# Another_ p^(2^k), Karzeddin's like Conjecture

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### bassam king karzeddin

May 25, 2006, 3:54:14 PM5/25/06
to
Dear All

I should like to state the following conjecture in number theory of prime numbers that depends more on feelings, without knowing if it is true or false, new or old

I will be glad if one of each two conjectures of mine stay as a challenge, and that only will give a good reason to prove them.

The Conjecture

"The ratio (of the sum (of two distinct non zero integers relatively prime to each other) of alike power of odd prime number that doesn't divide any of the integers) to their sum is equal to a positive integer which has all its prime factors exponents as a power of two to a natural number"

In symbols,
Let, (p) is odd prime number, and (x and y) are two none zero distinct coprime integers in their absolute value, where also (x and y) are not a multiple of (p),

If, n = (x^p + y^p) / (x + y),

then, all prime factor exponents of (n), are of the form of (2^k), that is a power of two)

I will be very gratfull for any counter example

Thanking you all

Bassam Karzeddin
Al-Hussein Bin Talal University
JORDAN

### Pubkeybreaker

May 25, 2006, 4:38:23 PM5/25/06
to

bassam king karzeddin wrote:
> Dear All
>
> I should like to state the following conjecture in number theory of prime numbers that depends more on feelings, without knowing if it is true or false, new or old

> In symbols,

> Let, (p) is odd prime number, and (x and y) are two none zero distinct coprime integers in their absolute value, where also (x and y) are not a multiple of (p),
>
> If, n = (x^p + y^p) / (x + y),
>
> then, all prime factor exponents of (n), are of the form of (2^k), that is a power of two)

It is unclear what you mean by "prime factor exponents". If you are
saying that
if a prime q divides n, then it must do so to an even power, then this
is false.
Or do you allow k = 0? You would then be saying that if q^s | n for
prime q, then
s is never 3,5,6,7,9,10,11, .......

This is almost certainly true on probabilistic grounds, and is not a
new conjecture.
You have no right to claim it as your own.

And the "power of 2" is a red herring. The conjecture would be
better stated
that if q^s | n, then s = 1 or 2 only. And I expect that s=2
will be VERY
rare. (if it ever occurs) Note that 23^2 does divide 3^11 -
2^11. I know of no
numerical examples where q^2 divides n as you have defined it. Nor
are any
examples known where q^2 | 2^p + 1.

This is a homogeneous form of the well-known conjecture that the
Mersenne
numbers and their "plus side" companions to M_p are always
squarefree.
Indeed, it seems even less likely that q^3 will ever divide n
than q^2 will divide a
Mersenne number.

Clearly, if you remove the restriction that p is prime, then there are
infinitely
many counter-examples.

### bassam king karzeddin

May 25, 2006, 5:38:27 PM5/25/06
to
>
> bassam king karzeddin wrote:
> > Dear All
> >
> > I should like to state the following conjecture in
> number theory of prime numbers that depends more on
> feelings, without knowing if it is true or false, new
> or old
>
> > In symbols,
> > Let, (p) is odd prime number, and (x and y) are
> two none zero distinct coprime integers in their
> absolute value, where also (x and y) are not a
> multiple of (p),
> >
> > If, n = (x^p + y^p) / (x + y),
> >
> > then, all prime factor exponents of (n), are of
> the form of (2^k), that is a power of two)
>
> It is unclear what you mean by "prime factor
> exponents". If you are
> saying that
I mean the prime factors of unique prime factorization of (n) as I defined will have exponent of the form 2^k, where k is non negative integer
previously

> if a prime q divides n, then it must do so to an
> even power, then this
> is false.
I clearly didn't say that

> Or do you allow k = 0? You would then be saying
> that if q^s | n for
> prime q, then
> s is never 3,5,6,7,9,10,11, .......
Yes Sir

>
> This is almost certainly true on probabilistic
> grounds, and is not a
> new conjecture.
> You have no right to claim it as your own.
Is what you say proven?
I will be thankful if you will kindly show me a reference link

>
> And the "power of 2" is a red herring. The
> conjecture would be
> better stated
> that if q^s | n, then s = 1 or 2 only. And I
> expect that s=2
> will be VERY
> rare. (if it ever occurs) Note that 23^2 does
> divide 3^11 -
> 2^11. I know of no
> numerical examples where q^2 divides n as you have
> defined it. Nor
> are any
> examples known where q^2 | 2^p + 1.

