FLT in a more general form & yet to be proved?

Question

Suppose
x^(n) = a^(m) - b^(m)
Here "n" is any number > 2 & "m" is any number greater then 2.
"n" can be equal to "m" then it is FLT.
when n=/=m
then also there is no natural solution for "x" , "a" & "b" to the above equation.
Can this marvellous truth be proved?

Answer 1

This would be a special case of the Fermat-Catalan Conjecture.

This conjecture is that there are only a finite number of solutions of x^p + y^q = z^r for 1/p + 1/q + 1/r < 1. Ten solutions are known, but all of them have the three exponents different (except for 1^p + 2^3 = 3^2) and one exponent equal to 2.

Your conjecture would be the special case that there are no solutions with p > 2 and q = r. It is already known that there are no solutions with p = q and r > 2 (Darmon and Merel, 1997).

Answer 2

I guess no ...

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