Alternative Proof: if $n$ is an integer, prove that $\frac{n ( n^4 - 1)}{5}$ is an integer

Question

I have proven this by the induction method but would like to know if it can be proven using an alternative method.

Answer 1

By Fermat's Little Theorem, we have that $$n^5 \equiv n \pmod 5 \Leftrightarrow \frac{n^5-n}{5} \in \mathbb{Z}$$

Answer 2

$$\frac{n(n^4-1)}{5}=\frac{n(n^2-1)(n^2+1)}{5}=\frac{(n-1)n(n+1)(n^2+1)}{5}$$ If $n=5k$, then $5|n$

If $n=5k+1$, then $5|n-1$

If $n=5k-1$, then $5|n+1$

If $n=5k\pm2$, then $n^2+1=(5k\pm2)^2+1=25k^2\pm10k+4+1=25k^2\pm10k+5=5(5k^2\pm2k+1)$, then $5|(n^2+1)$

Answer 3

Using a little linear algebra, we can write: $$ \frac{n(n^4-1)}{5}=24\binom{n}{5}+48\binom{n}{4}+30\binom{n}{3}+6\binom{n}{2}. $$ Since the binomial coefficients are always integers, this exhibits that your fraction is always an integer.

This method is kinda silly, I prefer the Fermat's little theorem approach. However, in general, any polynomial whose value on $\mathbb{Z}$ is always an integer can be written as an integer linear combination of binomial coefficients.