Prove $n\mid \phi(2^n-1)$

Question

If $2^p-1$ is a prime, (thus $p$ is a prime, too) then $p\mid 2^p-2=\phi(2^p-1).$

But I find $n\mid \phi(2^n-1)$ is always hold, no matter what $n$ is. Such as $4\mid \phi(2^4-1)=8.$

If we denote $a_n=\dfrac{\phi(2^n-1)}{n}$, then $a_n$ is A011260, but how to prove it is always integer?

Thanks in advance!

Answer 1

Consider $U(2^n-1)$. Clearly $2\in U(2^n-1)$. It can also be shown easily that the order of $2$ in the group $U(2^n-1)$ is $n$. By Lagrange's theorem $|2|=n$ divides $|U(2^n-1)|=\phi(2^n-1)$.

Remark: Thank you for letting me know about this fact. It's interesting!

Answer 2

$(\mathbb{Z}/\mathbb{(a^n-1)Z})^*$ is a group of order $\phi(a^n-1)$ and gcd$(a,a^n-1)=1\Rightarrow a\in (\mathbb{Z}/\mathbb{(a^n-1)Z})^*$

We have $a^n\equiv 1\mod (a^n-1)$ and $a^k-1<a^n-1$ when ever $k<n$ so there does not exist $k<n$ such that the above condition holds. So the order of $a$ in $(\mathbb{Z}/\mathbb{(a^n-1)Z})^*$ is $n$. And as the order of each element divides the order of the group so we have $n|\phi(a^n-1)$

Putting $a=2$ we have the required result asked in the question.

Answer 3

I will use Lifting the Exponent Lemma(LTE).

Let $v_p(n)$ denote the highest exponent of $p$ in $n$.

Take some odd prime divisor of $n$, and call it $p$. Let $j$ be the order of $2$ modulo $p$.

So, $v_p(2^n-1)=v_p(2^j-1)+v_p(n/j)>v_p(n)$ as $j\le p-1$.

All the rest is easy. Indeed, let's pose $n=2^jm$ where $m$ is odd.

Then $\varphi\left(2^{2^jm}-1\right)=\varphi(2^m-1)\varphi(2^m+1)\varphi(2^{2m}+1)\cdots\varphi\left(2^{2^{j-1}}m+1\right)$. At least $2^j$ terms in the right side are even.

Answer 4

Hint: Clearly $2$ has order $n$, modulo $2^n-1$.

Further Hint: We know that $2^{ \phi(2^n -1)} \equiv 1 \pmod{2^n-1}$.