Prove that $n^4$ is congruent to 0 or 1 modulo 5

Question

Prove that $n^4$ for all $n\in{\mathbb Z}$ is congruent to 0 or 1 modulo 5. Hint from professor: Do so using different cases.

I am confused on how to prove this for all $n$. I understand that you can test a few numbers but I am stuck on how to show it is true for all $n$.

Answer 1

Hint:$$(5n+k)^4=\color{red}{5^4}n^4+4\cdot\color{red}{5^3}n^3k+6\cdot\color{red}{5^2}n^2k^2+4\cdot\color{red}5nk^3+\color{blue}{k^4}$$

In general, we have

$$a^b\equiv(a\mod c)^b\pmod c$$

which follows from binomial expansion.

Answer 2

There are only five numbers modulo $5$. I recommend that you just try them all.

Answer 3

Hint:

$n\equiv 0,\pm 1,\pm2\mod5$, hence $\;n^2\equiv 0^2,1^2,2^2=0,1,-1$.