On the numerators of Bernoulli numbers

Question

Von Staudt-Clausen theorem implies that $pB_{2n} \in \mathbb{Z}_{p}$ for all primes $p$ and for all $n \in \mathbb{N}$. It means that the highest power of any prime that can occur in the denominator of any nonzero Bernoulli number is 1. But the same is true for the numerator any nonzero Bernoulli number i.e. $\frac{p}{B_{2n}} \in \mathbb{Z}_{p}$ for all primes $p$ and for all $n \in \mathbb{N}$. Does anybody know the proof of this?

Answer 1

Your claim is false...$B_{284}$ has numerator divisible by $37^2$.

See http://bernoulli.org/

In fact there is a txt file on that site that gives factorisations of Bernoulli numerators and the first square prime divisor appears in numerator of $B_{50}$.