How can I prove divisibility using congruence?

Question

I'm new to this, so please excuse me if I said something wrong or offended anyone. We're doing the number theory in class, and I came across this question, which I had no idea how to even begin..:

Use congruences to prove 5|(n5 − n) (by the way, we already proved this using induction)

Does anyone know how to do this? Thank you so much in advance.

Answer 1

If you meant exponent, try the following:

$$n^5-n=n(n^4-1)=n(n^2-1)(n^2+1)=n(n-1)(n+1)(n^2+1)$$

Now check that no matter what $\;n\pmod 5\;$ is, the above is always $\;0\pmod 5\;$

Answer 2

Notice that

$$0^5\equiv 0\mod 5$$ $$1^5\equiv 1\mod 5$$ $$2^5=32\equiv 2\mod 5$$ $$3^5\equiv (-2)^5\equiv -2\equiv 3\mod 5$$

$$4^5\equiv (-1)^5\equiv4\mod 5$$ so we conclude that for all $n$ $$n^5\equiv n\mod 5$$ and the result follows. Notice also that we can find this result using the Fermat's little theorem.