2. Given the quadratic trinomial $2 x^{2}-x-36$. Find all integer $x$ for which the values of this trinomial are equal to the square of a prime number.
Answer: $x=5$ or $x=13$.

Solution. We will find all pairs $(x, p)$ satisfying the equation $2 x^{2}-x-36=p^{2}$, where $x$ is an integer and $p$ is a prime. Factorize the quadratic polynomial:

$$
p^{2}=2 x^{2}-x-36=(2 x-9)(x+4) \text {. }
$$

If both brackets on the right-hand side are divisible by $p$, then $p$ will divide $2(x+4)-(2 x-9)=17$, that is, in this case $p=17$. Then either $2 x-9=x+4=17$, or $2 x-9=x+4=-17$. In the first case, we get $x=13$, and the second is impossible. If one of the brackets is not divisible by $p$, then either $x+4= \pm 1$ and $2 x-9= \pm p^{2}$, or $2 x-9= \pm 1$ and $x+4= \pm p^{2}$. The first case is impossible, and in the second case, $x=5$ fits.