Sklansky -Fermat Conjectures

[Zeno]

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-Zeno

[pzhon]

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Conjecture One: A to the nth plus B to the nth (when n is an integer, five or greater) cannot equal equal C to the nth plus q, for some if not most q's.

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I have a few comments about this conjecture.

[img]/images/graemlins/diamond.gif[/img] This is easy to prove for many values of q and n. For example, it is easy to prove that a^6 + b^6 = c^6 + 3 has no solutions, since every 6th power is of the form 7k or 7k+1, so when divided by 7, the left hand side would leave remainder 0, 1, or 2, while the right hand side would leave remainder 3 or 4.

[img]/images/graemlins/diamond.gif[/img] This is a generalization of Fermat's Last Theorem, but not in a direction that looks particulaly promising from the perspective of modern algebraic and analytic number theory. If you are interested in generalizations of Fermat's Last Theorem with more connections to deep parts of mathematics, see the ABC Conjecture:

For every epsilon greater than 0, there exists some k(epsilon)>0 so that for any positive integers A, B, C satisfying gcd(A,B)=1 and A+B=C,

C < k(epsilon) squarefree(A,B,C)^(1+epsilon),

where squarefree(A,B,C) is the product of the prime factors of A, B, and C (with repetition removed).

[img]/images/graemlins/diamond.gif[/img] Rather than conjecture specifics about particular values of q, perhaps it would be more interesting to conjecture that for any q and n, there are at most finitely many triples (a,b,c) satisfying a^n+b^n=c^n+q.

I'm not a number theorist, and it could be that both the specific cases and the general finiteness conjecture are settled.

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Conjecture Two: If there are in fact q's for which the conjecture holds, some will be formally unprovable.

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That doesn't look like a conjecture. That looks like a guess. I'd bet $1k against it, even money.

[David Sklansky]

Conjecture Two: If there are in fact q's for which the conjecture holds, some will be formally unprovable.


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That doesn't look like a conjecture. That looks like a guess. I'd bet $1k against it, even money.


My point is that I think there are qs for which the conjecture holds for no logical "reason" other than "sparseness".

[mybutthurts]

A+b is what n gd conjecture?
im a run of the mill player can someone tell me what this is in very simple idiot english

[Rotating Rabbit]

[ QUOTE ]
Conjecture One: A to the nth plus B to the nth (when n is an integer, five or greater) cannot equal equal C to the nth plus q, for some if not most q's.

Conjecture Two: If there are in fact q's for which the conjecture holds, some will be formally unprovable. In other words it might be true that (A to the n) + (B to the n) can never equal (C to the n) plus (lets just say) the number 846879032 (n greater than four), yet no proof of this fact is even theoretically findable.

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Have you ever had any formal mathematical training because it doesnt look like it; this is complete nonsense.

[drudman]

Is the basic idea: pick an integer q, and if it does satisfy (a^n)+(b^n)=(c^n)+q, there's no way to prove it?.

??

[EliteNinja]

What is the practical purpose of discovering q?
(I'm an engineering student)

Can you build anything useful from this concept?

[PairTheBoard]

[ QUOTE ]
Conjecture Two: If there are in fact q's for which the conjecture holds, some will be formally unprovable.


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That doesn't look like a conjecture. That looks like a guess. I'd bet $1k against it, even money.


My point is that I think there are qs for which the conjecture holds for no logical "reason" other than "sparseness".

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It might be much easier to show that such q's exist than to actually find them.

PairTheBoard

[ToneLoc]

Please fix your title to "Sklansky conjecture".
This is still a bit pompous, but shows a bit more respect to a great, dead mathematician, who might object to people using his name just for adding a twist to the original conjecture.

J.

[wmspringer]

Just thought I'd throw in - if anyone's interested in the general form (Fermat's Last Theorem), Simon Singh's book Fermat's Enigma is an excellent, relatively nonmathematical writeup.