The denominator of a Bernoulli number is always **an even** integer. Why?

Question

Apparently, the denominator of a Bernoulli number is always an even integer. Where does this come from?

Answer 1

This follows from the Von Staudt–Clausen theorem which states that for even Bernoulli numbers $B_{2n}$ (odd ones are almost all 0) we have

$$ B_{{2n}}+\sum _{{(p-1)|2n}}{\frac 1p}\in \mathbb{Z } $$

hence for $p = 2$ we have $p - 1 = 1$ which divides all $2n$ so $\frac1p$ always appears on the sum, hence the denominator of $B_{2n}$ must be divisible by 2 or else the total would not be an integer.

Maybe there is a simpler way but this theorem is a very general tool for understanding denominators of Bernoulli numbers.