proof of congruence of Clausen and von Staudt

Theorem 0.1

For $m\mathrm{\ge }\mathrm{1}$,

$$(m+1)S_m(n)=\sum _{k=0}^m\left(\genfrac{}{}{0pt}{}{m+1}{k}\right)B_k{n}^{m+1-k},$$

where

$$S_m(n)=1^m+2^m+\mathrm{\dots }+{(n-1)}^m.$$

Proof 1

In the equation

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

substitute $k\mathrm{=}\mathrm{0}\mathrm{,}\mathrm{1}\mathrm{,}\mathrm{2}\mathrm{,}\mathrm{\dots }\mathrm{,}n\mathrm{-}\mathrm{1}$ and add, obtaining,

${\displaystyle \sum _{m=0}^{\mathrm{\infty }}}{S}_m(n){\displaystyle \frac{t^m}{m!}}={\displaystyle \frac{e^{nt}-1}{e^t-1}}$ $=$ ${\displaystyle \frac{e^{nt}-1}{t}}\cdot {\displaystyle \frac{t}{e^t-1}}$

$=$ ${\displaystyle \sum _{k=1}^{\mathrm{\infty }}}{n}^k{\displaystyle \frac{t^{k-1}}{k!}}{\displaystyle \sum _{j=0}^{\mathrm{\infty }}}{B}_j{\displaystyle \frac{t^j}{j!}},$

since

$$\frac{t}{e^t-1}=\sum _{j=0}^{\mathrm{\infty }}{B}_j\frac{t^j}{j!}.$$

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

We will write this identity in the following form:

$S_m(n)$ $=$ ${\displaystyle \sum _{k=0}^m}\left({\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right)B_{m-k}{\displaystyle \frac{n^{k+1}}{k+1}}$ (1)

$=$ $B_{m}n+\left({\displaystyle \genfrac{}{}{0pt}{}{m}{1}}\right)B_{m-1}{\displaystyle \frac{n^2}{2}}+\mathrm{\dots }+{\displaystyle \frac{n^{m+1}}{m+1}}$ (2)

This follows by replacing $\left(\genfrac{}{}{0pt}{}{m+1}{k}\right)$ by $\frac{m+1}{m-k+1}\left(\genfrac{}{}{0pt}{}{m}{k}\right)$ and then switching $m-k$ and $k$ in the theorem.

Proposition 0.2

Let $p$ be prime and $m\mathrm{\ge }\mathrm{1}$. Then $p\mathit{}{B}_m$ is $p$-integral, and if $m\mathrm{\ge }\mathrm{2}$ is even, then $p\mathit{}{B}_m\mathrm{\equiv }{S}_m\mathit{}\mathrm{(}p\mathrm{)}\mathrm{(}\mathrm{mod}p\mathrm{)}$.

Proof 2

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

$$\left(\genfrac{}{}{0pt}{}{m}{k}\right)(p{B}_{m-k})\frac{p^k}{k+1}$$

is $p$-integral for $k\mathrm{=}\mathrm{1}\mathrm{,}\mathrm{2}\mathrm{,}\mathrm{\dots }\mathrm{,}m$. By induction $p\mathit{}{B}_{m\mathrm{-}k}$ is $p$-integral for $k\mathrm{\ge }\mathrm{1}$, and $\frac{p^k}{k\mathrm{+}\mathrm{1}}$ is $p$-integral since $k\mathrm{+}\mathrm{1}\mathrm{\le }{p}^k$ for all primes $p$. It follows that $p\mathit{}{B}_m$ is $p$-integral. To establish the congruence, we need to show that if $k\mathrm{\ge }\mathrm{1}$, then

$$\left(\genfrac{}{}{0pt}{}{m}{k}\right)(p{B}_{m-k})\frac{p^k}{k+1}\equiv 0(modp).$$

