For which natural numbers $n$ is the expression $P=\left(n^{2}-4\right)\left(n^{2}-1\right)\left(n^{2}+3\right)$ divisible by 2880?
$2880=2^{6} \cdot 3^{2} \cdot 5$, and here the three prime power factors are pairwise relatively prime. Therefore, $P$ is divisible by 2880 if and only if it is separately divisible by $2^{6}$, $3^{2}$, and 5.

In the divisibility test by $2^{6}$, we write $P$ as the product of two expressions:

$$
P=Q \cdot R, \quad \text{where} \quad Q=(n+2)(n-2), \quad R=(n+1)(n-1)\left(n^{2}+3\right)
$$

Focusing on even $n$, both factors of $Q$ are even, and all three factors of $R$ are odd, so $P$ is divisible by $2^{6}$ if and only if $Q$ is. The difference between the factors of $Q$ is 4, so either neither is divisible by 4, i.e., $2^{2}$, or both are, and the divisibility in question can only hold in the latter case. Then one of them is an odd multiple of $2^{2}$, and thus not divisible by $2^{3}$, so $Q$ can only be divisible by $2^{6}$ if the other factor is divisible by $2^{4}$. Therefore, $2^{6} \mid Q$ (read $2^{6}$ is a divisor of $Q$) if $n+2$ and $n-2$ are of the form $16k$, i.e., $n$ itself is

$$
16t-2 \quad \text{or} \quad 16t+2
$$

For odd $n$, $Q$ is odd because both of its factors are, while the factors of $R$ are even. Thus, we need to find which $n$ satisfy $2^{6} \mid R$. The factor $n^{2}+3$ is divisible by 4 but not by 8, because with $n=2k+1$, $n^{2}+3=4k(k+1)+4$, and the first term is divisible by 8 because one of $k$ and $k+1$ is even. Therefore, we seek the condition for $2^{4} \mid (n+1)(n-1)$. The difference between $n+1$ and $n-1$ is 2, and one of them is not divisible by a higher power of 2, so the other must be divisible by $2^{3}$, meaning $n$ must be of the form:

$$
8u-1 \quad \text{or} \quad 8u+1
$$

For the divisibility test by $3^{2}$ and 5, we write $P$ as follows:

$$
P=(n-2)(n-1)n \cdot n(n+1)(n+2)+3(n-2)(n-1)(n+1)(n+2)
$$

The first term is divisible by both $3^{2}$ and 5 because among its factors, there is a triplet of consecutive integers, one of which is divisible by 3, and a quintuplet of consecutive integers. The second term is only divisible by the first power of 3 if $n$ is divisible by 3, and in all other cases, two of the bracketed factors are divisible by 3, so the desired divisibility holds. Thus, the necessary and sufficient condition for $3^{2} \mid P$ is: $n \neq 3v$, and similarly, the condition for $5 \mid P$ is $n \neq 5w$.

According to all of this, $P$ is divisible by 2880 if and only if $n$ is not divisible by 3 or 5, and can be written in one of the following forms: $16t \pm 2, 8u \pm 1$.

Robert Freud (Budapest, Bolyai J. G. III. o. t.)

Note. If the divisibility holds for some $n$, then as can be easily seen, it holds for all numbers of the form $240k \pm n$. Among the first 120 numbers, the following $n$ satisfy the condition:

$$
\begin{array}{rrrrrrrrrrrr}
1, & 2, & 7, & 14, & 17, & 23, & 31, & 34, & 41, & 46, & 47, & 49, \\
62, & 71, & 73, & 79, & 82, & 89, & 94, & 97, & 98, & 103, & 113, & 119
\end{array}
$$