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

\[x^2 = m^2 - n^2, y^2 = 2m n, w = m^2 + n^2\]

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

\[x = r^2 - s^2, n = 2r s, m = r^2 + s^2\]

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 \(w\).

Ben Lynn blynn@cs.stanford.edu 💡