5. All prime numbers $p$ that make $2 p^{4}-p^{2}+36$ a perfect square are $\qquad$ .
5.2.

When $p=2$,
$$
2 p^{4}-p^{2}+36=64
$$

is a perfect square, thus, $p=2$ is the solution.
When $p=3$,
$$
2 p^{4}-p^{2}+36=189
$$

is not a perfect square.
When $p$ is an odd prime greater than 3, let
$2 p^{4}-p^{2}+36=k^{2}$ ( $k$ is an odd positive integer).
Then $p^{2}\left(2 p^{2}-1\right)=(k-6)(k+6)$
$\Rightarrow p \mid(k-6)$ or $p \mid(k+6)$.
However, $p$ cannot divide both $k-6$ and $k+6$. Therefore, one and only one of $k-6$ and $k+6$ is divisible by $p^{2}$.
If $p^{2} \mid(k-6)$, let $k-6=s p^{2}$ ( $s$ is an odd positive integer).
$$
\begin{array}{l}
\text { Then } p^{2}\left(2 p^{2}-1\right)=(k-6)(k+6) \\
=s p^{2}\left(s p^{2}+12\right) \\
\Rightarrow s\left(s p^{2}+12\right)=2 p^{2}-1 \\
\Rightarrow s^{2}<2 \Rightarrow s=1 \Rightarrow p^{2}=13,
\end{array}
$$

which is a contradiction.
If $p^{2} \mid(k+6)$, let $k+6=s p^{2}$ ( $s$ is an odd positive integer).
Similarly, there does not exist a prime $p$ greater than 3.