Find all pairs of primes $(P,Q)$ satisfying $$p^{3}-q^{5}=(p+q)^{2}$$
What I have been able to show until now is that
$$q+1<p<q^3+1$$ but I don't see how it's of any use here. Can anyone please suggest something? Thank you.
Apart from this, I can also prove that $p$ and $q$ both are odd. And they must leave the same remainder upon division by $4$.
P.S. i'm not looking for a complete solution…i just need a hint.
Best Answer
Note that
$$p^3 \equiv p \pmod{3}, \; q^5 \equiv q \pmod{3} \tag{1}\label{eq1A}$$
Consider neither $p$ or $q$ being $3$. If $p \equiv q \pmod{3}$, then the left side is a multiple of $3$ but the right side is not. Instead, if $p \not\equiv q \pmod{3}$, then $p + q \equiv 0 \pmod{3}$, so the left side is not divisible by $3$ but the right side is. This means that either $p$ or $q$ is $3$.
Since the question asks for a hint only, the rest of the solution is behind the spoiler below.