Exercise 7.8.6 ($n^2 + (n+1)^2 = m^2$ has infinitely many solutions)

Proof. For all pairs of integers $(n,m)$,

Let $(x,y)$ be any positive solution of $x^2-2y^2=-1$. Then $x$ is odd, so there is some integer $n\ge 0$ such that $x=2n+1$. Put $m=1$, then ${(2n+1)}^2-2m^2=-1$, so $n^2+{(n+1)}^2=m^2$.

Since $(1,1)$ is a solution of $x^2-2y^2=-1$, $x^2-2y^2=-1$ is solvable. By Problem 1, there are infinitely many solutions of $x^2-2y^2=-1$. For each such solution, $(n,m)=((x-1)∕2,y)$ is a solution of $n^2+{(n+1)}^2=m^2$.

Therefore $n^2+{(n+1)}^2$ is a perfect square for infinitely many values of $n$. □