5. Find all integers $n$ for which the equality $\frac{n^{2}+3 n+5}{n+2}=1+\sqrt{6-2 n}$ holds.
5. We have: $\frac{n^{2}+3 n+5}{n+2}-1=\sqrt{6-2 n} ; \frac{n^{2}+2 n+3}{n+2}=\sqrt{6-2 n} ; n+\frac{3}{n+2}=\sqrt{6-2 n}$. Since for integer $n$, the right-hand side can be either an integer or irrational, while the left-hand side is either an integer or rational, the given equality is possible only if $\frac{3}{n+2}$ is an integer. Then $n+2=\{-1 ; 1 ;-3 ; 3\} \rightarrow n=\{-3 ;-1 ;-5 ; 1\}$. Then we find that $6-2 n$ will be the square of an integer only for $n=-5$ and $n=1$. Upon verification, we find that only $n=1$ is a solution to the original equation. Answer $\{1\}$