# How does 1 not congruent imply Fermat n=4?

A natural number is said to be congruent if it is the area of a right triangle with rational sides. I've been told that Fermat actually proved his last theorem for $n=4$ by proving that number 1 is not congruent, but I can't seem to find the connection!

It is probably very easy, thanks in advance.

• – Willie Wong Nov 23 '16 at 3:27

Assume that $n=1$ is congruent. Then there is a right triangle with rational sides $a,b,c$ such that $$(1)\quad a^2+b^2=c^2,$$
$$(2) \quad 2ab=4.$$ Adding $(1)$ and $(2)$ we obtain $(a+b)^2=c^2+4$, substracting them gives $(a-b)^2=c^2-4$. Multiplying the new equations gives $$(a^2-b^2)^2=c^4-2^4,$$ which is a rational solution of $z^2=x^4-y^4$ with $xyz\neq 0$. Then $x^4+y^4=z^4$ has no integer solution with $xyz\neq 0$. The converse implication goes similarly.