Bourbaki

Exercise 6, § 2

Functions of a Real Variable, VI, 293

a) The rational numbers bnb_n having been defined by the formula (20) of VI, p. 275, one denotes by dnd_n the denominator >0> 0 of bnb_n written as an irreducible fraction. Show that no prime factor of dnd_n can be >n+1> n+1 (use the induction formula (23) of VI, p. 276)

b) Show that for every integer p>0p > 0 and every integer n>0n > 0

Sn(p)=bnp+(n1)p2bn−1p+⋯+(nr)p′r+1bn−rp+⋯+pn+1n+1. S_n(p) = b_n p + \binom{n}{1} \frac{p}{2} b_{n-1} p + \cdots + \binom{n}{r} \frac{p'}{r+1} b_{n-r} p + \cdots + \frac{p^{n+1}}{n+1}.

c) Deduce from b) by recursion on nn that, for every prime number pp the denominator of Sn(p)−bnpS_n(p) - b_n p written as an irreducible fraction, is not divisible by pp (observe that p′p' cannot divide r+1r+1).

d) Deduce from c) that the number

bn−∑pSn(p)p b_n - \sum_p \frac{S_n(p)}{p}

where pp runs through the set of prime numbers p≤n+1p \leq n + 1 and nn is even, is an integer. Deduce that

b2n+∑p1p b_{2n} + \sum_p \frac{1}{p}

where pp runs through the set of prime numbers such that p−1p - 1 divides 2n2n, is an integer (Eisenstein-von Staudt theorem; use exerc. 5 c).

Solution

No solution has been written yet.

Transcribed from the printed volume. Not translated.