We have $p|(q-1)(q+1)$ and $q|(p-1)(p+1)$ We have four cases First case. $p|q-1$ and $q|p-1$. Then $q$ is bigger than $p+1$ and smaller than $p-1$: absurd Second case. $p|q+1$ and $q|p+1$. Therefore $q$ is between $p-1$ and $p+1$ and we have to check the 3 cases, which is easy Third case (almost the same as fourth) $p|q-1$ and $q|p+1$. Let $q=pk+1$. Therefore $pk+1|pk+k=k(p+1)$. Thus $pk+1|k-1$, which is obviousy impossible. The fourth case is the same.