If a $\epsilon$ ℤ, show $a(a^m-1)B_m$ $\epsilon$ ℤ for all $m>0$

Question

For all $m=2k+1$, $k=1,2,...$, $a(a^m-1)B_m=0$ $\epsilon$ ℤ since $B_m=0$. However, I'm not really sure how to proceed for $B_{2k}.$

Answer 1

This follows from von Staudt-Clausen theorem which says that the denominator of $B_{2n}$ is given by $$\prod_{p,p-1|2n}p.$$ By Fermat's little theorem, for every $p$ given in the above product and every integer $a$, one has $$a^p\equiv a~({\rm mod~}p),$$ or equivalently $${\rm either~}p|a~{\rm or~}a^{p-1}-1\equiv 0~({\rm mod~}p),$$ where in the latter case, $p|(a^{2n}-1),$ since $a^{2n}=(a^{(p-1)})^k$ for some $k$. Accordingly all the factors in the denominator of $B_{2n}$ are being canceled when multiplying by $a(a^{2n}-1)$. QED