Show that if m> n, then $a^{2^n} + 1$ is a divisor of $a^{2^m} -1$. Show that if $ a, m, n \in \mathbb Z ^ + $ with $ m \neq n $, then $$\gcd(a^{2^m}+1,a^{2^n}+1)=\begin{cases} 1,& \text{ if } a\ \text{is even;} \\ 2 & \text{ if } a\ \text{is odd.} \end{cases}$$
Proof: Suppose without loss of generality that $ m> n $, and observe the following $$a^{2^m}-1=(a^{2^{m-1}}+1)(a^{2^{m-2}}+1)\cdots (a^{2^{n}}+1)(a^{2^{n}}-1).$$ Then, we have to $a^{2^m}+1=(a^{2^n}+1)q+2$, with $q\in\mathbb Z$. So, $$\gcd(a^{2^m}+1,a^{2^n}+1)=(a^{2^n}+1,2).$$ So, if $ a $ is even, then $ a ^ {2 ^ n} + 1 $ is odd, so the gcd is $ 1 $, and $ 2 $ otherwise.
There is a part that I do not understand, and it is the part where it says that $\gcd(a^{2^m}+1,a^{2^n}+1)=(a^{2^n}+1,2).$. Why the right side of the gcd becomes $2$, can someone explain that part to me?
