For what positive integers $p$ and $q$: $(p+1)!+(q+1)!=(pq)^2$

Question

I tried this problem using brute force and got the answers as $(3,4)$ and $(4,3)$,but is there a way to solve this question?

Answer 1

WLOG, Assume $p \ge q$. Then, $p^4 \ge (pq)^2 = (p+1)!+(q+1)! \ge (p+1)!$.

We can show by induction that $n^4 < (n+1)!$ for all $n \ge 5$.

Hence, we must have $q \le p \le 4$.

Now, just test all possible pairs with $1 \le p,q \le 4$.