Belarusian Mathematical Olympiad

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

We have $$q^3=p^2-p+1\iff (q-1)(q^2+q+1)=p(p-1).(1)$$ If $(q-1)\nmid p$, then $q\ge p+1$, so $q^3>p^2-p+1$. Hence, $(q^2+q+1)\nmid p$, i.e. $$q^2+q+1=kp(2)$$ for some $k\in \mathbb{N}$. It follows that $k(q-1)=p-1$, or $p=kq-k+1$. Substituting this expression in (2) we obtain $q^2+(1-k^2)q+(k^2-k+1)=0$. The discriminant of this quadratic on $q$ equation $$D=(k^2-1{)}^2-(4k^2-k+1)=k^4-6k^2+4k-3$$ must be a perfect square. But it is easy to check that $$(k^2-3{)}^2<k^4-6k^2+4k-3<(k^2-1{)}^2\text{for any }k>3.$$ Moreover, the equation $k^4-6k^2+4k-3=(k^2-2{)}^2$ has no solutions. Hence, $k\le 3$. For $k=1$ we have $D=-4$, and for $k=2$ we have $D=-3$, which is impossible. For $k=3$ we obtain $D=6^2$, so $q=(8\pm 6)/2$. Thus $q=7$ and $p=k(q-1)+1=19$. It remains to note the primes $p=19$, $q=7$ do satisfy the equation.