Find all integers $n$ for which $n^{2}+20 n+11$ is a perfect square.
Solution 1: We seek $m$ such that $n^{2}+20 n+11=m^{2}$. This equation can be rewritten as $(n+10)^{2}-89=$ $m^{2}$, or equivalently

$$
(n+10-m)(n+10+m)=89
$$

Since 89 is a prime number, we necessarily have $n+10+m=89$ and $n+10-m=1$. From this, we deduce that $n=35$ and $m=44$.

Solution 2: We aim to bound $n^{2}+20 n+11$ between two consecutive squares. We always have $n^{2}+20 n+11 < (n+9)^{2}$ if $n \geqslant 36$. Thus, for $n \leqslant 36$, $n^{2}+20 n+11$ is strictly between two consecutive squares and therefore cannot be a perfect square. We then handle the cases for $n \leqslant 35$ manually, potentially speeding up the work with parity arguments.

Solution 3: We aim to bound $n^{2}+20 n+11$ between two squares, not necessarily consecutive. For any integer $n$, we have:

$$
(n+3)^{2}<n^{2}+20 n+11<(n+10)^{2}
$$

Thus, if $n^{2}+20 n+11$ is a perfect square, it must be the square of $n+4, n+5, n+6, n+7, n+8$, or $n+9$. Each of these cases can be quickly solved by hand.