Note that, I'm not talking about a pair of (x,y)=(1,2)

>
> This is a homogeneous form of the well-known
> conjecture that the
> Mersenne
> numbers and their "plus side" companions to M_p
> are always
> squarefree.
> Indeed, it seems even less likely that q^3 will
> l ever divide n
> than q^2 will divide a
> Mersenne number.

Mersenne number is only a single case of my new conjecture hopefully

>
> Clearly, if you remove the restriction that p is
> prime, then there are
> infinitely
> many counter-examples.
>

But, I didn't remove the restriction

My regards
B.Karzeddin

### quasi

May 25, 2006, 6:57:24 PM5/25/06
to
On 25 May 2006 13:38:23 -0700, "Pubkeybreaker"
<Robert_s...@raytheon.com> wrote:

>bassam king karzeddin wrote:
>
>> Dear All
>>
>> I should like to state the following conjecture in number theory of prime numbers that depends more on feelings, without knowing if it is true or false, new or old
>
>> In symbols,
>> Let, (p) is odd prime number, and (x and y) are two none zero distinct coprime integers in their absolute value, where also (x and y) are not a multiple of (p),
>>
>> If, n = (x^p + y^p) / (x + y),
>>
>> then, all prime factor exponents of (n), are of the form of (2^k), that is a power of two)
>
>It is unclear what you mean by "prime factor exponents". If you are
>saying that
>if a prime q divides n, then it must do so to an even power, then this
>is false.
>Or do you allow k = 0? You would then be saying that if q^s | n for
>prime q, then
>s is never 3,5,6,7,9,10,11, .......
>
>This is almost certainly true on probabilistic grounds, and is not a
>new conjecture.

It's not true on any grounds.

For example, let p=3, x=17, y=53.

Then (x^p+y^p)/(x+y) equals 13^3.

However Karzeddin's has previously stated some similar conjectures
which appear to be correct.

For example, here is a version of one of his conjectures (edited):

If n,x,y are positive integers with n>2, then the prime factorization
of x^n+y^n includes at least one prime factor with exponent at most 2.

>You have no right to claim it as your own.

I don't see why "he has no right to claim it as his own."

It's a simple conjecture, stronger than FLT and weaker than the ABC
conjecture. It's easily testable and very appealing. So far, no
reference has been noted as having an essentially similar conjecture.

To my view, the conjecture is Karzeddin's for now, at least in
sci.math.

>And the "power of 2" is a red herring. The conjecture would be
>better stated
>that if q^s | n, then s = 1 or 2 only.

Yes, I agree -- powers of 2 mean nothing here. The real issue is
whether or not any of the prime power exponents can exceed 2.

>And I expect that s=2
>will be VERY
>rare. (if it ever occurs) Note that 23^2 does divide 3^11 -
>2^11. I know of no
>numerical examples where q^2 divides n as you have defined it.

Well, your expectation as stated above is wrong.

For example, let p=3, x=56, y=65.

Then (x^p+y^p)/(x+y) equals 61^2.

>Nor are any
>examples known where q^2 | 2^p + 1.
>
>This is a homogeneous form of the well-known conjecture that the
>Mersenne
>numbers and their "plus side" companions to M_p are always
>squarefree.

Um, sorry -- the statement you made above is very fuzzy. To say that
Karzeddin's conjecture is "a homogeneous form" of some well known
conjecture relating to the factorization of Mersenne numbers is
imprecise to the point of being misleading. While the conjectures
might be similar in spirit, they are obviously not equivalent.

> Indeed, it seems even less likely that q^3 will ever divide n
>than q^2 will divide a
>Mersenne number.

