Editing Bernoulli number (section)

==Arithmetical properties of the Bernoulli numbers==

The Bernoulli numbers can be expressed in terms of the Riemann zeta function as {{math|''B''<sub>''n''</sub> {{=}} −''nζ''(1 − ''n'')}} for integers {{math|''n'' ≥ 0}} provided for {{math|''n'' {{=}} 0}} the expression {{math|−''nζ''(1 − ''n'')}} is understood as the limiting value and the convention {{math|''B''<sub>1</sub> {{=}} {{sfrac|1|2}}}} is used. This intimately relates them to the values of the zeta function at negative integers. As such, they could be expected to have and do have deep arithmetical properties. For example, the [[Agoh–Giuga conjecture]] postulates that {{mvar|p}} is a prime number if and only if {{math|''pB''<sub>''p'' − 1</sub>}} is congruent to −1 modulo {{mvar|p}}. Divisibility properties of the Bernoulli numbers are related to the [[ideal class group]]s of [[cyclotomic field]]s by a theorem of Kummer and its strengthening in the [[Herbrand-Ribet theorem]], and to class numbers of real quadratic fields by [[Ankeny–Artin–Chowla congruence|Ankeny–Artin–Chowla]].

=== The Kummer theorems ===
The Bernoulli numbers are related to [[Fermat's Last Theorem]] (FLT) by [[Ernst Kummer|Kummer]]'s theorem,{{r|Kummer1850}} which says:

:If the odd prime {{mvar|p}} does not divide any of the numerators of the Bernoulli numbers {{math|''B''<sub>2</sub>, ''B''<sub>4</sub>, ..., ''B''<sub>''p'' − 3</sub>}} then {{math|''x''<sup>''p''</sup> + ''y''<sup>''p''</sup> + ''z''<sup>''p''</sup> {{=}} 0}} has no solutions in nonzero integers.

Prime numbers with this property are called [[regular prime]]s. Another classical result of Kummer are the following [[Modular arithmetic#Congruence|congruences]].{{r|Kummer1851}}

{{main|Kummer's congruence}}

:Let {{mvar|p}} be an odd prime and {{mvar|b}} an even number such that {{math|''p''&nbsp;−&nbsp;1}} does not divide {{mvar|b}}. Then for any non-negative integer {{mvar|k}}
:: <math> \frac{B_{k(p-1)+b}}{k(p-1)+b} \equiv \frac{B_{b}}{b} \pmod{p}. </math>

A generalization of these congruences goes by the name of {{math|''p''}}-adic continuity.

==={{math|''p''}}-adic continuity===

If {{mvar|b}}, {{mvar|m}} and {{mvar|n}} are positive integers such that {{mvar|m}} and {{mvar|n}} are not divisible by {{math|''p'' − 1}} and {{math|''m'' ≡ ''n'' (mod ''p''<sup>''b'' − 1</sup> (''p'' − 1))}}, then

:<math>(1-p^{m-1})\frac{B_m}{m} \equiv (1-p^{n-1})\frac{B_n} n \pmod{p^b}.</math>

Since {{math|''B''<sub>''n''</sub> {{=}} −''nζ''(1 − ''n'')}}, this can also be written

:<math>\left(1-p^{-u}\right)\zeta(u) \equiv \left(1-p^{-v}\right)\zeta(v) \pmod{p^b},</math>

where {{math|''u'' {{=}} 1 − ''m''}} and {{math|''v'' {{=}} 1 − ''n''}}, so that {{mvar|u}} and {{mvar|v}} are nonpositive and not congruent to 1 modulo {{math|''p'' − 1}}. This tells us that the Riemann zeta function, with {{math|1 − ''p''<sup>−''s''</sup>}} taken out of the Euler product formula, is continuous in the [[p-adic number|{{mvar|p}}-adic number]]s on odd negative integers congruent modulo {{math|''p'' − 1}} to a particular {{math|''a'' ≢ 1 mod (''p'' − 1)}}, and so can be extended to a continuous function {{math|''ζ''<sub>''p''</sub>(''s'')}} for all {{mvar|p}}-adic integers <math>\mathbb{Z}_p,</math> the [[p-adic zeta function|{{mvar|p}}-adic zeta function]].

=== Ramanujan's congruences ===

The following relations, due to [[Ramanujan]], provide a method for calculating Bernoulli numbers that is more efficient than the one given by their original recursive definition:

:<math>\binom{m+3}{m} B_m=\begin{cases}
\frac{m+3}{3}-\sum\limits_{j=1}^\frac{m}{6}\binom{m+3}{m-6j}B_{m-6j}, & \text{if } m\equiv 0\pmod 6;\\
\frac{m+3}{3}-\sum\limits_{j=1}^\frac{m-2}{6}\binom{m+3}{m-6j}B_{m-6j}, & \text{if } m\equiv 2\pmod 6;\\
-\frac{m+3}{6}-\sum\limits_{j=1}^\frac{m-4}{6}\binom{m+3}{m-6j}B_{m-6j}, & \text{if } m\equiv 4\pmod 6.\end{cases}</math>

=== Von Staudt–Clausen theorem ===
{{main|Von Staudt–Clausen theorem}}
The von Staudt–Clausen theorem was given by [[Karl Georg Christian von Staudt]]{{r|vonStaudt1840}} and [[Thomas Clausen (mathematician)|Thomas Clausen]]{{r|Clausen1840}} independently in 1840. The theorem states that for every {{math|''n'' > 0}},
: <math> B_{2n} + \sum_{(p-1)\,\mid\,2n} \frac1p</math>
is an integer. The sum extends over all [[prime number|primes]] {{math|''p''}} for which {{math|''p'' − 1}} divides {{math|2''n''}}.

