Failure of Fermat's Little Theorem for composite moduli, explained in terms of combinatorics? Notice that $$3^4 - 3 = 81 - 3 = 78,$$ which is divisible by $2$, but not by $4$. 
In your opinion, what "goes wrong" in this case? 
This is a question meant to be answered via combinatorics, and to me, what I was able to note that there may be $(i,j)$ repetitions, such that $3*3 - 3 = 6$, which $6$ is not divisible by $4$.
Can someone explain what goes wrong in terms of the combinatorics?
 A: The combinatorial proof that $p \mid a^p - a$ when $p$ is prime is that $a^p - a$ counts the number of acyclic necklaces of length $p$ with $a$ colors. For any $n$, the cyclic group $\mathbb{Z}/n\mathbb{Z}$ acts by cyclic permutation on the set of necklaces of length $n$ with $a$ colors. When $p$ is prime, we can split up the set of necklaces as follows:


*

*The necklaces all of whose colors are the same, of which there are $a$, and

*The acyclic necklaces, of which there are $a^p - a$.


The action of $\mathbb{Z}/p\mathbb{Z}$ on the first collection is trivial and the action on the second is free because $\mathbb{Z}/p\mathbb{Z}$ has no nontrivial subgroups, so the action on the second collection splits up into $\frac{a^p - a}{p}$ orbits of size $p$ by the orbit-stabilizer theorem. 
When $n$ is not prime, other orbit types may appear, since necklaces may have nontrivial cyclic structure such as $RBYRBY$, so there are orbits with nontrivial stabilizers. In this case the correct count of the number of acyclic necklaces (those on which $\mathbb{Z}/n\mathbb{Z}$ acts freely) is given, not by $a^n - a$, but by
$$\sum_{d \mid n}\; \mu(d) a^{n/d}$$
where $\mu$ is the Möbius function; the proof is a classical application of Möbius inversion and worth working out as an exercise. So the correct generalization to the non-prime case from the combinatorial point of view is that
$$n \mid  \sum_{d \mid n}\; \mu(d) a^{n/d}.$$

In the specific case $a = 3, n = 4$, the group $\mathbb{Z}/4\mathbb{Z}$ has $\mathbb{Z}/2\mathbb{Z}$ as its only nontrivial subgroup. Letting the colors be $R, B, Y$, the orbits with full stabilizer look like $RRRR$, the orbits with stabilizer $\mathbb{Z}/2\mathbb{Z}$ look like $RBRB$, and the orbits with trivial stabilizer look like $RBYB$. There are $3$ of the first kind and
$$\frac{\mu(1) 3^4 + \mu(2) 3^2 + \mu(4) 3^1}{4} = \frac{3^4 - 3^2}{4} = 18$$
of the third kind, so $3$ of the second kind. 
A: The point of Qiaochu Yuan's answer is that firstly that in order to be surprised the $a^p-a$ is not divisible by $p$ when $p$ is composite, you should first have a (combinatorial in your case) reason to find it natural that it does hold when $p$ is a prime number. One such reason is that $a^p-a$ counts words (which I find more convenient to talk about than necklaces, but it is the same thing) of length $p$ over an $a$-element alphabet that do not consist of $p$ identical letters. The reason this number is divisible by $p$ is that cyclic rotation partitions the set of such words into packets (orbits) of size $p$. That this is so requires a bit of thought: if there were any non-trivial cyclic rotation (not necessarily by a single place) that transforms a word into itself, then by iterating that one can obtain the cyclic rotation by a single place, and any word invariant by that rotation clearly has all its letters equal. This amounts to the fact that any non-identity element of a cyclic group of $p$ element generates the entire group.
Now the second point of the Qiaochu Yuan's answer is to analyze the situation when $p$ is composite (I'll henceforth call the number $n$ to avoid confusion). Now it is no longer true that any non-identity element of a cyclic group of $n$ element generates the entire group; notably it can generate a subgroup of any order dividing $n$. Consequently there are words that are invariant under some non-trivial cyclic rotation, but not under all of them, and in particular not under the rotation by a single place. The word RBYRBY is an example, invariant under rotation by $3$ places, but not under rotation by one place. You may very well stop here; it is clear why the argument breaks down, and the numerical examples you gave show that there is no other argument that will make $a^n-a$ divisible by $n$ for all $n$ either.
If you insist on finding something that is counted by a number always divisible by $n$, then you can still take all words that are not invariant under any non-trivial cyclic rotation, as these come in packets (orbits) of size $n$. Call this number $f(n)$, which will be useful later. Now rather than having $f(n)=a^n-a$, you need to exclude for every strict divisor $d$ of $n$ the words that come in packets of size $d$. These are harder to count than words that come in singleton packets (as we saw there were $a$ of them), but can be done by the following reasoning. Such words necessarily consist of a sub-word $s$ of $d$ letters that is repeated $d$ times: $w=s.s\ldots s$. However $s$ cannot just be any word, it must not have any internal cyclic symmetry, since if it did it would belong to a shorter orbit.
But words without any cyclic symmetry is just what we were out to count, except that it was for length $n$ and now we are confronted with length $d<n$. Hence the utility of a general function $f$: it is defined implicitly (by induction) by the relation
$$
  f(m)=a^m-\sum_{d\text{ strictly divides }m}f(d)
$$
for all $m\geq1$. Note that in particular $f(1)=a$ since $1$ has no strict divisors. This equation can be formulated more simply by moving the summation to the other side, which also avoid having to mention "strict":
$$
  \sum_{d\mid m}f(d)=a^m
$$
for all $m\geq1$. The general technique the will solve this is called Möbius inversion. Since you haven't heard about that, here's its secret. Think of a vectors $F(n)=(f(1),f(2),\ldots,f(n))$, the unknown we are after, and $G(n)=(a,a^2,\ldots,a^n)$ which tabulates the right-hand-sides of our last equation. This equation then says $F(n)=\mathbf D(n)\cdot F(n)$, where $\mathbf D(n)$ is the lower triangular matrix with entries $\mathbf D(n)_{i,j}=1$ whenever $j$ divides $i$, and $\mathbf D(n)_{i,j}=0$ otherwise. Actually the equation gives us the result of multiplying row $m$ of of the matrix $\mathbf D(n)$ and the vector $G(n)$, so the full matrix equation takes care of "for all $m$" (upto $n$). Now all we need is the inverse matrix $\mathbf D(n)$, and it can (rather easily) be shown to be given by
$$
 \mathbf D(n)^{-1}_{i,j} = \begin{cases}(-1)^c & 
  \text{if }i/j\text{ is integer and a product of }c\text{ distinct primes}\\
  0 & \text{otherwise}\\ \end{cases}
$$
which is usually written as
$$
 \mathbf D(n)^{-1}_{i,j} = \begin{cases}\mu(i/j)&\text{if }j\mid i\\
  0 & \text{otherwise}\\ \end{cases}
$$
where $\mu$ is the so-called (arithmetic) Möbius function. Now one can write $F(n)=\mathbf D(n)^{-1}\cdot G(n)$, giving
$$
  f(m)=\sum_{d\mid m}\mu(m/d)a^d
$$
which is the formula Qiaochu gave (with $m=n$) for the number of entirely non-symmetric necklacesof length $n$. (Actually he interchanged $n/d$ and $d$, which can be seen to make no difference; personally I prefer to always put $n/d$ in the argument of $\mu$.)
