how to find all integer $n$ such that $ n\mid2^{n!}-1$

I find:

Of course $2 \nmid n$. We prove that, if $2 \nmid n$ then $n \mid 2^{n!}-1$. $2 \nmid n \iff n = 2k+1 , k \ge 0$, we'll prove:

$2^{(2k+1)!} \equiv 1\pmod{2k+1}$

Let $n = p_1^{a_1}\cdot p_2^{a_2} \cdot ... \cdot p_s^{a_s}$, we'll prove that:

$2^{(2k+1)!} \equiv 1 \pmod{p_1^{a_1}}$

Let $t = ord_{p_1^{a_1}}2 \iff 2^t \equiv 1\pmod{p_1^{a_1}}$, so:

$t \mid (2k+1)! \iff (2k+1)! = l\cdot t \iff l = \frac{(2k+1)!}{t} \in \mathbb{Z}_{+}$

And $2^t \equiv 1\pmod{p_1^{a_1}}/^l \Rightarrow 2^{(2k+1)!} \equiv 1\pmod{p_1^{a_1}}$

Analogously we show divisibility $2^{(2k+1)!}-1$ by $p_2^{a_2} , ... , p_s^{a_s}$


2 Answers 2


We know that $2^{\phi(n)} \equiv 1 \pmod n$ and $\phi(n) < n$.
So, $\phi(n)$ must be a factor of $n!$ that is $x\phi(n)=n!$ for some $x.$
$2^{\phi(n)x} \equiv 2^{n!} \equiv 1 \pmod n$
Hence, any odd $n$ will satisfy the given relation.


$n$ needs to be odd as $2^{n!}-1$ is odd.Let $n=2k+1$ for some $k\in \Bbb Z.$ Now, $2^{\phi(2k+1)}\equiv1 \pmod {2k+1}$ as $\gcd(2,2k+1)=1$. Therefore, $2^{(2k+1)!}\equiv 1\pmod{2k+1}$ to hold $\phi(2k+1)$ must divide $(2k+1)!$. As we know $\phi(2k+1)\lt 2k+1$, therefore, it exists somewhere in the product $1\times 2\times ... \times (2k+1)$. Hence, $\phi(2k+1)$ divides $(2k+1)!\implies $$\frac{2^{n!}-1}{n}$ is an integer for all odd $n\in\Bbb Z$.


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