## Task 32/82

We are looking for all prime numbers $p$ for which $z=2^{p}+p^{2}$ is also a prime number.
For $p=2$, $z>2$ is obviously even, thus not a prime number.

For $p=3$, we have $z=2^{3}+3^{2}=17$. Thus, one of the desired numbers is found.

For $p>3$, it always holds that $p \equiv \pm 1(\bmod 3)$, hence $p^{2} \equiv 1(\bmod 3)$. Furthermore, for $p>3$, $p$ is odd, thus can be represented in the form $p=2 k+1, k \in N, k>1$. Then it holds that

$$
2^{p}=2^{2 k+1}=2 \cdot 2^{2 k}=2 \cdot 4^{k} \equiv 2(\bmod 3)
$$

due to $4 \equiv 1(\bmod 3), 4^{k} \equiv 1^{k} \equiv 1(\bmod 3)$. Therefore, for $p>3$,

$$
z \equiv 2+1 \equiv 0(\bmod 3)
$$

thus not a prime number. Consequently, $p=3$ is the only prime number with the required property.