Solve equation in prime numbers

Question

Solve the equation in prime number $$p^3+q^3+1=p^2 q^2.$$

I have found the solutions $(2,3), (3,2)$ and need to prove that there are no other solutions. I think that there is an inequality $p^3+q^3+1\leq p^2 q^2$ for $p,q >2 $ but how to prove it.

Answer 1

If $p=q$ we get $2p^3+1=p^4$ so $p\mid 1$ which is impossible. So we can assume $p\ne q$.

Because of simmetry we can also assume that $p>q$.

Now we have $$p^2\mid q^3+1 = (q+1)(q^2-q+1)$$

Case 1: $p\not{\mid}\;q+1$, then $$p^2\mid q^2-q+1\implies p^2\leq q^2-q+1<q^2$$ so $p<q$, a contradiction.

Case 2: $p\mid q+1$ then $p\leq q+1$ so $p=q+1$ (remember that $p>q$, so $p\geq q+1$). So $p$ and $q$ are consecutive primes so one is even, so $q=2$ and $p=3$.

Answer 2

My solution

Case 1. Let $p \mod 3=q \mod 3=1$. Then $p^3+q^3+1=3=0 \mod 3 \neq p^2 q^2=1 \mod 3$

Case 2 $p \mod 3=q \mod 3=2$. Then $p^3+q^3+1=2+2+1=2 \mod 3 \neq p^2 q^2=1 \mod 3$

Сase 3. $p \mod 3=1, q \mod 3=2$. Then $p^3+q^3+1=1+2+1=1 \mod 3 \neq p^2 q^2=1 \cdot 2 =2 \mod 3$

So, must be, say $p=0 \mod 3 \implies p=3$. We get the equation $q^3+28=9 q^2 $ or $q^3-9 q^2+28=\left( q-2 \right) \left( {q}^{2}-7\,q-14 \right) $ and $q=2.$ By symmetry another solution is $p=2, q=3.$