Von Staudt–Clausen theorem

A proof of the Von Staudt–Clausen theorem follows from an explicit formula for Bernoulli numbers which is:

${\displaystyle B_{2n}=\sum _{j=0}^{2n}{\frac {1}{j+1}}\sum _{m=0}^{j}{(-1)^{m}{j \choose m}m^{2n}}}$

and as a corollary:

${\displaystyle B_{2n}=\sum _{j=0}^{2n}{\frac {j!}{j+1}}(-1)^{j}S(2n,j)}$

where ${\displaystyle S(n,j)}$ 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,

${\displaystyle \sum _{m=0}^{p-1}{(-1)^{m}{p-1 \choose m}m^{2n}}\equiv {-1}{\pmod {p}}}$

2. If p-1 does not divide 2n then,

${\displaystyle \sum _{m=0}^{p-1}{(-1)^{m}{p-1 \choose m}m^{2n}}\equiv 0{\pmod {p}}}$

Proof of (1) and (2): One has from Fermat's little theorem,

${\displaystyle m^{p-1}\equiv 1{\pmod {p}}}$

for ${\displaystyle m=1,2,...,p-1}$.
If p-1 divides 2n then one has,

${\displaystyle m^{2n}\equiv 1{\pmod {p}}}$

for ${\displaystyle m=1,2,...,p-1}$.
Thereafter one has,

${\displaystyle \sum _{m=1}^{p-1}{(-1)^{m}{p-1 \choose m}m^{2n}}\equiv \sum _{m=1}^{p-1}{(-1)^{m}{p-1 \choose m}}{\pmod {p}}}$

from which (1) follows immediately.
If p-1 does not divide 2n then after Fermat's theorem one has,

${\displaystyle m^{2n}\equiv m^{2n-(p-1)}{\pmod {p}}}$

If one lets ${\displaystyle \wp =[{\frac {2n}{p-1}}]}$ (Greatest integer function) then after iteration one has,

${\displaystyle m^{2n}\equiv m^{2n-\wp (p-1)}{\pmod {p}}}$

for ${\displaystyle m=1,2,...,p-1}$ and ${\displaystyle 0<2n-\wp (p-1)<p-1}$.
Thereafter one has,

${\displaystyle \sum _{m=0}^{p-1}{(-1)^{m}{p-1 \choose m}m^{2n}}\equiv \sum _{m=0}^{p-1}{(-1)^{m}{p-1 \choose m}m^{2n-\wp (p-1)}}{\pmod {p}}}$

Lemma (2) now follows from the above and the fact that S(n,j)=0 for j>n.
(3). It is easy to deduce that for a>2 and b>2, ab divides (ab-1)!.
(4). 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 (3), j+1 divides j!.
For j=3,

${\displaystyle \sum _{m=0}^{3}{(-1)^{m}{3 \choose m}m^{2n}}=3\cdot 2^{2n}-3^{2n}-3\equiv 0{\pmod {4}}}$

If j+1 is prime then we use (1) and (2) and if j+1 is composite then we use (3) and (4) to deduce:

${\displaystyle B_{2n}=I_{n}-\sum _{(p-1)|2n}{\frac {1}{p}}}$

where ${\displaystyle I_{n}}$ is an integer, which is the Von-Staudt Clausen theorem.[1][2]

• Kummer's congruence

• Clausen, Thomas (1840), "Theorem", Astronomische Nachrichten, 17 (22): 351–352, doi:10.1002/asna.18400172204
• Rado, R. (1934), "A New Proof of a Theorem of V. Staudt", J. London Math. Soc., 9 (2): 85–88, doi:10.1112/jlms/s1-9.2.85
• von Staudt, Ch. (1840), "Beweis eines Lehrsatzes, die Bernoullischen Zahlen betreffend", Journal für die Reine und Angewandte Mathematik, 21: 372–374, ISSN 0075-4102, ERAM 021.0672cj

• Weisstein, Eric W. "von Staudt-Clausen Theorem". MathWorld.