Determine all integers $n$ for which $n^{2}+15$ is the square of an integer.

Remark. Because the problem asks you to "determine all integers $n$ ", you must verify that all the $n$ you find have the desired property, and moreover prove that these are the only such integers $n$.
We may limit the search to nonnegative integers $n$, since $n^{2}+15=(-n)^{2}+15$. Suppose there is a nonnegative integer $m$ such that

$$
\begin{aligned}
n^{2}+15 & =m^{2} \\
15 & =m^{2}-n^{2} \\
& =(m+n)(m-n) .
\end{aligned}
$$

Note that the factors $m+n$ and $m-n$ are not both negative, since their sum $2 m$ is nonnegative, so they are both positive. Now the only ways to factor 15 into two positive integer factors are $15 \cdot 1$ and $5 \cdot 3$. Also $m+n \geq m-n$ since $n \geq 0$. If $m+n=15$ and $m-n=1$, then

$$
n=\frac{(m+n)-(m-n)}{2}=\frac{15-1}{2}=7
$$

and $m=8$. If $m+n=5$ and $m-n=3$, then

$$
n=\frac{(m+n)-(m-n)}{2}=\frac{5-3}{2}=1
$$

and $m=4$. Both of these solutions indeed work in the original problem. Finally, allowing for negative $n$, we compute the answers to be $-7,-1,1$, and 7 .