Coprime numbers - Need help with proof Let $a \in \mathbb{Z}$ be an odd number. Prove that the numbers $$a^{2^n} + 2^{2^n}, a^{2^m} + 2^{2^m}$$
are relatively prime (coprime) for all $m.n\in\mathbb{Z}^+$ $(m\neq n)$.
Any tipps?
 A: suppose the they are not coprime and suppose $n<m$ and take $F_n=a^{2^n}+2^{2^n}$

*

*First we have $a^{2^n}\equiv -2^{2^n} \mod F_n$  and use it to find  $$F_m \equiv \left(a^{\displaystyle 2^n}\right)^{\displaystyle 2^{m-n}}+2^{2^m}\equiv \left(-2^{\displaystyle 2^n}\right)^{\displaystyle 2^{m-n}}+2^{2^m}\equiv \left(2^{\displaystyle 2^n}\right)^{2^{m-n}}+2^{2^m} \equiv 2\cdot 2^{2^{m}}\mod F_n $$

*Now if we take $d=\gcd(F_n,F_m)$ using the first quote we have $d$ divides $2\cdot 2^{2^m}$ and because $a$ is odd we have $F_n,F_m$ are odd, so $d$ must be odd hence $d$ is an odd divisor of $2\cdot 2^{2^m}$ so $d=1$
A: Below let $\ b = 2\ $ and $\ f(n) = 2^n$. Here, since all $\,c_i\,$ are odd, the gcd $\, d = 1.$
Theorem $\ $ Let $\,a,b\in\Bbb Z\,$ be coprime, and $\,c_n = a^{f(n)}+b^{f(n)}\,$ for all integers $\,n> 0,\,$ where  the function $\,f(n)\,$  satisfies $\ 0<f(n)\in \Bbb N\ $ and $\ n< m\,\Rightarrow\,\color{#90f}{2f(n)\mid f(m)}\ .\ $ Then $$\,{\rm the\ gcd}\,\  d  = (c_m,c_n)\,\ {\rm is}\,\ 1\ {\rm or}\,\  2,\ \ {\rm for\ all}\,\ m,n>0\qquad $$
Proof $\ \ $ ${\rm mod}\ d\!:\ c_n\equiv 0\,\Rightarrow\,\color{#c00}{a^{f(n)}\equiv -b^{f(n)}}\ $  By symmetry, wlog  $\ m > n$
${\rm mod}\ d\!:\ -b^{f(m)}\equiv a^{f(m)}\equiv (\color{#c00}{a^{f(n)}})^{\large \frac{f(m)}{f(n)}}\equiv (\color{#c00}{-b^{f(n)}})^{\large\color{#90f}{\frac{f(m)}{f(n)}}}\equiv b^{f(m)} $
So $\ d\mid 2b^{f(m)}.\ $  $\, (c_i,b) = (a^{f(i)}\!+b^{f(i)},b)= (a^{f(i)},b) = 1\ $ by $\ (a,b) = 1,\, $  $\, f(i) > 0$
So $\ (c_m,b)=1=(c_n,b)\,\Rightarrow\,((c_m,c_n),b)=1,\:$ so $\,d=(c_m,c_n)\mid 2b^{f(m)}\Rightarrow\, d\mid 2$
