## Task 31/80

Given the quadratic equation in $x$, $x^{2}+p x+2=0$ with $p \in N$, and let $x_{1}$ and $x_{2}$ be the solutions of this equation. All $p$ are to be determined for which $x_{1}^{2}+x_{2}^{2}$ is a prime number.
According to Vieta's root theorem, for the solutions $x_{1}$; $x_{2}$ of the quadratic equation $x^{2}+p x+2=0$: $x_{1}+x_{2}=-p$ and $x_{1} x_{2}=2$. Now,

$$
x_{1}^{2}+x_{2}^{2}=\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}=(-p)^{2}-4=p^{2}-4=(p-2)(p+2)
$$

This expression can only be a prime number if the smaller of the two factors is equal to 1, which is when $p=3$. Indeed, then

$$
x_{1}^{2}+x_{2}^{2}=(3-2)(3+2)=5
$$

a prime number (the solutions $x_{1 ; 2}$ are then

$$
x_{1 ; 2}=-\frac{3}{2} \pm \sqrt{\frac{9}{4}-2} \rightarrow x_{1}=-1 ; x_{2}=-2
$$