Prove that if $n$ is not divisible by $5$, then $n^4 \equiv 1 \pmod{5}$ Suppose $n$ is an integer which is not divisible by $5$. Prove that $n^4 \equiv 1 \pmod{5}$.

Here's one particularly pretty way of proving this statement. It's nice because it uses combinatorics to prove an essentially number theoretic result. We start with the problem:

Suppose you want to make a necklace with $5$ beads, and you can paint each one of them with one of $n$ available colors. What's the number of different necklace you can have? (Necklace that can be obtained from other necklaces under rotation are considered the same)

We could stay that there are $n$ different ways to color the first bead, $n$ different ways to color the second, etcetera. But there are $5$ different rotations for each collar. Therefore the answer must be $\frac{n^5}{5}$, right? No, but close enough.
For instance, if all of the beads have only one color, then this necklace won't generate any other necklace under rotation. On the other hand, if a necklace has more than one color, then we affirm that it generates $5$ different necklaces (including itself). This is not an immediately obvious result so we shall prove it formally.
Indeed, let $(a_1,a_2,a_3,a_4,a_5)$ be a necklace, where $a_k$ is the color of the bead $k$. By assumption, there are $i,j \in \Bbb{Z}$ such that $a_i\neq a_j$ (consider the indices modulo $5$). Suppose we rotate the necklace by $r$ places. We affirm that $(a_1,a_2,...,a_5)=(a_{1+r},a_{2+r},...,a_{5+r})$ if and only if $r$ is a multiple of $5$. Suppose it's not, and $a_{i}=a_{i+r}$ for all $i$. Inductively, $a_i=a_{i+mr}$, for $m\in\Bbb{Z}$. But since we're looking at the indices modulo $5$, let $m=(j-i)\cdot r^{-1}$ where $r^{-1}$ is the inverse of $r$ modulo $5$ (it exists since $5$ is prime and $5\nmid r$). Then $a_i=a_{i+mr}=a_{i+(j-i)r^{-1}r}=a_j$ for all $i,j$. But we're assuming  there are $i,j \in \Bbb{Z}$ such that $a_i\neq a_j$. Absurd.
Therefore the number of ways is actually $\frac{n^5-n}{5}+n$. The nice fact is that this number must be an integer (quite obviously), so in particular $\frac{n^5-n}{5}=\frac{n(n^4-1)}{5}$ is an integer. If $5\nmid n$, then $\gcd(5,n)=1$ since $5$ is prime, therefore $5\mid n^4-1\Rightarrow n^4\equiv 1 \pmod 5$. $\,\,\blacksquare$
