Problem 1054

The expression to be examined

$$n^{2}\left(n^{4}-1\right)=\left(n^{2}-1\right) n^{2}\left(n^{2}+1\right)=(n-1) n(n+1) n\left(n^{2}+1\right)$$

If $n$ is an integer, then the first three factors are three consecutive integers. Among these, one is divisible by 3.

If $n$ is odd, then the first and third factors, and if $n$ is even, then the second and fourth factors are divisible by 2, making the product divisible by 4.

Among the first three factors, one is also divisible by 5, either if $n$ is divisible by 5, or if it is adjacent to a number divisible by 5. If, however, $n$ is the second neighbor of a number divisible by 5: $n=5 k+2$, then the last factor

$$n^{2}+1=25 k^{2}+20 k+4+1=5\left(5 k^{2}+4 k+1\right)$$

is divisible by 5.

Thus, the expression $n^{6}-n^{2}$ is divisible by 3, 4, and 5 for every integer $n$. This implies that it is also divisible by 60, because any number can be written in the form

$$60 h+m$$

where $0 \leq m \leq 59$ is the remainder. Since the first term is divisible by 3, 4, and 5, the sum can only be divisible by all three if $m$ is also divisible by all three.

Only numbers ending in 0 or 5 are divisible by 5, and among these, only those ending in 0 can be divisible by 4, so the possible values of $m$ are

$$0,10,20,30,40,50$$

Among these, only 0 and 30 are divisible by 3, but the latter is not divisible by 4, so $m=0$ and thus any number divisible by 3, 4, and 5 is also divisible by 60. This proves the statement of the theorem.

Remark. The fact that we can conclude divisibility by 60 from divisibility by 3, 4, and 5 is based on the fact that no two of these numbers have a common divisor greater than 1. It would not be difficult to prove the divisibility of the expression by 4, 5, and 6, but from this we could not conclude divisibility by $4 \cdot 5 \cdot 6=120$, as the expression gives 60 for $n=2$ and this is not divisible by 120. Many have fallen into this error, concluding divisibility by the product from divisibility by two numbers, without considering the common divisors of the factors.[^0]

[^0]: ${ }^{1}$ We could simply refer to the theorem that if an integer is divisible by integers that are pairwise coprime, then it is also divisible by their product. However, the following consideration shows that this can be easily proven directly for specific numbers.