A consequence of this is that the denominator of {{math|''B''<sub>2''n''</sub>}} is given by the product of all primes {{math|''p''}} for which {{math|''p'' − 1}} divides {{math|2''n''}}. In particular, these denominators are [[square-free]] and divisible by&nbsp;6.

=== Why do the odd Bernoulli numbers vanish? ===
The sum

:<math>\varphi_k(n) = \sum_{i=0}^n i^k - \frac{n^k} 2</math>

can be evaluated for negative values of the index {{math|''n''}}. Doing so will show that it is an [[odd function]] for even values of {{math|''k''}}, which implies that the  sum has only terms of odd index. This and the formula for the Bernoulli sum imply that {{math|''B''<sub>2''k'' + 1 − ''m''</sub>}} is 0 for {{math|''m''}} even and {{math|2''k'' + 1 − ''m'' > 1}}; and that the term for {{math|''B''<sub>1</sub>}} is cancelled by the subtraction.  The von Staudt–Clausen theorem combined with Worpitzky's representation also gives a combinatorial answer to this question (valid for ''n'' > 1).

From the von Staudt–Clausen theorem it is known that for odd {{math|''n'' > 1}} the number {{math|2''B''<sub>''n''</sub>}} is an integer. This seems trivial if one knows beforehand that the integer in question is zero. However, by applying Worpitzky's representation one gets

: <math> 2B_n =\sum_{m=0}^n (-1)^m \frac{2}{m+1}m! \left\{{n+1\atop m+1} \right\} = 0\quad(n>1 \text{ is odd})</math>

as a ''sum of integers'', which is not trivial. Here a combinatorial fact comes to surface which explains the vanishing of the Bernoulli numbers at odd index. Let {{math|''S''<sub>''n'',''m''</sub>}} be the number of surjective maps from {{math|1={1, 2, ..., ''n''}}} to {{math|1={1, 2, ..., ''m''}}}, then {{math|''S''<sub>''n'',''m''</sub> {{=}} ''m''!<big><big>{</big></big>{{su|p=''n''|b=''m''|a=c}}<big><big>}</big></big>}}. The last equation can only hold if

: <math> \sum_{\text{odd }m=1}^{n-1} \frac 2 {m^2}S_{n,m}=\sum_{\text{even } m=2}^n \frac{2}{m^2} S_{n,m} \quad (n>2 \text{ is even}). </math>

This equation can be proved by induction. The first two examples of this equation are

:{{math|1=''n'' = 4: 2 + 8 = 7 + 3}},
:{{math|1=''n'' = 6: 2 + 120 + 144 = 31 + 195 + 40}}.

Thus the Bernoulli numbers vanish at odd index because some non-obvious combinatorial identities are embodied in the Bernoulli numbers.

=== A restatement of the Riemann hypothesis ===
The connection between the Bernoulli numbers and the Riemann zeta function is strong enough to provide an alternate formulation of the [[Riemann hypothesis]] (RH) which uses only the Bernoulli numbers. In fact [[Marcel Riesz]] proved that the RH is equivalent to the following assertion:{{r|Riesz1916}}

:For every {{math|''ε'' > {{sfrac|1|4}}}} there exists a constant {{math|''C''<sub>''ε''</sub> > 0}} (depending on {{math|''ε''}}) such that {{math|{{abs|''R''(''x'')}} < ''C''<sub>''ε''</sub>''x''<sup>''ε''</sup>}} as {{math|''x'' → ∞}}.

Here {{math|''R''(''x'')}} is the [[Riesz function]]

: <math> R(x) = 2 \sum_{k=1}^\infty
\frac{k^{\overline{k}} x^{k}}{(2\pi)^{2k}\left(\frac{B_{2k}}{2k}\right)}
= 2\sum_{k=1}^\infty \frac{k^{\overline{k}}x^k}{(2\pi)^{2k}\beta_{2k}}. </math>

{{math|''n''<sup>{{overline|''k''}}</sup>}} denotes the [[Pochhammer symbol#Alternate notations|rising factorial power]] in the notation of [[D. E. Knuth]]. The numbers {{math|''β''<sub>''n''</sub> {{=}} {{sfrac|''B''<sub>''n''</sub>|''n''}}}} occur frequently in the study of the zeta function and are significant  because {{math|''β''<sub>''n''</sub>}} is a {{math|''p''}}-integer for primes {{math|''p''}} where {{math|''p'' − 1}} does not divide {{math|''n''}}.  The {{math|''β''<sub>''n''</sub>}} are called ''divided Bernoulli numbers''.