Your statement above is invalidated by the same counterexample as
mentioned earlier, namely: p=3, x=17, y=53.

>Clearly, if you remove the restriction that p is prime, then there are
>infinitely
>many counter-examples.

Well, we are talking about an invalidated conjecture, so let's focus
Let me state it again ...

One of Karzeddin's conjectures (edited by quasi):

If n,x,y are positive integers with n>2, then the prime factorization
of x^n+y^n includes at least one prime factor with exponent at most 2.

quasi

### Gerry Myerson

May 25, 2006, 8:24:14 PM5/25/06
to
quasi <qu...@null.set> wrote:

> For example, let p=3, x=56, y=65.
>
> Then (x^p+y^p)/(x+y) equals 61^2.

Ha! these are exactly the numbers that turn up in the counterexample
I gave to another problem in another thread.

If u(x) = x^2 + 4 x - 8 and f(x) = x^3 + 1,
then u^2 = q f - r,
where q(x) = x + 8 and r(x) = 65 x - 56.

In the notation of the other thread, a / b = 56 / 65,
and f(a/b) = (56/65)^3 + 1 = (56^3 + 65^3) / 65^3
= (11^2) (61^2) / 65^3 is not a square in the rationals.

--
Gerry Myerson (ge...@maths.mq.edi.ai) (i -> u for email)

### bassam king karzeddin

May 26, 2006, 5:03:04 PM5/26/06
to
> >It is unclear what you mean by "prime factor
> exponents". If you are
> >saying that
> >if a prime q divides n, then it must do so to an
> even power, then this
> >is false.
> >Or do you allow k = 0? You would then be saying
> that if q^s | n for
> >prime q, then
> >s is never 3,5,6,7,9,10,11, .......
> >
> >This is almost certainly true on probabilistic
> grounds, and is not a
> >new conjecture.
>
> It's not true on any grounds.
>
> For example, let p=3, x=17, y=53.
>
> Then (x^p+y^p)/(x+y) equals 13^3.

I should like to thank you alot quasi for a very nice counter example, as this will help me alot to state it in a better version

>
> However Karzeddin's has previously stated some
> similar conjectures
> which appear to be correct.
>
> For example, here is a version of one of his
> conjectures (edited):
>
> If n,x,y are positive integers with n>2, then the
> prime factorization
> of x^n+y^n includes at least one prime factor with
> exponent at most 2.

I think, you are developing my previous conjecture to make (k) as finite as equal to (1 or 2), only, and this is a very important step of yours

>
> >You have no right to claim it as your own.

Indeed, I have no right to make a misleading conjectures, that was discovered by quasi

>
> I don't see why "he has no right to claim it as his
> own."
>
> It's a simple conjecture, stronger than FLT and
> weaker than the ABC

I still coudn't get the connection with ABC

> conjecture. It's easily testable and very appealing.
> So far, no
> reference has been noted as having an essentially
> similar conjecture.
>
> To my view, the conjecture is Karzeddin's for now, at
> least in
> sci.math.

I really don't mind if its not mine, as now it has been pointed out and developed by quasi, and I will not hesitate a second to mention its original reference if I KNOW.
The important issue is to prove it as new version of quasi or older one of mine, and hence make FLT available to school students only, and then will be a vectory

### bassam king karzeddin

May 26, 2006, 5:46:38 PM5/26/06
to
Dear All

As quasi found a very fast counter example to my conjecture, I was inspired by quasi to restate my previous conjecture in a better version based on quasi's notes

The Conjecture

"The ratio (of the sum (of two distinct non zero integers relatively prime to each other) of alike power of odd prime number that doesn't divide any of the integers) to their sum is equal to odd positive integer which has one "at most" of its prime factors exponents as not equal to one or two "

In symbols,
Let, (p) is odd prime number, and (x and y) are two none zero distinct coprime integers in their absolute value, where also (x and y) are not a multiple of (p),

If, n = (x^p + y^p) / (x + y),

then, one "at most" of prime factor exponents of the odd integer (n), is not equal to one or two.

