Find all integers $n$ such that $n(n+1)$ is a perfect square.
If $n$ is strictly positive, we see from the prime factorization (for example) that $n$ and $n+1$ being coprime, they are both squares. There exist integers $u$ and $v$ such that $n=u^{2}$ and $n+1=v^{2}$, so $u^{2}+1=v^{2}$, hence $(v-u)(u+v)=1$, so $v-u=u+v$, so $v=0$ and thus $u=1$, from which $n=1$, and $1 \times 2$ is not a perfect square. Finally, $n=0$, which works. If $n$ is negative, we reduce the problem by considering $n-1$ instead of $n$, and we find that only -1 works, and -1 works well.