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.
Note that for $n\geq 2$ $$0=2\sum_{k=0}^{2n}\binom {2n+1}kB_k=1-2n+\binom{2n+1}{2}\frac{1}{3}+\sum_{k=2}^{n-1}\binom{2n+1}{2k}(2B_{2k})+(2n+1)(2B_{2n}).$$ Now if $2B_{2m}\equiv 1\pmod{4}$ for $m=2,\dots,n-1$ then $$(2n+1)(2B_{2n})\equiv -1+2n-3\binom{2n+1}{2} -\sum_{k=2}^{n-1}\binom{2n+1}{2k}\\=-4^n-4n^2+2n+1\equiv 2n+1\pmod{4}.$$ and, since $(2n+1)$ is odd, it follows that $2B_{2n}\equiv 1\pmod{4}$.