Fermat’s Last Theorem, proved in 1995, is that the diophantine equation

has no solution in positive integers x, y, z when n > 2. Knowing the solution for n = 2 (see previous post) we can show that there is no solution for n = 4. If

…………………………… (1)

has a solution x_{1}, y_{1}, z_{1} then

……………………………. (2)

has a solution x_{1}, y_{1}, z_{1}^{2} and a solution x_{1}/d, y_{1}/d, z_{1}^{2}/d^{2} where d = (x_{1}, y_{1}), the greatest common divisor of x_{1} and y_{1}. This implies that (2) has a solution where (x, y) = 1. If we can show there is no such solution then by the contrapositive there is no solution to (1).

Suppose to the contrary there is such a solution x_{1}, y_{1}, z_{1}. It is easy to see that these numbers have greatest common divisor 1. We use the method of infinite descent by showing that there exist x_{2}, y_{2}, z_{2} such that z_{2} < z_{1}.

By the properties of Pythagorean Numbers (previous post) we know that

; ;

with a > b> 0, (a,b) = 1, and a, b of opposite parity.

x_{1}^{2} + b^{2} = a^{2} and it is not hard to show that that x_{1}, b and a have the remaining properties of Pythagorean numbers. Thus

; ; and

with (p,q) = 1, p > q > 0, and p,q of opposite parity. Consequently

Furthermore p, q and p ^{2} + q^{2} have greatest common divisor one, so by the Unique Factorisation Theorem each is a square and we can write:

; ; and and thus

with x_{2}, y_{2}, z_{2} > 0 and z_{1} = a^{2}+b^{2} = (p^{2}+q^{2})^{2}+b^{2}=z_{2}^{4}+b^{2} > z_{2}

To prove Fermat’s Last Theorem it suffices to prove it for every prime index. If n > 2 is composite and has no prime factor other than 2 then it is a multiple of 4 and the Theorem is true for this index because if

then

which is impossible. If n is composite and has an odd prime factor p and if

then

thus by the contrapositive if the Theorem is true for p it is true for n.

A copy of this post in pdf form is here.

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