Prove that $n^4-n$ is divisible by 2 for all integer

Question

Here n is an integer. And for any integer number $n^4-n$ is divisible by $2$.

Answer 1

If $n$ is an integer, then $$n^4-n=n(n^3-1)=n(n-1)(n^2+n+1)$$

But there is an even intger amoung any two consecutive integers. ($n$ and $n-1$)

Therefore $2|n^4-n$.

Answer 2

Newton's interpolation formula gives $$ n^4-n=14 \binom{n}{2}+ 36 \binom{n}{3}+ 24 \binom{n}{4}, $$ a sum of even numbers.

A simpler solution is to use induction. Let $a_n=n^4-n$. Then $$ a_{n+1}-a_n = 4 n^3 + 6 n^2 + 4 n $$ implies that $a_n$ is even for all $n$ because $a_0=0$ is even.