For $k\mathrm{\ge }\mathrm{2}$, $\frac{p^k}{k\mathrm{+}\mathrm{1}}\mathrm{\equiv }\mathrm{0}\mathrm{(}\mathrm{mod}p\mathrm{)}$, since . For $k\mathrm{=}\mathrm{1}$, we have

$$\frac{m}{2}(p{B}_{m-1})p\equiv 0(modp),$$

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

Lemma 1

Let $p$ be prime. Then

Proof 3

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

$S_m(p)$ $=$ $1^m+2^m+\mathrm{\dots }+{(p-1)}^m$

$\equiv$ $1^m+g^m+g^{2m}+\mathrm{\dots }+g^{(p-2)m}(modp)$

Hence,

$$(g^m-1)S_m(p)\equiv g^{m(p-1)}-1\equiv 0(modp).$$

If $p\mathrm{-}\mathrm{1}\mathrm{\nmid }m$, then $g^m\mathrm{\not\equiv }\mathrm{1}\mathrm{(}\mathrm{mod}p\mathrm{)}$, and $S_m\mathit{}\mathrm{(}p\mathrm{)}\mathrm{\equiv }\mathrm{0}\mathrm{(}\mathrm{mod}p\mathrm{)}$. If $p\mathrm{-}\mathrm{1}\mathrm{\mid }m$, then $S_m\mathit{}\mathrm{(}p\mathrm{)}\mathrm{\equiv }\mathrm{1}\mathrm{+}\mathrm{1}\mathrm{+}\mathrm{\dots }\mathrm{+}\mathrm{1}\mathrm{\equiv }p\mathrm{-}\mathrm{1}\mathrm{\equiv }\mathrm{-}\mathrm{1}\mathrm{(}\mathrm{mod}p\mathrm{)}$.

We are now ready to prove the congruence.

Proof 4 (Proof of von Staudt-Claussen congruence)

Assume $m$ is even. Then by the proposition, $p\mathit{}{B}_m$ is $p$-integral and $p\mathit{}{B}_m\mathrm{\equiv }{S}_m\mathit{}\mathrm{(}p\mathrm{)}\mathrm{(}\mathrm{mod}p\mathrm{)}$. Therefore, by the lemma, if $p\mathrm{-}\mathrm{1}\mathrm{\nmid }m$, then $B_m$ is a $p$-integer and if $p\mathrm{-}\mathrm{1}\mathrm{\mid }m$, then $p\mathit{}{B}_m\mathrm{\equiv }\mathrm{-}\mathrm{1}\mathrm{(}\mathrm{mod}p\mathrm{)}$. 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 $q\mathrm{-}\mathrm{1}\mathrm{\nmid }m$, then $B_m$ is $q$-integral and hence $A_m$ is as well, since the sum contributes no negative power of $q$. Otherwise, $q\mathrm{-}\mathrm{1}\mathrm{\mid }m$ and

$A_m$ $=$ $B_m+{\displaystyle \frac{1}{q}}+{\displaystyle \sum _{\begin{array}{c}p-1\mid m\\ p\ne q\end{array}}}{\displaystyle \frac{1}{p}}$

$=$ ${\displaystyle \frac{q{B}_m+1}{q}}+{\displaystyle \sum _{\begin{array}{c}p-1\mid m\\ p\ne q\end{array}}}{\displaystyle \frac{1}{p}}$

$\equiv$ ${\displaystyle \sum _{\begin{array}{c}p-1\mid m\\ p\ne q\end{array}}}{\displaystyle \frac{1}{p}}(mod\mathbb{Z}),$

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

$$B_m\equiv -\sum _{\begin{array}{c}pprime\\ p-1\mid m\end{array}}\frac{1}{p}(mod\mathbb{Z})$$

Title	proof of congruence of Clausen and von Staudt
Canonical name	ProofOfCongruenceOfClausenAndVonStaudt
Date of creation	2013-03-22 15:33:58
Last modified on	2013-03-22 15:33:58
Owner	slachter (11430)
Last modified by	slachter (11430)
Numerical id	5
Author	slachter (11430)
Entry type	Proof
Classification	msc 11B68