Conjectured analogue of Fermat's Little Theorem for Bernouli numbers

Question

Is the following analogue of Fermat's Little Theorem for Bernouli numbers true?

Let $D_{2n}$ be the denominator of $\frac{B_{2n}}{4n}$ where $B_n$ is the $n$-th Bernoulli number. If $\gcd(a, D_{2n}) = 1$ then

$$ a^{2n} \equiv 1 \;(\bmod\; D_{2n}) $$

Update 18-Jan-2022: Posted in MO

Answer 1

I found a partial proof that $a^{2n} - 1$ is divisible by the denominator of $D_{2n}$. Haven't been able to account for the additional factor $4n$.

The Von Staudt–Clausen theorem is that $D_{2n}$ is given by the product of all primes $p$ for which $p − 1|2n$. In particular, $D_{2n}$ is square free. Thus if $p_1, p_2, \ldots, p_k$ are the distinct prime factors of $D_{2n}$ then $$p_1p_2 \ldots p_k = D_{2n}$$ $$(p_1-1)(p_2-1)\ldots (p_k-1) = 2n$$

If $\gcd(a,D_{2n}) = 1$ then by Fermat's Little theorem,

$$ a^{(p_1-1)(p_2-1)\ldots (p_k-1)} \equiv 1 \;(\bmod\; p_1) $$ $$ a^{(p_1-1)(p_2-1)\ldots (p_k-1)} \equiv 1 \;(\bmod\; p_2) $$ $$\cdots$$ $$ a^{(p_1-1)(p_2-1)\ldots (p_k-1)} \equiv 1 \;(\bmod\; p_k) $$

Since the moduli are all distinct primes, they have no common factor. Thus

$$ a^{(p_1-1)(p_2-1)\ldots (p_k-1)} \equiv 1 \;(\bmod\; p_1 p_2 \ldots p_k) $$

i.e. $a^{2n}-1$ is divisible by the denominator of $D_{2n}$.

Answer 2

If you work out the powers of a prime dividing $D_{2n}$ you get the following which is a well-known stronger version of the Clausen-von Staudt result.

If $p=2$, $\nu_2(D_{2n})=\nu_2(n)+3$. If $p$ is odd and $(p-1)\mid 2n$ then $\nu_p(D_{2n})=\nu_p(n)+1$, otherwise $\nu_p(D_{2n})=0$.

Now for the congruence to hold you just need to check it modulo these powers of each prime and this is easy because of the structure of the group of units module a prime power.