Divisor Of Mersenne Number

Let's solve this problem by dividing it into cases.

Case 1: $n$ is even .
If $n$ is even then $2^n - 1$ must also be even since an even number cannot divide an odd number, but it is not possible for $2^n - 1$ to be even when $n > 1$ is a positive integer.
Hence, $n$ cannot be even .

Case 2: $n$ is odd .

Let's define $p$ to be the smallest prime factor of $n$ and clearly $p \mid 2^n - 1 \Rightarrow 2^n \equiv 1 \pmod{p}$ .

Now let $k$ be the multiplicative order of $2$ modulo $p$ .
Clearly $k \mid n$ , but now by Fermat's Little Theorem $k \leq p-1$ .
This implies that $n$ has a factor $k$ which is positive and smaller than $p$ which means that $k = 1$ .
Now $2 \equiv 1 \pmod{p}$ which clearly is impossible for any prime $p$ .
Hence, $n$ cannot be odd .

Since $n$ can be neither even nor odd , the answer must be $\boxed{0}$

Since for any integer $n\geq 2$ , the number $2^n-1$ is always an odd integer, it can't be divisible by an even integer. Thus, such an $n$ , if exists at all, must be odd. Now, we have by Fermat's little theorem, for any odd integer $n$ $2^{\phi(n)}=1 \mod (n)$ This implies (via division theorem), that such an integer $n$ must be divisible by $\phi(n)$ . But we know that, $\phi(n)= n \prod_{p} (1-\frac{1}{p})= n \prod_{p} \frac{p-1}{p},$ where the product is over all prime factors of $n$ . Hence we require the following to be an integer $\frac{n}{\phi(n)}= \prod_{p} \frac{p}{p-1}$ But remember that $p$ 's are distinct primes, so can't be divisible by numbers of the form $p-1$ , unless $p-1=1$ , i.e., $p=2$ , but then $n$ would be even, which is not the case. $\blacksquare$

A solution can be found here