Find all integers $n$ such that $n^{2}+8 n+44$ is a perfect square.
We proceed by bounding between two squares. Suppose already $n \geq 0$, since we know how to handle it.

We have $n^{2}+8 n+44>n^{2}+8 n+16=(n+4)^{2}$. Moreover, we have $n^{2}+8 n+44<n^{2}+10 n+25 \Longleftrightarrow$ $19<2 n \Longleftrightarrow n \leq 9$. By verifying each value for $0 \leq n \leq 9$, we find that $n=2$ is the only solution, for which we get $n^{2}+8 n+44=64$. For the negatives, there are several ways to proceed. Or we can bound in the same way, but this time by $(n+3)^{2}$, which involves a lot of verifications behind.

Or we can say that since $n^{2}+8 n+44=(n+4)^{2}+28$, if the term in the square takes a negative sign, we get the same result. In other words, $f(n)=f(-8-n)$. Thus, any value that is a square for $n \geq-4$ corresponds to a value for $n \leq-4$, and vice versa. We obtain $n=-10$ as a solution, and we verify that for $-4 \leq n \leq 0$, there are no solutions.

A final method is to write $n^{2}+8 n+44=(n+4)^{2}+28=m^{2}$, factorize and solve.

In all cases, we find $n=2,-10$ as the only solutions.