I will be very gratfull for any counter example

If this conjecture comes true, you may call it as
"quasi-karzeddin" conjecture

### quasi

May 26, 2006, 6:01:51 PM5/26/06
to

No, it's essentially your conjecture. All I did was make a very minor
modification and then edit the statement slightly to improve the
wording.

>> I don't see why "he has no right to claim it as his
>> own."
>>
>> It's a simple conjecture, stronger than FLT and

>> weaker than the ABC conjecture. It's easily testable

>> and very appealing. So far, no reference has been
>> noted as having an essentially similar conjecture.

>I still coudn't get the connection with ABC

I'm not so clear about that either but, apparently, the truth of one
of the versions of the ABC conjecture would imply the truth of yours
(the one I stated above).

>> To my view, the conjecture is Karzeddin's for now, at
>> least in sci.math.
>
>I really don't mind if its not mine, as now it has been pointed out and developed by quasi, and I will not hesitate a second to mention its original reference if I KNOW.

Yes, there might be a reference with essentially the same conjecture
-- maybe, maybe not.

>The important issue is to prove it as new version of quasi or older one of mine, and hence make FLT available to school students only, and then will be a vectory

Yes, but you have to realize, just because you can discover a property
that appears to always be true, that doesn't mean that an elementary
proof is feasible.

You keep asking for a proof or a counterexample, but the reality is
that if the conjecture is true, today's math is probably not powerful
enough to prove it.

Besides stating the conjecture, you have provided no reason as to why
the conjecture should be true. You have not suggested any method of
attacking the problem. Why do you think others should suddenly devote
their energy to this problem when there's no real reason to believe
its truth and even if one does believe it, the task of proving it is
most likely hopeless.

quasi

### quasi

May 26, 2006, 6:19:47 PM5/26/06
to

No, it's not my conjecture.

Also, it's not true.

Counterexample:

Let p=3, x=701, y=971

Then (x^p+y^p)/(x+y) = (7^3)*(13^3)

This conjecture is not the same as the one I stated in my earlier

quasi

### bassam king karzeddin

May 26, 2006, 6:42:10 PM5/26/06
to

Ok, I shall give up conjecturing now
Thanks Quasi again for your nice and fast counter examples

### bassam karzeddin

Aug 9, 2023, 10:07:59 PMAug 9
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### bassam karzeddin

Aug 30, 2023, 5:36:08 PMAug 30
to
On Thursday, May 25, 2006 at 10:54:14 PM UTC+3, bassam king karzeddin wrote:

### bassam karzeddin

Sep 8, 2023, 7:47:27 PMSep 8
to
On Thursday, May 25, 2006 at 10:54:14 PM UTC+3, bassam king karzeddin wrote:

### bassam karzeddin

Sep 21, 2023, 10:40:51 AMSep 21
to
On Thursday, May 25, 2006 at 10:54:14 PM UTC+3, bassam king karzeddin wrote:
When I used to post my puzzles on public for the whole world 🌎, I used to think that people are generally clever & decent especially the mathematicians, but the years have tought me otherwise, where generally, every thing about the academic people especially the mathematicians is more than horrible, FOR SURE

BKK

### Python

Sep 21, 2023, 5:01:36 PMSep 21
to
There is no one more horrible than you here, for SURER bASSam !

### bassam karzeddin

Sep 22, 2023, 6:35:15 AMSep 22
to
I do realize now that unbreakable conjectures in number theory in an era of supercomputer & Artificial Intelligence wherever sourced from free public sources then they are too insulting for the mainstream academic proffessional mathematicians sheeples since they injure their so sensitive feelings & make them feel perpetually inferior with inferiority complex

But when the same conjectures are sourced from their history or their alleged top-most Journals and Universities, then they would become extremely adorable & too wonderful as well, FOR SURE

BKK

### bassam karzeddin

Oct 8, 2023, 10:43:36 PMOct 8
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On Thursday, May 25, 2006 at 10:54:14 PM UTC+3, bassam king karzeddin wrote:

Good luck

BKK

### bassam karzeddin

Oct 12, 2023, 4:27:33 AMOct 12
to