathematicians for

sure

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

Journals, (not learning any lesson from

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)

simplest challenge? wonder!

also? wonder!

mathematicians

dignity that make you pretend to be deaf and so

blind? wonder!

vast majority of the professional mathematicians

acquire? no wonder!

self-being as real traitors to the science that

feeds, drinks, and cloths you? wonder!

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!

need to worry at all, because you are dancing alone

in the dark

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

find yourself facing a mirror mathematicians, but

wait and don't break the mirror...

another little category, and there are much higher

categories for sure

people who generally think that they were born genius

and very distinct from others, but the fact is really

shocking, but why?

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

those helpless wouldn't change their inevitable

destiny unless they learn the biggest unforgottable

lesson for sure

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

characters

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

anyone to turn it down (for sure)

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

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

https://en.wikipedia.org/wiki/Euler%27s_sum_of_powers_

conjecture

world Peer-reviewed (Top Journals and Universities)

to prove me wrong in any of my PUBLISHED theorems

mainly for school students and another

non-professional-mathematicians, especially for

footballers and in the best area they feel so proud

(for sure)

were serious. It is not solved. If you have a

solution; give it.

y why one shouldn't be serious here

And if it isn't a truly unsolvable problem, then why

y 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

T source or a reference for mathematical problems?

wonder!

Then, they should have to get ready for a stage

e 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

y 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

o 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

s 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

a that was announced repeatedly and in other sites

too

However, the formula was done (in 1990) by the

e formal mathematical analysis that mathematicians

usually are familiar

BKK

Read it so carefully heretirsdag 26. september 2017 11.14.18 UTC+2 skrev

primitive Pythagorean triplet (in positive integers)On Sunday, September 24, 2017 at 7:17:00 PM

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

being with all terms as powerful numbers?

Just imagine how little and how so tiny and

sure

A new theorem had been stated since a long

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

the history)

But just imagine if this (half line theorem

But just imagine if this (half line theorem

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

Don't we live in an era of supercomputers

I also announced a modest prize for you genius

Is it your so negligible and so unnoticeable

blind? wonder!

Or is it the inherited dishonesty that the

acquire? no wonder!

Aren't you so shameful of your little

feeds, drinks, and cloths you? wonder!

Or is it that your inferiority complex that

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

in the dark

And just imagine the football world where 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

wait and don't break the mirror...

And I was only comparing you with a class of

categories for sure

But why all this is happening with a class of

and very distinct from others, but the fact is really

shocking, but why?

It is because mathematicians had been deceived

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

destiny unless they learn the biggest unforgottable

lesson for sure

BKK

And observe carefully here the complete absencesci.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

And I know that they would run away and hide

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

And not because that I can’t prove it, of

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

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.

conjecture

I also challenge all those (themselves) called

to prove me wrong in any of my PUBLISHED theorems

In fact, I want to make a great fun of them

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 youBassam King Karzeddin

Sep. 26, 2017

solution; give it.

KON

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

And if it isn't a truly unsolvable problem, then why

y 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

T source or a reference for mathematical problems?

wonder!

Then, they should have to get ready for a stage

e 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

y 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

o 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

s 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

a that was announced repeatedly and in other sites

too

However, the formula was done (in 1990) by the

e formal mathematical analysis that mathematicians

usually are familiar

BKK

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

BKK

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