## Task A-2.3.

Determine all prime numbers $p$ for which $2^{p}+p^{2}$ is also a prime number.
## Solution.

For $p=2, 2^{p}+p^{2}=8$.

(1 point)

(1 point)

For every odd prime number $p$, the number $2^{p}$ gives a remainder of 2 when divided by 3.

(3 points)

$\left(2^{p} \equiv(3-1)^{p} \equiv(-1)^{p} \equiv-1 \equiv 2(\bmod 3)\right)$

For every prime $p>3$, its square $p^{2}$ gives a remainder of 1 when divided by 3.

(3 points) $\left(p \equiv \pm 1(\bmod 3) \Rightarrow p^{2} \equiv 1(\bmod 3)\right)$

Therefore, for every prime $p>3$, the number $2^{p}+p^{2}$ is divisible by 3

$\left(2^{p}+p^{2} \equiv 2+1 \equiv 3 \equiv 0(\bmod 3)\right)$

and thus cannot be prime.

(1 point)

Therefore, the only prime number that satisfies the condition of the problem is $p=3$.