Number Fields - Fermat's Last Theorem: n=4

Fermat's Last Theorem: n=4

We prove Fermat's Last Theorem for this case by showing $x^{4} + y^{4} = w^{2}$ has no solutions in the positive integers.

Suppose there is a solution. Then let $x, y, w$ be a solution with the smallest possible $w$ . First note $x^{2}, y^{2}, w$ form a Pythagorean triple. Without loss of generality assume $x$ is odd, so write

x^{2} = m^{2} - n^{2}, y^{2} = 2 m n, z^{2} = m^{2} + n^{2}

for coprime $m, n$ that are not both odd.

Then the first equation implies that $x, n, m$ also form a Pythagorean triple with $x$ odd, so we may write

x = r^{2} - s^{2}, n = 2 r s, m = r^{2} + s^{2}

for coprime integers $r, s$ that are not both odd.

The last of these three equations implies $r, s, m$ are pairwise coprime (otherwise $r, s$ could not be coprime) and from $y^{2} = 4 r s m$ we deduce that $r = a^{2}, s = b^{2}, m = c^{2}$ for some integers $a, b, c$ .

But substituting these in the equation for $m$ implies that $a^{4} + b^{4} = c^{2}$ , contradicting the minimality of $w$ .