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Is the following analogue of Fermat's Little Theorem for Bernoulli 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

$$ a^{2n} \equiv 1\pmod{D_{2n}}.$$

This question was posted in MSE 3 weeks back but it is still open. Hence posting in MO.

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    $\begingroup$ For the denominator of $B_{2n}$ itself this follows immediately from en.m.wikipedia.org/wiki/Von_Staudt%E2%80%93Clausen_theorem $\endgroup$
    – Fedor Petrov
    Commented Jan 18, 2022 at 6:15
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    $\begingroup$ I think this is true, see the comment by Peter J. Cameron to the OEIS entry A006863. This comment implies that $2n$ is a multiple of the exponent of the group of units of $D_{2n}$, which immediately implies the statement in your question. $\endgroup$
    – pregunton
    Commented Jan 18, 2022 at 9:24
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    $\begingroup$ Also this follows from the comment by T. Khovanova: "Michael Lugo (see link) conjectures, and Peter McNamara proves, that a(n) = gcd_{ primes p > 2n+1 } (p^(2n) - 1)." (take a prime congruent to $a$ modulo $D_n$) $\endgroup$
    – Fedor Petrov
    Commented Jan 18, 2022 at 12:37
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    $\begingroup$ @MaxAlekseyev Post it as an answer, I will accept it. $\endgroup$
    – Nilotpal Kanti Sinha
    Commented Jan 19, 2022 at 4:14

1 Answer 1

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A proof is essentially given in Section 5.1 of Notes on primitive lambda-roots by P. J. Cameron and D. A. Preece.

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