Von Staudt–Clausen theorem


In number theory, the von Staudt–Clausen theorem is a result determining the fractional part of Bernoulli numbers, found independently by
and.
Specifically, if n is a positive integer and we add 1/p to the Bernoulli number B2n for every prime p such that p − 1 divides 2n, we obtain an integer, i.e.,
This fact immediately allows us to characterize the denominators of the non-zero Bernoulli numbers B2n as the product of all primes p such that p − 1 divides 2n; consequently the denominators are square-free and divisible by 6.
These denominators are

Proof

A proof of the Von Staudt–Clausen theorem follows from an explicit formula for Bernoulli numbers which is:
and as a corollary:
where are the Stirling numbers of the second kind.
Furthermore the following lemmas are needed:

Let p be a prime number then,

1. If p-1 divides 2n then,
2. If p-1 does not divide 2n then,
Proof of and : One has from Fermat's little theorem,
for.
If p-1 divides 2n then one has,
for.
Thereafter one has,
from which ' follows immediately.
If p-1 does not divide 2n then after Fermat's theorem one has,
If one lets then after iteration one has,
for and.
Thereafter one has,
Lemma ' now follows from the above and the fact that S=0 for j>n.
. It is easy to deduce that for a>2 and b>2, ab divides !.
. Stirling numbers of second kind are integers.
Proof of the theorem: Now we are ready to prove Von-Staudt Clausen theorem,
If j+1 is composite and j>3 then from, j+1 divides j!.
For j=3,
If j+1 is prime then we use and and if j+1 is composite then we use and to deduce:
where is an integer, which is the Von-Staudt Clausen theorem.