Kummer’s congruence

Let $B_k$ denote the $k$th Bernoulli number:

$$B_0=1,B_1=-\frac{1}{2},B_2=\frac{1}{6},B_3=0,B_4=-\frac{1}{30},\mathrm{\dots },B_{10}=\frac{5}{66},\mathrm{\dots }$$

In fact, $B_k=0$ for all odd $k\ge 3$, so we will only consider $B_k$ for even $k$. The following congruence is due to Ernst Eduard Kummer:

Theorem (Kummer’s congruence).

Let $p$ be a prime. Suppose that $k\mathrm{\ge }\mathrm{2}$ is an even integer which is not divisible by $\mathrm{(}p\mathrm{-}\mathrm{1}\mathrm{)}$. Then the quotient $B_k\mathrm{/}k$ is $p$-integral, that is, as a fraction in lower terms, $p$ does not divide its denominator. Furthermore, if $h$ is another even integer with $\mathrm{(}p\mathrm{-}\mathrm{1}\mathrm{)}\mathrm{\nmid }k$ and $k\mathrm{\equiv }h\mathrm{mod}\mathrm{(}p\mathrm{-}\mathrm{1}\mathrm{)}$ then

$$\frac{B_k}{k}\equiv \frac{B_h}{h}modp.$$

The interested reader should see also the congruence of Clausen and von Staudt for a similar result. As an example of Kummer’s congruence, let $p=7$ and $k=4$. Then:

$$\frac{B_4}{4}=\frac{-\frac{1}{30}}{4}=-\frac{1}{120}\equiv 6mod7$$

If we pick $h=10$ (so that $10\equiv 4mod(p-1)$) then:

$$\frac{B_{10}}{10}=\frac{\frac{5}{66}}{10}=\frac{1}{132}\equiv 6mod7$$

which is what the theorem predicted.