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

On Fri, 26 May 2006 17:03:04 EDT, bassam king karzeddin
<bassam@xxxxxxxxxx> wrote:

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

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
.

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