Primes, Cubes And Squares

Case 1 - Both p and q are Odd

$Odd^3 + Odd^3 +1 = Odd^2 \times Odd^2$

$Odd + 1 = Odd$

$Even \neq Odd$ )

Case 2 - Both p and q are Even

$Even^3 + Even^3 + 1 = Even^2 \times Even^2$

$Even + 1 = Even$

$Odd \neq Even$

Case 3 - P is odd and q is even

As the only even prime is 2 , Putting p = 2 we get

$9 + p^3 = 4 \times p^2$

$p^3 - 4p^2 = -9$

$p^2(p-4) = -9$

As $p^2$ is always positive that means $p-4$ term will be negative . So ,

$p-4 < 0$

$p < 4$

The only odd prime less than 4 is "3"

So $p = 3$

$p + q = \boxed{5}$

Relevant wiki: Diophantine Equations - Solve by Factoring

WLOG, $q \leq p$ . We may rewrite $p^3 + q^3 + 1 = p^2q^2$ as $q^3 + 1 = p^2q^2 - p^3$ . We now factor both sides to obtain $(q+1)(q^2-q+1) = p^2(q^2-p)$ . Hence $p^2$ divides $(q+1)(q^2-q+1)$ . Since $q \leq p$ , $0 ˂ q^2-q+1 ˂ p^2$ . Therefore, $p=q+1$ so that the only solution is $p=3$ and $q=2$ . The maximum value of $p+q$ is $5$ .

We have $p^{3} + q^{3} + 1 = p^{2}q^{2} \Rightarrow q^{3} + 1 = p^{2}q^{2} - p^{3} \Rightarrow q^{3} + 1 = p^{2}(q^{2} - p) \Rightarrow \frac{q^{3} + 1}{q^{2} - p} = p^{2}$

Since $p$ is a prime, this means $\frac{q^{3} + 1}{q^{2} - p}$ is an integer, so $q^{2} - p$ divides $q^{3} + 1$ .

We know that $(q^{2} - p) \times q = q^{3} - pq$ .

As $q^{3} + 1 = q^{3} - pq + (pq + 1)$ , we know that $q^{2} - p | (q^{2} - p)q + (pq + 1) \Rightarrow q^{2} - p | pq + 1 ----------- (1)$

Since the given equation is symmetric, we may assume $q \geq p$ . We will now consider 3 cases :

(i) $q \geq p + 2$

In this case, we have $\frac{pq+1}{q^{2} - p} \leq \frac{q(q-2) + 1}{q^{2} - q} = \frac{q(q-2) + 1}{q(q-1)} \leq 1$ since $q$ is a prime and thus larger than 1. So $\frac{pq+1}{q^{2} - p}$ cannot be an integer, a contradiction to (1). Thus there is no solution in this case.

(ii) $q = p + 1$

Since one of $p, q$ must be even, $p$ must be 2 since 2 is the only even prime and $p \leq q$ . Thus $(p, q) = (2, 3)$ which satisfies the given equation, so this is one possible solution.

(iii) $q = p$

We immediately have $2q^{3} + 1 = q^{4} \Rightarrow 1 = q^{4} - 2q^{3}$ which is impossible since the RHS is divisible by $q$ , but the LHS is not.

Thus, there exists only one possible solution (down to permutations) which is $(p, q) = (2, 3)$ . So we know that the maximum value of $p + q$ is $\boxed{5}$ .