63rd Czech and Slovak Mathematical Olympiad

We look for $n=p\cdot q\cdot r$, with primes $p\le q\le r$ satisfying $$(p+1)(q+1)(r+1)=pqr+963.(1)$$ If $p=2$, the right-hand side of (1) is odd, hence the factors $q+1,r+1$ on the left must be odd too. This implies that $p=q=r=2$, which contradicts to (1). Thus we have proved that $p\ge 3$. Now we will show that $p=3$. Suppose on the contrary that $3<p\le q\le r$. Then the right-hand side of (1) is not divisible by $3$. The same must be true for the product $(p+1)(q+1)(r+1)$. Consequently, all the primes $p,q,r$ are congruent to $1$ modulo $3$, and hence $(p+1)(q+1)(r+1)-pqr$ is congruent to $2\cdot 2\cdot 2-1\cdot 1\cdot 1=7$, which contradicts to $(p+1)(q+1)(r+1)-pqr=963$. Therefore, the equality $p=3$ is established. Putting $p=3$ into (1) we get $4(q+1)(r+1)=3qr+963$, which can be rewritten as $(q+4)(r+4)=975$. In view of the prime factorization $975=3\cdot 5^2\cdot 13$ and inequalities $7\le q+4\le r+4$, we conclude that $q+4\le \sqrt{975}<32$ and hence $q+4\in \{13,15,25\}$. Since $q$ is a prime, it holds that $q=11$. Then $r+4=65$, and hence $r=61$ (which is a prime indeed). Consequently, the problem has a unique solution $$n=3\cdot 11\cdot 61=2013.$$