Romanian Mathematical Olympiad

Let the three odd consecutive numbers be $2p + 1$, $2p + 3$ and $2p + 5$, where $p$ is a positive integer. Then one of these numbers is divisible by $3$:

If $p = 3k$, with integer $k$, then $2p + 3 = 2(3k) + 3 = 6k + 3 = 3(2k + 1)$;

If $p = 3k + 1$, with integer $k$, then $2p + 1 = 2(3k + 1) + 1 = 6k + 3 = 3(2k + 1)$;

If $p = 3k + 2$, with integer $k$, then $2p + 5 = 2(3k + 2) + 5 = 6k + 9 = 3(2k + 3)$.

In all the cases the product $P$ is a multiple of $3$, therefore $P = 3a = (a-1) + a + (a+1)$, with integer $a$.