Number Theory: The ultimate weapon

$2^{n}$ is always even for all positive integers.

Therefore, $2^{n} -1$ is always odd, irrespective of n.

Now if $n$ is even, there is no solution.

Therefore $n$ is odd.

But $2^{n} -1 \equiv 0 \pmod{n}$ ---required to prove

$\Rightarrow 2^{n} \equiv 1 \pmod{n}$ ---(1)

But $2^{n-1} \equiv 1 \pmod{n}$ ---- By Fermat's Little Theorem as $n$ is odd ---(2)

Condition (1) and Condition (2) simultaneously cannot be satisfied.

Therefore, $n=\boxed{1}$ (as all integers are divisible by 1)

Just did by trial and error. Put n=1,then value comes out to be 2-1=1. Put n =2,then value comes out to be 4-1=3 which is not equal to 1. As we increase the value for n,the difference between n and 2^n-1 increases. Therefore only 1 solution.