Exercise 9, § 2

To say that a rational number $r$ is a $p$-adic integer (\emph{Gen. Top.}, III, p. 322, exerc. 1) for $p$ a prime number, means that when $r$ is expressed as an irreducible fraction its denominator is not divisible by $p$; one writes $r \equiv 0$ (mod. $p$) to express the fact that $r / p$ is a $p$-adic integer, and $r \equiv r'$ (mod. $p$) is equivalent by definition to $r - r' \equiv 0$ (mod. $p$).

a) Let $m$ be a rational integer; the function $F_m(z) = \frac{m}{e^{mz} - 1} - \frac{1}{e^z - 1}$ is analytic for $|z|$ sufficiently small, so can be written as an entire series convergent on a neighbourhood of $z = 0$:

Show that also on a neighbourhood of $0$

where the $c_n$ are $p$-adic integers; deduce that the numbers $a_n = (m^n - 1) \frac{b_n}{n}$ are $p$-adic integers; further, one has $a_{n+p-1} \equiv a_n$ (mod. $p$).

b) Deduce from a) that if $p - 1$ does not divide $n$, then $b_n / n$ is a $p$-adic integer and that

(Kummer’s congruences). (Take for m an integer whose class mod. p is a generator of the multiplicative group of the invertible elements of $\mathbf{Z}/p\mathbf{Z}$ (Alg., VII. 12).)