Prime solutions to $p^2(p^3-1) = q(q+1)$

Question

Solve: $$p^2(p^3-1)=q(q+1)$$ where $p$ and $q$ are prime numbers.

Could someone help me with this question? I tried several things but I couldn't get it. I can show what I already did about that question, but I didn't go too far. I don't have the answers.

Answer 1

$p=q$ and $p^{2} = q$ are impossible, so we must have $p^{2}\mid q + 1$. Let $q + 1 = kp^{2}$ for some $k > 0$. Note that $q(q+1)=kp^{2}(kp^{2}-1)=k^{2}p^{4}-kp^{2}\equiv -kp^{2}\pmod{p^{4}}$. However, $p^{2}(p^{3}-1)\equiv -p^{2}\pmod{p^{4}}$, so $k\equiv 1\pmod{p^{2}}$.

$k=1$ has no solutions, so $k\geq p^{2} + 1$. Then, $q = kp^{2} - 1\geq p^{4}+p^{2} - 1$. However, because $q\neq p, p^{2}$ we must have $q\mid p^{3}-1$, so $q\leq p^{3}-1$, contradiction because $p\geq 1$. Thus, there are no solutions.

Answer 2

Suppose that $p$ and $q$ are prime numbers such that $$q(q+1)=p^2(p^3-1).$$ Clearly $q>p$ so $q$ divides $p^3-1$, but does not divide $p-1$. This implies $q\equiv1\pmod{3}$ and so $$p^2(p^3-1)\equiv q(q+1)\equiv2\pmod{3},$$ which is impossible. Hence no such primes $p$ and $q$ exist.

Answer 3

We are given that $p$ and $q$ are prime, with $$p^2(p^3-1)=q(q+1).\qquad(1)$$ First note that $2^2(2^3-1)=28$ cannot be factorized in the form $q(q+1)$. Hence $p\geqslant3$. Also observe that $q\neq p\,$; otherwise eqn $1$ becomes $p^2(p^3-1)=p(p+1)$ and so $p(p^3-2)=1\,$: a falsehood since $p>2$. It follows that $p^2$ and $q$ are coprime. Therefore $q$ divides $p^3-1=(p-1)(p^2+p+1)$. Now $q$ cannot divide $p-1$, because then $q\leqslant p-1$, and this would give $p^2(p^3-1)\leqslant (p-1)p$ and so $p(p^2+p+1)\leqslant 1$: an impossibility. Consequently $q$ divides $p^2+p+1$. It follows that $$q\leqslant p^2+p+1.$$ From this inequality and eqn $1$, dividing out the factor $p^2+p+1$ gives $p^2(p-1)\leqslant p^2+p+2$, which may be written $$(p-3)^3+7(p-3)^2+14(p-3)+4\leqslant0.$$ But this must be false, since $p\geqslant3$. Hence the given equation has no solution.