if $2^n+1$ is prime, then n = $2^r$ and $(x - y) \mid(x^k - y^k$) only with natural numbers I need to prove that if $2^n + 1$ is prime, then $n = 2^r$ for a natural number $r$.
I do know how to prove it with the lemma $(x - y) \mid (x^k - y^k)$, but in order to prove it with this lemma, I need to substitute $y$ with $-1$ and my problem is, that in my case, this lemma is only given if $x$, $y$, $k$ are natural numbers.
Could you please tell me how to prove it without needing a number from $Z$ to prove this theorem?!
Thank you!
 A: Suppose that $p$ is odd prime divisior of $n$. Then $n=p\cdot m$, 
$$ 2^n +1 = (2^m)^p + 1^p = (2^m + 1)\cdot ((2^m)^{p-1}-(2^m)^{p-2}+ \cdots + 1)$$
and each factor greater than $1$. So $2^n +1 $ isn't prime which gives a contradiction.
Some Notes:
1) We use the identity $x^{2k+1}+y^{2k+1}=(x+y)\cdot (x^{2k} - x^{2k-1}y + x^{2k-2}y^2 - \cdots + x^2y^{2k-2}- xy^{2k-1}+y^{2k})$. Proof of it is not difficult. By distribution property
$(x+y)\cdot (x^{2k} - x^{2k-1}y + x^{2k-2}y^2 - \cdots + x^2y^{2k-2}- xy^{2k-1}+y^{2k}) \\ =\left(x^{2k+1}-x^{2k}y + x^{2k-1}y^2 - \cdots -x^2y^{2k-1}+xy^{2k}\right)+\left(x^{2k}y-x^{2k-1}y^2 + x^{2k-2}y^3 - \cdots -xy^{2k}+y^{2k+1}\right) \\ =x^{2k+1}+y^{2k+1}$.
2) We can see that $x^{2k} - x^{2k-1}y + x^{2k-2}y^2 - \cdots + x^2y^{2k-2}- xy^{2k-1}+y^{2k} >1$. Otherwise,if $$x^{2k} - x^{2k-1}y + x^{2k-2}y^2 - \cdots + x^2y^{2k-2}- xy^{2k-1}+y^{2k} =1$$
then 
$$2^n +1=2^m +1 $$
and $n=m$ which gives a contradiction.
A: In any commutative ring and for any $k\in\Bbb N_0$, 
$$ (x-y)\cdot\sum_{i=0}^{k}x^iy^{k-i}=x^{k+1}-y^{k+1},$$
as follows by induction.
