ok so if $p\ne 2$ then $2\mid n$ which says that either $r=2$ or $q=2$ which is impossible. now we have that that $3\nmid q+r$ so $q$ and $r$ are congruent modulo 3 which says that $3\mid n$ because $qr-1\mid n$ therefore of of $q$ and $r$ must be 3. contradiction