Strange primes

Note first that for $p = 5$ we have both $4p^{2} + 1 = 101$ and $6p^{2} + 1 = 151$ being prime, so $p = 5$ is one such value.

Primes other than $5$ are equivalent to one of $1,2,3,4$ modulo $5$ . Looking at these case by case:

• $p \equiv 1 \pmod{5} \Longrightarrow 4p^{2} + 1 \equiv 0 \pmod{5}$ , so $4p^{2} + 1$ is not prime;

• $p \equiv 2 \pmod{5} \Longrightarrow 6p^{2} + 1 \equiv 0 \pmod{5}$ , so $6p^{2} + 1$ is not prime;

• $p \equiv 3 \pmod{5} \Longrightarrow 6p^{2} + 1 \equiv 0 \pmod{5}$ , so $6p^{2} + 1$ is not prime;

• $p \equiv 4 \pmod{5} \Longrightarrow 4p^{2} + 1 \equiv 0 \pmod{5}$ , so $4p^{2} + 1$ is not prime.

So only if $p \equiv 0 \pmod{5}$ will it be even possible for both $4p^{2} + 1$ and $6p^{2} + 1$ to be prime, so for both these expressions and $p$ itself to be prime, $p = 5$ is the only possibility. As we have shown already that $p = 5$ satisfies the conditions, the sum of all possible values for $p$ is $\boxed{5}$ .