nLab freshman's dream

Idea

Over a ring whose characteristic is a prime number $p$ it is true that

$$(a+b)^p = a^p + b^p \,,$$

because all the non-trivial binomial coefficients are multiples of $p$.

Primality

Lemma

An integer $n \geq 2$ is prime if and only if the binomial coefficients satisfy

$$\binom{n}{k} \equiv 0 \pmod{n}$$

whenever $0 \lt k \lt n$.

Proof

If $p$ is prime, then $p$ divides neither $k!$ nor $(p-k)!$ if $0 \lt k \lt p$. So, since $p$ divides

$$p! = \binom{p}{k}k! \cdot (p-k)!,$$

it must divide $\binom{p}{k}$.

If $n$ is composite, write $n = p q$ where $p$ is any prime dividing $n$, so that $0 \lt p \lt n$. If the factor $p$ appears with multiplicity $j$ in $n$, then it appears with multiplicity $j$ in $n(n - 1) \ldots (n - p + 1)$ and with multiplicity $1$ in $p! = p(p-1)\ldots 1$, and so it appears with multiplicity $j-1$ in $\binom{n}{p}$. It follows that $n$ cannot divide $\binom{n}{p}$, i.e., $\binom{n}{p} \nequiv 0 \pmod{n}$.

Related concepts

• Frobenius morphism

References

• Wikipedia, Freshman’s dream