This problem is from Borevich&Shafarevich' book.
Let $P=\{p_1,\ldots,p_s\}$ be a set of prime numbers greater than 3 and let $q$ be a positive integer such that $q\equiv1\pmod M$ where $M=(p_1-1)\cdots(p_s-1).$ Show that none of the prime numbers $p_1,\ldots,p_s$ divide the numerator of the fraction $B_{2q}/(2q)$.
To be honest, i was not able to solve this even for $s=1.$ Altough I see some blurred pattern behind the problem and known properties like $B_{2q}/(2q)$ is $p$-integral or $1^{2q}+2^{2q}+\cdots+p^{2q}\equiv B_{2q}\cdot p\pmod{p^2}$ for $p\in P$ since $p-1\nmid2q$.
It seems to be related to irregular primes, since Carlitz's proof is also using (almost) the same product (see (2.4)). (It is mentioned before that the statemnet of the problem and one another should be used for proof, that there exist infinitely many irregular primes of the form $4n+3$.)
For any $s$, we have $$2q \equiv 2 \pmod {p_s-1}$$ Moreover, $2$ is non divisible by $p_s-1$, since by hypothesis $p_s-1 >2$.
Then, by Kummer's congruence, for any $s$, $$\frac{B_{2q}}{2q}-\frac{B_{2}}{2}\equiv 0 \pmod {p_s}.$$ Now, if $\frac{B_{2q}}{2q} \equiv 0 \bmod p_s$, we would have $-\frac{1}{12} \equiv 0 \bmod p_s$, which is impossible.