proof of congruence of Clausen and von Staudt

Theorem 0.1

For m1,



Proof 1

In the equation


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

m=0Sm(n)tmm!=ent-1et-1 = ent-1ttet-1
= k=1nktk-1k!j=0Bjtjj!,



Now by comparing coefficients of tm and then multiplying by (m+1)!, we obtain the result.

We will write this identityPlanetmathPlanetmath in the following form:

Sm(n) = k=0m(mk)Bm-knk+1k+1 (1)
= Bmn+(m1)Bm-1n22++nm+1m+1 (2)

This follows by replacing (m+1k) by m+1m-k+1(mk) and then switching m-k and k in the theorem.

Proposition 0.2

Let p be prime and m1. Then pBm is p-integral, and if m2 is even, then pBmSm(p)(modp).

Proof 2

The first statement is equivalentMathworldPlanetmathPlanetmathPlanetmathPlanetmath to showing if pb then p2b, where b is the denominator of pBm. This is clear for pB1=-p/2. We proceed by inductionMathworldPlanetmath. Suppose m>1 and let n=p in (2). Since Sm(p)Z, it suffices to prove that


is p-integral for k=1,2,,m. By induction pBm-k is p-integral for k1, and pkk+1 is p-integral since k+1pk for all primes p. It follows that pBm is p-integral. To establish the congruenceMathworldPlanetmathPlanetmathPlanetmath, we need to show that if k1, then


For k2, pkk+10(modp), since k+1<pk. For k=1, we have


since m is even. In fact, since Bm-1=0 for m4 even, it suffices to check it for m=2, which is obvious.

Lemma 1

Let p be prime. Then

Sm(p){0(modp),if p-1m-1(modp),if p-1m
Proof 3

Let g be a primitive rootMathworldPlanetmath modulo p. Then

Sm(p) = 1m+2m++(p-1)m



If p-1m, then gm1(modp), and Sm(p)0(modp). If p-1m, then Sm(p)1+1++1p-1-1(modp).

We are now ready to prove the congruence.

Proof 4 (Proof of von Staudt-Claussen congruence)

Assume m is even. Then by the propositionPlanetmathPlanetmath, pBm is p-integral and pBmSm(p)(modp). Therefore, by the lemma, if p-1m, then Bm is a p-integer and if p-1m, then pBm-1(modp). Hence,


is a p-integer for all primes p. For if q is a prime, and q-1m, then Bm is q-integral and hence Am is as well, since the sum contributes no negative power of q. Otherwise, q-1m and

Am = Bm+1q+p-1mpq1p
= qBm+1q+p-1mpq1p

which is clearly q-integral. Since Am is p-integral for all primes p, it must be the case that AmZ. That is,

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