BxMO Team Selection Test

We show that the only solution is $(p,q)=(13,31)$. First suppose that $p=q$. Then $p^2-p-1=2q+3=2p+3$, therefore $(p-4)(p+1)=0$. As neither $4$ and $-1$ are prime numbers, there are no solutions $(p,q)$ with $p=q$.

Hence $p\ne q$, so from the equation, it follows that $p\mid 2q+3$ and $q\mid p^2-p-1$. As $2q+3$ and $p^2-p-1$ are positive, it follows that $p\le 2q+3$ and $q\le p^2-p-1$. To find a better lower bound for $p$, we multiply these relations: $$\begin{array}{rl}pq\mid & (2q+3)(p^2-p-1)\\ & 2q{p}^2-2qp-2q+3(p^2-p-1)\\ & 3(p^2-p-1)-2q.\end{array}$$ Note that $3(p^2-p-1)-2q\ge 3q-2q=q>0$. Therefore the relation above yields $$\begin{array}{rl}pq& \le 3(p^2-p-1)-2q\\ & =3p^2-3p-(2q+3)\\ & \le 3p^2-3p-p\\ & =3p^2-4p.\end{array}$$ Adding $4p$ to both sides, then dividing both sides by $p$, we find that $q+4\le 3p$, so $$\frac{2q+3}{6}<\frac{q+4}{3}\le p.$$ As $p$ is a divisor of $2q+3$, we deduce that $2q+3=kp$ with $k\in \{1,2,3,4,5\}$.

If $k=1$, then $2q+3=p$ and therefore also $q=p^2-p-1$. This implies that $p=2q+3=2(p^2-p-1)+3=2p^2-2p+1$. This in turn implies that $(2p-1)(p-1)=0$, but this equation does not have prime solutions. Note that $k=2$ and $k=4$ are not possible either, since then $kp$ would be even, while $2q+3$ is odd. * If $k=3$, then it follows from $2q+3=3p$ that $3\mid q$, and therefore that $q=3$. Hence $p=\frac{2q+3}{3}=3$. However, this does not give a solution of the given equation.

Therefore $k=5$. Then we have $5p=2q+3$, and therefore $5q=p^2-p-1$ as well. Substituting this gives $25p=5(2q+3)=2(p^2-p-1)+15=2p^2-2p+13$, and therefore $(p-13)(2p-1)=0$. As $p$ is prime, it follows that $p=13$, and that $q=\frac{5p-3}{2}=31$. Note that $(p,q)=(13,31)$ is indeed a solution of the given equation, so it is the only solution of the given equation. $\square$