Exercise 9, § 2

Saying that a rational number $r$ is a $p$-adic integer (TG, III, p. 84, exerc. 23) for a prime number $p$ means that, when $r$ is put in the form of an irreducible fraction, its denominator is not divisible by $p$; one writes $r \equiv 0 \pmod{p}$ to express that $r/p$ is a $p$-adic integer, and $r \equiv r' \pmod{p}$ is equivalent by definition to $r' - r \equiv 0 \pmod{p}$. a) Let $m$ be a rational integer; the function $F_m(z) = \frac{m}{e^{mx} - 1} - \frac{1}{e^x - 1}$ is analytic for $|z|$ sufficiently small and can therefore be written in the form of a convergent power series in a neighbourhood of 0

Now prove that one also has in 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; moreover one has $a_{n+p-1} \equiv a_n \pmod{p}$.

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

(Kummer congruences). (Take for $m$ an integer whose class mod. $p$ is a generator of the multiplicative group of invertible elements of $\mathbf{Z}/p\mathbf{Z}$ (A, VII, § 2, n° 4).)