
1. A number $n$ is a product of three (not necessarily distinct) prime numbers. Adding 1 to each of them, after multiplication we get a larger product $n+963$. Determine the original product $n$.

(Pavel Novotný)


Solution. We look for $n=p \cdot q \cdot r$, with primes $p \leqslant q \leqslant r$ satisfying

$$
(p+1)(q+1)(r+1)=p q r+963 .
$$

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 $p \geqslant 3$.

Now we will show that $p=3$. Suppose on the contrary that $3<p \leqslant q \leqslant r$. 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 $(p+1)(q+1)(r+1)-p q r$ is congruent to $2 \cdot 2 \cdot 2-1 \cdot 1 \cdot 1=7$, which contradicts to $(p+1)(q+1)(r+1)-p q r=963$. Therefore, the equality $p=3$ is established.

Putting $p=3$ into (1) we get $4(q+1)(r+1)=3 q r+963$, 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 $7 \leqslant q+4 \leqslant r+4$, we conclude that $q+4 \leqslant \sqrt{975}<32$ 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

$$
n=3 \cdot 11 \cdot 61=2013 .
$$
