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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.