1. Find all positive integers $n$, such that $n^{2}+32 n+8$ is a perfect square.
1. Let $n^{2}+32 n+8=m^{3}(m>0)$. Then
$$
\begin{array}{l}
(n+m+16)(n-m+16) \\
=248=2^{3} \times 31 .
\end{array}
$$

Notice that, $n+m+16$ and $n-m+16$ have the same parity. Therefore, $n+m+16=124$ or 62. Accordingly, $n-m+16=2$ or 4.
Solving these, we get $n=47$ or 17.