TITLE: Congruence involving Bernoulli numbers
QUESTION [5 upvotes]: I need to show that $$2B_{2m}\equiv1\pmod{4}$$ for $m\ge2.$ This should be something easy using (at most) Claussen and von Staudt's theorem, but I haven't been successful yet. Writing $B_{2m}=U_{2m}/V_{2m},(U_{2m},V_{2m})=1,V_{2m}>0$ I know $$V_{2m}\equiv2U_{2m}\pmod4,$$
but since $2\mid V_{2m},$ this is not very useful. I also wanted to try Hensel's lifting lemma for $f(x)=2x-1$ (because $2B_{2m}\equiv1\pmod2$ holds), but $f'(x)\equiv0\pmod2,$ so this is not the right way as well. Any other tips?
Remark. I needed this result for finishing the following problem.

Let $q$ be a prime number such that $2q+1$ is composite. Show that the numerator of the Bernoulli number $B_{2q}$ is divisible by a prime of the form $4n+3$.

The congruence mentioned above is a final step in my solution.

REPLY [2 votes]: By Faulhaber's formula for the $2m$-th powers,$$2^{2m}+1=\frac1{2m+1}\left(2^{2m+1}+\frac{2^{2m}}2+\cdots
+(2m+1)2B_{2m}\right).$$
All terms but the term are multiples of $4$ so we get $1\equiv 2B_{2m}
\pmod4$.