home/LANG:EN/CAT:Mathematics/ARTICLE:/Discrete mathematics/ Number theory/

von Staudt-Clausen Theorem

The von Staudt-Clausen Theorem

The von Staudt-Clausen theorem is a result in number theory that relates the Bernoulli numbers to the denominators of certain Bernoulli polynomials. It was first conjectured by Ernst Eduard Kummer and proven by Georg von Staudt and Thomas Clausen in the mid-1800s.

Bernoulli Numbers and Polynomials

The Bernoulli numbers are a sequence of rational numbers that arise in many areas of mathematics, including number theory, combinatorics, and calculus. They are defined by the generating function:

\frac{t}{e^t - 1} = \sum_{n=0}^{\infty} \frac{B_n}{n!} t^n

where $B_n$ is the $n$ th Bernoulli number. The first few Bernoulli numbers are:

B_0 = 1, \quad B_1 = -\frac{1}{2}, \quad B_2 = \frac{1}{6}, \quad B_3 = 0, \quad B_4 = -\frac{1}{30}, \quad B_5 = 0, \quad B_6 = \frac{1}{42}, \quad \dots

The Bernoulli polynomials are a sequence of polynomials that depend on a parameter $t$ . They are defined by the generating function:

\frac{te^{xt}}{e^t - 1} = \sum_{n=0}^{\infty} \frac{B_n(x)}{n!} t^n

where $B_n(x)$ is the $n$ th Bernoulli polynomial. The first few Bernoulli polynomials are:

B_0(x) = 1, \quad B_1(x) = x - \frac{1}{2}, \quad B_2(x) = x^2 - x + \frac{1}{6}, \quad B_3(x) = x^3 - \frac{3}{2} x^2 + \frac{1}{2} x, \quad \dots

The von Staudt-Clausen Theorem

The von Staudt-Clausen theorem states that for any prime $p$ and any positive integer $m$ , the denominator of $B_{mp}$ is divisible by $p$ . In other words, if we write $B_{mp}$ as a fraction in lowest terms:

B_{mp} = \frac{a}{b}

where $a$ and $b$ are integers with no common factors, then $p$ divides $b$ .

This theorem has several important consequences in number theory. For example, it implies that the numerator of $B_{mp}$ is divisible by $p$ if $m$ is not divisible by $p$ , and that $B_{p-1}$ is congruent to $-1$ modulo $p$ for any prime $p > 2$ .

Proof

The proof of the von Staudt-Clausen theorem is quite involved and uses a combination of algebraic and combinatorial techniques. Here, we will only sketch the main ideas of the proof.

The first key observation is that the Bernoulli polynomials satisfy a certain recurrence relation:

B_n(x+1) - B_n(x) = nx^{n-1}

This relation can be used to express $B_{mp}(x)$ in terms of the lower-degree Bernoulli polynomials $B_0(x), B_1(x), \dots, B_{mp-1}(x)$ . Then, we can use the fact that the coefficients of the Bernoulli polynomials are rational numbers with denominators that are powers of $2$ , to show that the denominator of $B_{mp}(x)$ is divisible by $2^m$ .

The second key observation is that the Bernoulli numbers can be expressed as certain sums of binomial coefficients:

B_n = \sum_{k=0}^{n} (-1)^k \binom{n}{k} \frac{T_k}{k+1}

where $T_k$ is the $k$ th triangular number. Using this formula, we can show that the denominator of $B_{mp}$ is divisible by $p$ by examining the powers of $p$ that appear in the denominators of the binomial coefficients.

Conclusion

The von Staudt-Clausen theorem is a beautiful result in number theory that relates the Bernoulli numbers to the denominators of certain Bernoulli polynomials. Its proof involves a combination of algebraic and combinatorial techniques, and has important consequences in many areas of mathematics.

Links

クラウゼン・フォンシュタウトの定理[JA]