$5 \cdot 21$ Find all $x \in Z$, such that the polynomial
$$2 x^{2}-x-36$$

is the square of some prime number.
［Solution］Let $2 x^{2}-x-36 \doteq p^{2}, p$ be a prime number, then
$$p^{2}=(x+4)(2 x-9)=a b,$$

where $a=x+4, b=2 x-9, a, b \in \mathbb{Z}, 2 a-b=17$.
Since $a$ is an integer and divides $p^{2}$, there are only the following 6 cases:
(1) $a=p^{2}, b=1$, then $2 p^{2}-1=17$, i.e., $p=3$, thus
$$x=a-4=p^{2}-4=5$$
(2) $a=p, b=p$, then $2 p-p=17$, i.e., $p=17$. Thus
$$x=a-4=p-4=13 .$$
(3) $a=1, b=p^{2}$, then $2-p^{2}=17$, i.e., $p^{2}=-15$, which is impossible.
(4) $a=-p^{2}, b=-1$, then $-2 p^{2}+1=17$, i.e., $p^{2}=-8$, which is impossible.
(5) $a=-p, b=-p$, then $-2 p+p=17$, i.e., $p=-17$, then $x=a-$ $4=-p-4=13$
(6) $a=-1, b=-p^{2}$, then $-2+p^{2}=17$, i.e., $p^{2}=19$, which is impossible.

Therefore, the values sought are $x=5$ and $x=13$.