Thursday, May 04, 2006

My quadratic residue result may offer the route to a tiny proof of Fermat's Last Theorem.

The result is that given naturals n_1, p a prime factor of n_1, n_2, C = n_1 + n_2, and k a difference of factors of 2*C, it must be true that

(8*n_2 + k^2) is a square modulo p

so with x^n + y^n = z^n, you can let C = z^n, and k is a difference of factors and there is a set of all such possibles.

The thing is, n_1 and n_2 can be ANY naturals that add to give z^n.

It's trivial to show then that

(8*z^n + k^2) would have to be a square modulo p

for all p, primes less than z^n.

A simple check of combinations using f_1 f_2 = z, with n=3, gives an indication why that's impossible, as I think it actually gets blocked by 5, or if not 5, 7, which has only three quadratic residues.

(f_1^3 - f_2^3), (f_1^2 f_2 - f_1 f_2^2), (f_1^3 f_2^3 - 1)

are just some combinations I can think of immediately with n=3, and if FLT were false, for like p=7, it'd have to be true that all of those squared would add to 8*z^2 mod 7 to give a quadratic residue of 7, i.e. give 0, 1, 2 or 4.

This quadratic residue result of mine may be one of the most powerful in number theory, offering routes to proving Goldbach's conjecture, and probably all kinds of Diophantine equations.

A tiny proof of FLT that is just about as trivially easy as you can get. Kind of strange.

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