## Monday, March 8, 2010

### Simple Method for Fermat's Last Theorem and Beyond

Fermat's Last Theorem states that no three positive integers a, b, and c can satisfy the Diophantine equation an + bn = cn for any integer value of n greater than two.

All variables denote positive integers.

Where p is a prime number, n = 2, 1 < a < b < p,
(an + bn) can be evenly divided by pn-1.
Examples:
(22 + 42) / 51 = 4
(32 + 42) / 51 = 5
(22 + 102) / 131 = 8

Therefore, since (a2 + b2) can be evenly divided by p1, a2 + b2 = c2 has solutions.

Where p is a prime number, n > 2, 1 < a < b < p,
(an + bn) is not evenly divisible by pn-1. Therefore, it is reasonable to assume an + bn = cn cannot be satisfied when n > 2.

Adding terms to the left of the equation:

Let the variables in equation an + bn = cn be renamed to a1n + a2n = bn.
Adding two additional terms to the left of the equation would be:
a2n + a2n + a3n + a4n = bn.

Conjecture:

Where p is prime, 1 < a1 < a2 +...+ < am < p, if there exists
{n, a1, a2, ..., am, p, x} that can satisfy (a1n + a2n +...+ amn) / pn-1 = x, then there exists, with the same number of terms (a) and exponent (n), {n, a1, a2, ... , am, b} that can satisfy a1n + a2n +...+ amn = bn.

An example of a {n, a1, a2, a3, a4, p, x} that satisfies
(a15 + a25 + a35 + a45) / p4 = x, is {5, 31, 40, 43, 51, 61, 45}.

(315 + 405 + 435 + 515) / 614 = 45

Therefore, there exists {n, a1, a2, a3, a4, b} that can satisfy
a15 + a25 + a35 + a45 = b5.