$n^2 (n^4 - 1)$ is divisible by 60

Question

How does one prove that $n^2 (n^4 - 1)$ is divisible by 60 for any positive integer $n$? I tried mathematical induction method without success so far.

Answer 1

Hint: Use that $$n^2(n^4-1)=n^2(n-1)(n+1)(n^2+1)$$

Answer 2

Hint Factor the expression $N := n^2 (n^4 - 1)$ as $(n + 1) n^2 (n - 1) (n^2 + 1)$. Then, for example, $N$ has $n - 1$, $n$, $n + 1$ as factors; one of these is a multiple of $3$, whence so is $N$. Since $60 = 3 \cdot 20$ and $3$ and $20$ are coprime, it now suffices to show that $N$ is divisible by $20$.

Answer 3

Hint:

Since $60=2^2\cdot 3\cdot 5$, use congruences mod. $4$, $3$ and $5$ and the Chinese remainder theorem to show either $n^2$ or $n^4-1\equiv 0$ modulo each of them.

Answer 4

Here is a way to prove one case using induction (using the Binomial Theorem): $$n^5=(n+1-1)^5=(n+1)^5+5\left(-(n+1)^4+2(n+1)^3-2(n+1)^2+(n+1)) \right)-1\\n^5-n=(n+1)^5+5K-1-n$$

So if $5\mid n(n^4-1)\Rightarrow 5\mid (n+1)\left((n+1)^4-1\right )$.

Now you only have to show that $3,4\mid n^2(n^4-1)$.