SELECTION EXAMINATION

The given equation can be written as $$\begin{gathered} p^4+p^3+p^2+p=q^2+q \\ \iff p(p+1)(p^2+1)=q(q+1)(1) \end{gathered}$$ $$\iff p(p^2-1)(p^2+1)=q(q+1)(p-1)(2)$$ For $q\le p$ is not possible. Hence we should have $q>p$. Therefore from (2) we conclude that: $$q|(p^2-1)(p^2+1).(3)$$ Next we distinguish the cases:: If $q\le p^2$, then $q^2\le p^4$ and $q^2+q\le p^4+p^2<p^4+p^2+p^3+p$. Hence the equation has no solutions.. If $q>p^2$, then $q\ge p^2+1$, and hence (3), since $q$ is prime, gives $q=p^2+1$. Thus from equation (1) we obtain $$p(p+1)=q+1\iff p^2+p=p^2+2\iff p=2$$ Therefore, since $q=p^2+1=5$, we have the solution $(p,q)=(2,5)$.