Fermat: Prove $a^4-b^4=c^2$ impossible

Question

Prove by infinite descent that there do not exist integers $a,b,c$ pairwise coprime such that $a^4-b^4=c^2$.

Answer 1

Lemma. If $a,b,c$ are coprime and $a^2+b^2=c^2$, then $$ \{a,b\}=\{2pq,p^2-q^2\},\quad c=p^2+q^2 $$ with $p,q$ coprime and not both odd.

This follows from the fact that we are looking for rational points on the curve $x^2+y^2=1$, that allows the parametrization $\left(\frac{2t}{1-t^2},\frac{1-t^2}{1+t^2}\right)$. Assuming that $P=(x_0,y_0)$ is a rational point, we consider the lines through $P$ with a rational slope. They all must intersect the circle in a second point with rational coordinates, due to Viète's theorem.

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The lemma hence gives two possibilities:

1. $a^2=p^2+q^2,\quad c=2pq,\quad b^2=p^2-q^2.$

2. $a^2=p^2+q^2,\quad c=p^2-q^2,\quad b^2=2pq.$

The infinite descent starts as soon as we notice that in 1. $p$ must be the sum of two squares, since $p^2=b^2+q^2$, while in 2. $p+q$ must be the sum of two squares since $a^2+b^2=(p+q)^2$.