(5) Let $n \in \mathbf{N}^{*}$, and $n^{4}+2 n^{3}+5 n^{2}+12 n+5$ is a perfect square, then $n=$
(5) 1 or 2 Hint: Since
$$
\left(n^{2}+n+2\right)^{2}<n^{4}+2 n^{3}+5 n^{2}+12 n+5<\left(n^{2}+n+4\right)^{2},
$$

so $n^{4}+2 n^{3}+5 n^{2}+12 n+5=\left(n^{2}+n+3\right)^{2}$, solving it yields $n=1,2$.