Prime Time

We must first determine the parity of $P$ and $Q$ to solve this problem.

It is known that all primes are odd except $2$ .

For my solution, I used modulo $2$ and established $4$ cases.

It is already given that for any prime $P$ and $Q$ , $2^{P - 1}$ and $4P^2$ are both even and $7$ is an odd number.

Case $1$ : $P$ and $Q$ are odd primes.

$5PQ$ and $Q^2$ will be both odd numbers.

Thus if we substitute these into the original equation,

$even + odd = odd + even + odd$

$odd = even$

which is a contradiction.

Therefore, the solution set for this case is null set.

Case $2$ : $P = 2$ and $Q$ is an odd prime.

$5PQ$ will be even and $Q^2$ will be odd.

$even + even = odd + even + odd$

$even = even$

Therefore, we may have solutions in this case. (I will go back to this case later)

Case $3$ : $P$ is an odd prime and $Q = 2$

$5PQ$ and $Q^2$ are both even.

$even + even = even + even + odd$

$even = odd$

Contradiction. There are no possible solutions.

Case $4$ : $P = Q = 2$

Same with Case $3$ where $5PQ$ and $Q^2$ are both even.

Therefore, there are no possible solutions.

After checking all the cases, only case $2$ satisfies the equation.

Therefore, $P = 2$ .

Solving for $Q$ ,

$2^{P - 1} + 5PQ = Q^2 + 4P^2 + 7$

$2^{2 - 1} + 5(2)Q = Q^2 + 4(2)^2 + 7$

$2 + 10Q = Q^2 + 16 + 7$

$0 = Q^2 - 10Q + 21$

$0 = (Q - 7)(Q - 3)$

$Q = 3$ or $Q = 7$

By checking, we can prove that these two solutions for $Q$ satisfies the equation.

Thus the ordered pairs of solutions for $(P, Q)$ are $(2, 3)$ and $(2, 7)$

$2 + 2 + 3 + 7 = 14$