66th Belarusian Mathematical Olympiad

Answer: $q=7$, $p=19$.

Note that $q\le p$, otherwise LHS $<0$, RHS $>0$. We have $$\begin{array}{rl}{p}^3-q^3& =p{q}^3-1\iff p^3+1=p{q}^3+q^3\\ & \iff (p+1)(p^2-p+1)=q^3(p+1)\\ & \iff p(p-1)=q^3-1\\ & \iff p(p-1)=(q-1)(q^2+q+1).\end{array}$$ Since $p(p-1)\nmid p$, we have $(q-1)(q^2+q+1)\nmid p$. But $q-1\nmid p$ since $q-1<p$, so $q^2+q+1\nmid p$, i.e., $q^2+q+1=kp$, $k\in \mathbb{N}$. Therefore $$p(p-1)=(q-1)kp\iff p-1=k(q-1)\iff p=k(q-1)+1.$$ Then $$q^2+q+1=kp\iff q^2+q+1=k^2(q-1)+k.(1)$$ If $k>3$ we have $$\begin{array}{rl}{q}^2+q+1=k^2(q-1)+k& ⟹(q-1)(q+2)+3=k^2(q-1)+k\\ & ⟹(q-1)(q+2)-k^2(q-1)=k-3.\end{array}$$ Hence, $$\begin{array}{c}(k-3)\mid (q-1)⟹k\ge q-1+3=q+2⟹\\ (q-1)(q+2)=k^2(q-1)+k-3\ge (q+2{)}^2(q-1)+q-1>(q-1)(q+2),\\ \text{which is impossible. So, we have no solutions when }k>3.\text{ It remains to}\\ \text{consider three cases: }k=1,2,3.\end{array}$$ If $k=1$ from (1) we have $q^2+q+1=q$. Hence $q^2+1=0$, which is impossible.

If $k=2$ we have $q^2+q+1=4(q-1)+2$, which gives $q^2-3q+3=0$. It is easy to see that this equation has no real solution.

If $k=3$ from (1) we have $$q^2+q+1=9(q-1)+3\iff q^2-8q+7=0.$$ It is easy to see that either $q=1$ or $q=7$. But 1 is not prime, so $q=7$, then $p=3(q-1)+1=19$.