Find other factors of the euler function of $2^k-1$ where $k\ge 2$ is an integer, except $2$.

Question

Euler function of $x$ is denoted by $\varphi(x)$, which is defined in here (https://en.wikipedia.org/wiki/Euler%27s_totient_function ).

When $k$ is prime. @egreg comments that there does not exist a formula to obtain $\varphi(2^k-1)$. @Dietrich Burde comments that there need not be other factors of $\varphi(2^k-1)$ different from $2$ and $k$. However, if $k$ is not prime, it seems that finding factors of $\varphi(2^k-1)$ is not obvious.

1. Motivation ($k$ is prime.) Let $n=2^k-1$, where $k$ is prime. In $\mathbb{Z}_{n}$, the group $G=\langle 2 \rangle$ is of order $k$. Since the order of the multiplicative group $$\mathbb{Z}_{n}^*=\{ x \in \mathbb{Z}_{n} \mid \gcd(x,n)=1\}$$ is $\varphi(n)$, it should have $$k \mid \varphi(2^k-1).$$

   Proposition 1: If $k$ is prime, then $2 \mid \varphi(2^k-1)$ and $k \mid \varphi(2^k-1)$.

2. Problem

   Find other factors of $\varphi(2^k-1)$ where $k\ge 2$ is an integer, except $2$. Or proof some conjectures below.

3.My trying($k$ is not prime.)

[ <k, EulerPhi(2^k-1)> : k in [2..200] | (EulerPhi(2^k-1) mod 2 ne 0) or ( EulerPhi(2^k-1) mod k ne 0) ];

By the above magma program, it seems that $2$ and $k$ are two factors of $\varphi(2^k-1)$ for $2\le k \le 200$.

Proposition 2: $2 \mid \varphi(2^k-1)$ for $k\ge 2$.

Proof. Note that $2^k-1$ is odd. It follows that $\varphi(2^k-1)$ is even. QED.

For $k$ is not prime, it only has $2^k\equiv 1 \pmod{2^k-1}$ which is not sufficient to show that the group $G=\langle 2 \rangle$ is of order $k$.

conjecture 1: $k \mid \varphi(2^k-1)$ for $k\ge 2$.

Moreover, by the following magma program, it conjectures that $3\mid \varphi(2^k-1)$ if $5 \mid k$ or $7 \mid k$. It would be better if the sufficient and necessary condition of $3 \mid \varphi(2^k-1)$ can be given.

conjecture 2: $3\mid \varphi(2^k-1)$ if $5 \mid k$ or $7 \mid k$.

[ <k, Factorization(EulerPhi(2^k-1))> : k in [5..500 by 5] | (EulerPhi(2^k-1) mod 2 ne 0) or ( EulerPhi(2^k-1) mod 3 ne 0) ];
[ <k, Factorization(EulerPhi(2^k-1))> : k in [7..210 by 7] | (EulerPhi(2^k-1) mod 2 ne 0) or ( EulerPhi(2^k-1) mod 3 ne 0) ];

Answer 1

Conjecture 3 is clearly true, and here is a generalization: if $d \mid k$ and $2^d-1$ is a Mersenne prime, then $2^d - 2$ divides $\varphi(2^k -1)$.

Proof: if $d \mid k$ then there exists $l \in \mathbb N$ such that $k = dl$, therefore

$$\varphi(2^k - 1) = \varphi \left( (2^d)^l - 1 \right) = \varphi \left( (2^d - 1) \sum_{i=0} ^{l-1} (2^d)^i \right) = \varphi(2^d - 1) \ \varphi \left( \sum_{i=0} ^{l-1} 2^{di} \right) = (2^d - 2) \ \varphi \left( \sum_{i=0} ^{l-1} 2^{di} \right) \ .$$

Now, notice that if $d \in \{5,7\}$ then $2^d - 2 \in \{30,126\}$, and these numbers are both divisible by $3$.