Exercise 5, § 2

If one sets $S_n(x) = \frac{1}{n+1} (B_{n+1}(x) - B_{n+1}(0))$, one has, for every integer $a > 0$

a) Now prove that for every integer $n \geq 0$ and every integer $a > 0$, one has $2S_{2n+1}(a) \equiv 0$ (mod. $a$) (consider the sum $k^{2n+1} + (a-k)^{2n+1}$).

b) If $r$ and $s$ are any two integers $0$, prove that

c) Let $p$ be a prime number. Prove that if $n$ is divisible by $p - 1$, one has $S_n(p) \equiv -1 \pmod{p}$, and if $n$ is not divisible by $p - 1$, $S_n(p) \equiv 0 \pmod{p}$ (if $p$ does not divide the integer $g$, note that $S_n(p) \equiv g^n S_n(p) \pmod{p}$).