How many prime numbers ?

It's easy to see that $p = 2$ is not a solution. Thus, $p$ must be odd.

With a odd prime, its last digit when it got square is $1;5;9$ .

Or we can say $p^{2}$ 's last degit is $1;5;9$ .

● But except when $p = 5$ , there's no way $p^{2}$ 's last degit is $5$ . Checking for $p = 5$ , we can see it's one of our solutions.

● If $p^{2}$ 's last degit is $1$ , then $4p^{2} +1$ always divisible by 5.

● If $p^{2}$ 's last degit is $9$ , then $6p^{2} +1$ always divisible by 5.

So, $p = 5$ is our only solution

If $p \equiv \pm 1 \pmod{5}$ then $4p^2 + 1 \equiv 0 \pmod{5}$ , so $4p^2+1$ is not prime. If $p \equiv \pm 2 \pmod{5}$ then $6p^2+1 \equiv 0 \pmod{5}$ , so $6p^2+1$ is not prime. Thus the only possible prime is $p=5$ , and that one works.