A well-known fallacy committed by students is the so-called “Law of Universal Linearity” (the link is to a discussion of this phenomenon on Mathematics Stack Exchange). The most famous example of this is the statement
$$\left(x+y\right)^n = x^n + y^n,$$
known as the Freshman’s dream.
This is clearly false, as $4=2^2=\left(1+1\right)^2\neq 2 = 1^2+1^2$. However, the dream comes true when one finally gets into a course which discusses rings, particularly those of prime characteristic.
Definition. A (commutative) ring is a set $R$ on which compatible addition $+$ and multiplication $\cdot$ operations are defined. That is, for every $a,b,c\in R$, we have that:
(A1) $a+b\in R$,
(A2) $a+b=b+a$,
(A3) $\left(a+b\right)+c=a+\left(b+c\right)$,
(A4) there exists $0\in R$ with $0+a\in R$,
(A5) there exists $-a\in R$ with $-a+a=0$,
(M1) $ab\in R$,
(M2) $ab=ba$,
(M3) $a\left(bc\right)=\left(ab\right)c$, and
(Distributive Property) $a\left(b+c\right)=ab+ac$.
Definition. The characteristic of a ring $R$ is the lowest integer $n>0$ such that $n\cdot1=\underbrace{1+1+\cdots+1}_{n\text{ terms}}=0$. If no such $n$ exists, then the characteristic is $0$.
Note: An immediate consequence of this is that any integer which is a multiple of $n$ in a ring of characteristic $n$ is treated as if it is $0$. This is because in, for example, a ring of characteristic $2$, we have that, say, $8=2\cdot4=\left(2\cdot1\right)\cdot4=0\cdot4=0$, and so any element which is added up an even number of times is $0$. The same goes for any other characteristic, replacing $2$ by $n$ and $8$ by $nq$ for some integer $q$.
Definition. Let $n,k$ be positive integers. The number of ways to choose $k$ things from a set of $n$ things is denoted ${n \choose k}$, read $n$ choose $k$, and is called the binomial coefficient (for reasons which will be illuminated shortly).
Theorem. (Binomial Theorem for Commutative Rings). Let $R$ be a commutative ring. Then, For any two elements $x,y\in R$, we have that $\displaystyle{\left(x+y\right)^n = \sum_{k=0}^n {n \choose k}x^ky^{n-k}}$.
Proof. There are really two ways to prove this. One way is more about symbol manipulation followed by an application of a result known as Pascal’s Recursive Formula for Binomial Coefficients along with induction, whereas a more combinatorial proof can be made. I like the combinatorial proof, so that is what we will be using here.
When we see the expression $\left(x+y\right)^n$, we, of course, mean
$$\underbrace{\left(x+y\right)\cdots\left(x+y\right)}_{n\text{ terms}}$$
Now, if you imagine distributing terms over each other over and over again, then we can see that we will get a sum in terms of the form $x^ky^{n-k}$ for $k=0,1,2,\ldots,n$. There is only one $x^n$ term and only one $y^n$ term, as these are of maximal degree. Notice that there is exactly one way to choose zero things and one way to choose $n$ things in a collection of $n$ items. Therefore, we may say that the coefficient of $y^n$ is ${n \choose 0}$ and of $x^n$ is ${n \choose n}$. In general, there are ${n \choose k}$ terms of the form $x^ky^{n-k}$ for every $k$, as there are ${n \choose k}$ different ways to choose $k$ $x$’s to multiply together. Thus, the equality is proven, that is, we have shown that
$$\left(x+y\right)^n = \sum_{k=0}^n {n \choose k}x^ky^{n-k},$$
as required. QED
Now, there is an important formula for ${n \choose k}$ that we must prove. It will be from this formula that we will see the main result.
Theorem. For any integers $n,k$, with $0\leq k\leq n$, we have that ${n \choose k} = \frac{n!}{k!\left(n-k\right)!}$.
Proof. Given $n$ items, there are $n!=n\left(n-1\right)\cdots3\cdot2\cdot1$ ways of rearranging them, as there are $n$ different ways to choose the first item, leaving $n-1$ of them so that there are $n-1$ ways to choose the second, and so on. If, however, we only stopped after choosing $k$ of them, we would have $n\left(n-1\right)\cdots\left(n-k+1\right)$ ways to choose $k$ items, but since $\frac{n!}{\left(n-k\right)!}=\frac{n\left(n-1\right)\cdots3\cdot2\cdot1}{\left(n-k\right)\cdots3\cdot2\cdot1}=n\left(n-1\right)\cdots\left(n-k+1\right)$, we now know how many rearrangements we may get when we choose $k$ items. Finally, since we don’t care about the order we choose our items in, we only care about which items we have, then each $k!$ rearrangements of the $k$ items we’ve chosen are essentially the same to us. Thus, to get the number of ways to choose $k$ items from $n$, we must divide this by $k!$ to get
$${n \choose k}=\frac{n!}{k!\left(n-k\right)!},$$
as required. QED
Theorem. An integer $p$ is prime if and only if $p\mid {p \choose k}$ for all $1\leq k\leq p-1$.
Proof.
First, suppose that $p$ is prime. Then, since ${p \choose k}=\frac{p!}{k!\left(p-k\right)!}$, we have that $k!{p \choose k}=\frac{p!}{\left(p-k\right)!}=p\left(p-1\right)\cdots\left(p-k+1\right)=pq$, where $q=\left(p-1\right)\cdots\left(p-k+1\right)$ is an integer. Note that $p\nmid k!$ since its only factors are the integers up to $k\lt p$. Therefore, it must be the case that ${p \choose k}$ is divisible by $p$ whenever $1\leq k\leq p-1$.
Now, suppose that $p$ is composite, and let $q$ be the smallest prime factor of $p$ and let $n=p/q$. Then,
$${p \choose q} = \frac{p!}{q!\left(p-q\right)!} = \frac{p\left(p-1\right)\ldots\left(p-q+1\right)}{q!} =$$
$$\frac{n\left(p-1\right)\ldots\left(p-q+1\right)}{\left(q-1\right)!},$$
but $p$ cannot divide this as this would imply that $p$ divides $n\left(p-1\right)\ldots\left(p-q+1\right)$, which would mean that $q$ divides $\left(p-1\right)\ldots\left(p-q+1\right)$. However, $q$ cannot divide this, as these are the $q-1$ integers before a multiple of $q$ and, hence, will have nonzero remainders when divided by $q$. Therefore, $p\nmid {p \choose q}$.
QED
Corollary. If a ring has characteristic $p$, then $p$ is prime if and only if $\left(x+y\right)^p = x^p + y^p$.
Proof.
$$\left(x+y\right)^p=\sum_{k=0}^p {p \choose k}x^ky^{p-k} = x^p + y^p$$
$$\iff {p \choose k}\equiv 0\,\left(\text{mod }p\right)\text{ for }k\text{ between }1\text{ and }p-1$$
$$\iff p\text{ is prime.}$$
QED
An example of a ring of characteristic $p$ is the finite field of order $p$, namely, $\mathbb{Z}/p\mathbb{Z}$. For $p=3$, we check $\left(1+1\right)^3=2^3=2=1+1=1^3+1^3$. Remember that each equality means remainder when dividing by $3$, and since $8=2\cdot3+2$, we have that $8\equiv 2\,\left(\text{mod }3\right)$.