23.3. (Czechoslovakia, 62). Find all values of $x \in \mathbf{Z}$ for which the polynomial $2 x^{2}-x-36$ takes values equal to the squares of prime numbers.
23.3. Let $2 x^{2}-x-36=p^{2}$, where $p$ is a prime number. Then $p^{2}=$ $=(x+4)(2 x-9)=a b$, where it is denoted $a=x+4, b=2 x-9$, and $a, b \in \mathbb{Z}$ with $2 a-b=17$. Since $a$ is an integer that divides $p^{2}$, there are only the following 6 cases:

1) $a=p^{2}, b=1$, then $2 p^{2}-1=17$ and $p=3$, so $x=a-4=$ $=p^{2}-4=5$
2) $a=p, b=p$, then $2 p-p=17$ and $p=17$, so $x=a-4=$ $=p-4=13$
3) $a=1, b=p^{2}$, then $2-p^{2}=17$ and $p^{2}=-15$, which is impossible;
4) $a=-p_{2}^{2}, b=-1$, then $-2 p^{2}+1=17$ and $p^{2}=-8$, which is impossible
5) $a=-p, b=-p$, then $-2 p+p=17$ and $p=-17$, which is impossible;
6) $a=-1, b=-p^{2}$, then $-2+p^{2}=17$ and $p^{2}=19$, which is impossible.

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