How to prove that $2^n$ has $(2\cdot n + 2)$ divisors So we have $2^n$ and we can see that it has $(2\cdot n+2)$ divisors.
I tried to prove it with induction:
base:
$$2^1 \text{has divisors: } \pm 1,\pm 2 $$
$$2^2 \text{has divisors: }\pm 1, \pm 2, \pm 4$$
$$\text{...and so on...}$$
However I do not know how to do the induction step. My idea would be to do a recursive sequence of the number of divisors and then show that it converges to $2\cdot n +2$, but still I do not know how to do it correctly. I would need help with this problem.
 A: This is an approach that does not use induction.
In the general case for $a=\prod p_i^{k_i}$, the number of (positive) divisors is given by $\tau(a)=\prod (k_i+1)$. This is because each divisor can be created by taking the first prime factor of a divisor in any of $k_1+1$ ways (i.e. $0,1,2,\dots,k_1$), and so on for each succeeding prime factor, yielding $k_2+1,k_3+1$, etc.
If there is only one prime factor, as in $2^n$, then the number of (positive) divisors is simply $n+1$
Your question indicates you wish to include negative divisors as well, each of which occur once for each positive divisor, so the total number of divisors of positive and negative divisors of $2^n$ is just $2(n+1)=2n+2$, which is what you wish to show.
A: Consider trying to prove these three statements individually.
If $0 \le n \le M$ then $\pm 2^n$ is a divisor of $2^M$.
If $n > M$ then $\pm 2^n$ are neither divisors of $2^M$.
If $k$ is not a power of $2$ then $k$ is not a divisor of $2^M$.
If you can can prove that you are basically done.
The divisors of $2^n$ will be $2^k; 0< k \le n$  which are precisely $\pm 2^0; \pm 2^1; \pm 2^2,...., \pm 2^{n-1}, \pm 2^n$.  And as there are $n+1$ values of $k: k \le n$ (namely $k = 0, 1,2,......,n$) and we have $2$ divisors for each value of $k$ (that is we have $2^k$ and $-2^k$ are two divisors) we have $2(n+1) = 2n+ 2$ divisors.
Now just prove those three statements.
A: Proof without induction:
Clearly $2^n$ is a power of a prime 2,so it has divisors only of the form $±2^k$ where $k ≤ n$
Then $k$ can vary from $0$ to $n$, and for each value of $k$ there are two divisors - one positive and another negative
Then the divisors will be,
$$-2^n , -2^{n-1},..., -2^0, 2^0, 2^1,..., 2^{n-1} , 2^n$$
Total no of divisors $2n+2$
