Find $\gcd(a^{2^n}+1,a^{2^m}+1)$ where $m,n$ are two distinct positive integers.
Let $\gcd(a^{2^n}+1,a^{2^m}+1)=d$. Then $d\mid a^{2^n}+1$ and $d\mid a^{2^m}+1$. So, $a^{2^n}\equiv -1 \pmod d$ and $a^{2^m}\equiv -1 \pmod d$.
Let $m>n$. Then $m-n>0$. From $a^{2^n}\equiv -1 \pmod d$ we get $(a^{2^n})^{2^{m-n}}\equiv1 \pmod d$. But $(a^{2^n})^{2^{m-n}}=a^{2^n\cdot 2^{m-n}}=a^{2m}\equiv -1 \pmod d$. Then, $1\equiv -1\pmod d$, i.e., either $d=1$ or $d=2$.
How do I find the value of $a$ for $d=1$ or $d=2$?
