Prime Time 2: Random 96

We have two possible cases for $p$ .

Case 1: $p = 5$

$625 = p^4 = 96q^2 + 1$ $q^2 = 6.5$

There is no solution for this case since $q$ must be a prime.

Case 2: $p \neq 5$

Working in modulo 5,

$p^4$ will always be $1 ($ mod $5)$ .

$p^4 = 96q^2 + 1$ $1 = 96q^2 + 1$ $0 = 96q^2$

This implies that $q$ should be divisible by $5$ . Since $q$ must be prime, $q = 5$ and $p = 7$ .

$(p, q) = (5, 7)$ is the only solution for the given solution. Thus the required answer should be $\boxed{35}$ .

We see that $p$ is odd and $q$ cannot be $2$ or $3$
So, letting $p=2m+1$ for some integer $m$ , we see that the equation turns into

$\displaystyle \left(m^2+m\right)\left(2m^2+2m+1\right)=12q^2$

Putting $m^2+m=n$ we see that the equation changes to

$\displaystyle n\left(2n+1\right)=2^2\cdot3\cdot q^2$

Since $q$ is prime, the RHS has only 18 factors. Listing them out and checking, we easily find that the only solution exists for $n=12$ and hence, $q=5$ giving the only ordered pair satisfying the equation being $\left(p,q\right)=\left(7,5\right)$