## Task $4 / 89$

Let $p ; q ; p^{2}+q^{2} ; 2 p+q^{2}$ all be prime numbers. Determine $p$ and $q$ as well as the product

$$
q^{2}\left(p^{2}+q^{2}\right)\left(2 p^{2}+q^{2}\right)
$$
If $q=2$, then $2 p^{2}+q^{2} \equiv 0(\bmod 2)$ and because $2 p^{2}+q^{2}>2$ it would not be a prime number. Therefore, $q \geq 3$, and thus $q \equiv q^{2} \equiv 1(\bmod 2)$. If $p \geq 3$, then $p \equiv p^{2} \equiv 1(\bmod 2)$ and $p^{2}+q^{2} \equiv 0(\bmod 2)$, and because $p^{2}+q^{2}>2$ it would not be a prime number. Therefore, $p=2$ must be true. For $q=3$, we get

$$
p^{2}+q^{2}=13 ; \quad 2 p^{2}+q^{2}=17 ; \quad q^{2}\left(p^{2}+q^{2}\right)\left(2 p^{2}+q^{2}\right)=1989
$$

If $q>3$, then $q \equiv \pm 1(\bmod 3), q^{2} \equiv 1(\bmod 3)$, and thus $2 p^{2}+q^{2} \equiv 8+1 \equiv 0(\bmod 3)$, and because $2 p^{2}+q^{2}>3$ it would not be a prime number. Therefore, the found solution is the only one.