# proof of congruence of Clausen and von Staudt

###### Theorem 0.1

For $m\geq 1$,

 $(m+1)S_{m}(n)=\sum_{k=0}^{m}\binom{m+1}{k}B_{k}n^{m+1-k},$

where

 $S_{m}(n)=1^{m}+2^{m}+\ldots+(n-1)^{m}.$
###### Proof 1

In the equation

 $e^{kt}=\sum_{m=0}^{\infty}k^{m}\left(\frac{t^{m}}{m!}\right),$

substitute $k=0,1,2,\ldots,n-1$ and add, obtaining,

 $\displaystyle\sum_{m=0}^{\infty}S_{m}(n)\frac{t^{m}}{m!}=\frac{e^{nt}-1}{e^{t}% -1}$ $\displaystyle=$ $\displaystyle\frac{e^{nt}-1}{t}\cdot\frac{t}{e^{t}-1}$ $\displaystyle=$ $\displaystyle\sum_{k=1}^{\infty}n^{k}\frac{t^{k-1}}{k!}\sum_{j=0}^{\infty}B_{j% }\frac{t^{j}}{j!},$

since

 $\frac{t}{e^{t}-1}=\sum_{j=0}^{\infty}B_{j}\frac{t^{j}}{j!}.$

Now by comparing coefficients of $t^{m}$ and then multiplying by $(m+1)!$, we obtain the result.

 $\displaystyle S_{m}(n)$ $\displaystyle=$ $\displaystyle\sum_{k=0}^{m}\binom{m}{k}B_{m-k}\frac{n^{k+1}}{k+1}$ (1) $\displaystyle=$ $\displaystyle B_{m}n+\binom{m}{1}B_{m-1}\frac{n^{2}}{2}+\ldots+\frac{n^{m+1}}{% m+1}$ (2)

This follows by replacing $\binom{m+1}{k}$ by $\frac{m+1}{m-k+1}\binom{m}{k}$ and then switching $m-k$ and $k$ in the theorem.

###### Proposition 0.2

Let $p$ be prime and $m\geq 1$. Then $pB_{m}$ is $p$-integral, and if $m\geq 2$ is even, then $pB_{m}\equiv S_{m}(p)\pmod{p}$.

###### Proof 2

The first statement is equivalent     to showing if $p\mid b$ then $p^{2}\nmid b$, where $b$ is the denominator of $pB_{m}$. This is clear for $pB_{1}=-p/2$. We proceed by induction  . Suppose $m>1$ and let $n=p$ in (2). Since $S_{m}(p)\in\mathbb{Z}$, it suffices to prove that

 $\binom{m}{k}(pB_{m-k})\frac{p^{k}}{k+1}$

is $p$-integral for $k=1,2,\ldots,m$. By induction $pB_{m-k}$ is $p$-integral for $k\geq 1$, and $\frac{p^{k}}{k+1}$ is $p$-integral since $k+1\leq p^{k}$ for all primes $p$. It follows that $pB_{m}$ is $p$-integral. To establish the congruence    , we need to show that if $k\geq 1$, then

 $\binom{m}{k}(pB_{m-k})\frac{p^{k}}{k+1}\equiv 0\pmod{p}.$

For $k\geq 2$, $\frac{p^{k}}{k+1}\equiv 0\pmod{p}$, since $k+1. For $k=1$, we have

 $\frac{m}{2}(pB_{m-1})p\equiv 0\pmod{p},$

since $m$ is even. In fact, since $B_{m-1}=0$ for $m\geq 4$ even, it suffices to check it for $m=2$, which is obvious.

###### Lemma 1

Let $p$ be prime. Then

 $S_{m}(p)\equiv\begin{cases}0&\pmod{p},~{}\mbox{if p-1\nmid m}\\ -1&\pmod{p},~{}\mbox{if p-1\mid m}\end{cases}$
###### Proof 3

Let $g$ be a primitive root  modulo $p$. Then

 $\displaystyle S_{m}(p)$ $\displaystyle=$ $\displaystyle 1^{m}+2^{m}+\ldots+(p-1)^{m}$ $\displaystyle\equiv$ $\displaystyle 1^{m}+g^{m}+g^{2m}+\ldots+g^{(p-2)m}\pmod{p}$

Hence,

 $(g^{m}-1)S_{m}(p)\equiv g^{m(p-1)}-1\equiv 0\pmod{p}.$

If $p-1\nmid m$, then $g^{m}\not\equiv 1\pmod{p}$, and $S_{m}(p)\equiv 0\pmod{p}$. If $p-1\mid m$, then $S_{m}(p)\equiv 1+1+\ldots+1\equiv p-1\equiv-1\pmod{p}$.

We are now ready to prove the congruence.

###### Proof 4 (Proof of von Staudt-Claussen congruence)

Assume $m$ is even. Then by the proposition  , $pB_{m}$ is $p$-integral and $pB_{m}\equiv S_{m}(p)\pmod{p}$. Therefore, by the lemma, if $p-1\nmid m$, then $B_{m}$ is a $p$-integer and if $p-1\mid m$, then $pB_{m}\equiv-1\pmod{p}$. Hence,

 $A_{m}=B_{m}+\sum_{p-1\mid m}\frac{1}{p}$

is a $p$-integer for all primes $p$. For if $q$ is a prime, and $q-1\nmid m$, then $B_{m}$ is $q$-integral and hence $A_{m}$ is as well, since the sum contributes no negative power of $q$. Otherwise, $q-1\mid m$ and

 $\displaystyle A_{m}$ $\displaystyle=$ $\displaystyle B_{m}+\frac{1}{q}+\sum_{\begin{subarray}{c}p-1\mid m\\ p\neq q\end{subarray}}\frac{1}{p}$ $\displaystyle=$ $\displaystyle\frac{qB_{m}+1}{q}+\sum_{\begin{subarray}{c}p-1\mid m\\ p\neq q\end{subarray}}\frac{1}{p}$ $\displaystyle\equiv$ $\displaystyle\sum_{\begin{subarray}{c}p-1\mid m\\ p\neq q\end{subarray}}\frac{1}{p}\pmod{\mathbb{Z}},$

which is clearly $q$-integral. Since $A_{m}$ is $p$-integral for all primes $p$, it must be the case that $A_{m}\in\mathbb{Z}$. That is,

 $B_{m}\equiv-\sum_{\begin{subarray}{c}p~{}prime\\ p-1\mid m\end{subarray}}\frac{1}{p}\pmod{\mathbb{Z}}$
Title proof of congruence of Clausen and von Staudt ProofOfCongruenceOfClausenAndVonStaudt 2013-03-22 15:33:58 2013-03-22 15:33:58 slachter (11430) slachter (11430) 5 slachter (11430) Proof msc 11B68