Exercise 5, § 2

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

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

b) If $r$ and $s$ are any two integers show that

c) Let $p$ be a prime number. Show that if $n$ is divisible by $p-1$ one has $S_n(p) \equiv -1$ (mod. $p$), and if $n$ is not divisible by $p-1$ then $S_n(p) \equiv 0$ (mod. $p$) (if $p$ does not divide the integer $g$ remark that $S_n(p) \equiv g^n S_n(p)$ (mod. $p$)).