SELECTION and TRAINING SESSION

(Solution by V. Vityaz.) First, $q\ne r$ otherwise from $$q(q^2-q-1)=r(2r+3)(1)$$ we would have $r^2-r-1=2r+3$ which gives $r=4$, which is not prime.

Hence (1) implies $q^2-q-1=kr$, so $q^2-q-1=kr$, $2r+3=kq$. Eliminating $r$ we obtain $$2q^2-(2+k^2)q+3k-2=0.(2)$$ The discriminant of this equation is equal to $D=k^4+4k^2-24k+20$ and $D$ must be a perfect square. Note that $(k^2{)}^2<D<(k^2+2{)}^2$ for $k>5$. Hence $$D=(k^2+1{)}^2⟺2k^2-24k+19=0.$$ But the last equation has no integer roots.

If $k=1$, then (2) becomes $2k^2-3q+1=0$, $q=1$ which is not prime. Note that $k$ is odd since $kr=q^2-q-1$ is odd. If $k=3$, then (2) has no integer solutions. If $k=5$, then (2) becomes $2q^2-27q+13=0$, then $q=13$, $r=31$, and $n=r(2r+3)=2015$.