# congruence of Clausen and von Staudt

Let $B_{k}$ denote the $k$th Bernoulli number:

 $B_{0}=1,\quad B_{1}=-\frac{1}{2},\quad B_{2}=\frac{1}{6},\quad B_{3}=0,\quad B% _{4}=-\frac{1}{30},\ldots$

In fact, $B_{k}=0$ for all odd $k\geq 3$, so we will only consider $B_{k}$ for even $k$. The following is a well-known congruence, due to Thomas Clausen and Karl von Staudt.

###### Theorem (Congruence of Clausen and von Staudt).

For an even integer $k\geq 2$,

 $B_{k}\equiv-\sum_{p\text{ prime},\ (p-1)|k}\frac{1}{p}\mod\mathbb{Z}$

where the sum is over all primes $p$ such that $(p-1)$ divides $k$. In other words, there exists an integer $n_{k}$ such that

 $B_{k}=n_{k}-\sum_{p\text{ prime},\ (p-1)|k}\frac{1}{p}.$

For example:

 $B_{2}=\frac{1}{6}=1-\frac{1}{2}-\frac{1}{3},\quad B_{4}=-\frac{1}{30}=1-\frac{% 1}{2}-\frac{1}{3}-\frac{1}{5}.$

Sometimes the theorem is stated in this alternative form:

###### Corollary.

For an even integer $k\geq 2$ and any prime $p$ the product $pB_{k}$ is $p$-integral, that is, $pB_{k}$ is a rational number $t/s$ (in lowest terms) such that $p$ does not divide $s$. Moreover:

 $pB_{k}\equiv\begin{cases}-1\mod p,\text{ if (p-1) divides k;}\\ 0\mod p,\text{ if (p-1) does not divide k}.\end{cases}$
Title congruence of Clausen and von Staudt CongruenceOfClausenAndVonStaudt 2013-03-22 15:11:58 2013-03-22 15:11:58 alozano (2414) alozano (2414) 4 alozano (2414) Theorem msc 11B68 Staudt-Clausen theorem von Staudt-Clausen theorem KummersCongruence OddBernoulliNumbersAreZero