Problem 999

Solution. Let $a_{n}$ denote the number of distinct odd prime divisors of the number $n(n+3)$. Suppose that there are only finitely many numbers for which $a_{n}$ is divisible by three. Then for some $m$, when $n \geqslant m$, the number $a_{n}$ will not be divisible by three.

Consider the product

$$n(n+1)(n+3)(n+4)=n(n+4) \cdot(n+1)(n+3)=n(n+4) \cdot(n(n+4)+3)$$

Let's see what common prime divisors the numbers $n(n+3)$ and $(n+1)(n+4)$ can have. The numbers $n$ and $n+1$, as well as the numbers $n+3$ and $n+4$, are coprime. The numbers $n+1$ and $n+3$, as well as the numbers $n$ and $n+4$, can have only two as a common prime divisor. Therefore, $a_{n(n+4)}=a_{n}+a_{n+1}$. If $n \geqslant m$, then $a_{n}$, $a_{n+1}$, and $a_{n(n+4)}$ do not divide by three. But this will not be the case if the remainders of the numbers $a_{n}$ and $a_{n+1}$ are different. This means that the remainders of the numbers $a_{n}$ and $a_{n+1}$ for $n \geqslant m$ are the same. Then all remainders of the numbers $a_{n}$ modulo three for $n \geqslant m$ are the same and not equal to zero. But this contradicts the equality $a_{n(n+4)}=a_{n}+a_{n+1}$.