proving that elements of a sequence are pairwise coprime For the sequence $a_n = 6^{2^n} + 1$: 
How can I prove that the elements of this sequence are pairwise coprime, i.e. prove that if $m$ is not equal to $n$ then $\gcd(a_m, a_n) = 1$.
(begin by proving that $a_n∣ (a_{n+1} - 2)$)
 A: Suppose the prime $p$ divides $a_{n}$ and $a_{m}$, for $n > m$.
Then $p$ divides the remainder $r$ of the division of $a_{n}$ by $a_{m}$, as $r = a_{n} - a_{m} q$ for some $q$.
The remainder $r$ can be calculated as
$$
6^{2^{n}} + 1 = (6^{2^{m}})^{2^{n-m}} + 1 \equiv (-1)^{2^{n-m}} + 1 = 2 \pmod{6^{2^{m}} + 1}.
$$
So $r = 2$, and $p$ could only be $2$, whereas the $a_{n}$ are all odd.
A: Note that $\, \ a_{\large n+1}\! = (6^{\large 2^{\Large n}})^{\large 2}+1 = (a_{\large n}\!-\!1)^{\large 2}+1= f(a_{\large n}),\ \  f(x) = (x\!-\!1)^2+1$
${\rm mod}\ a_{\large n}\!:\,\ \color{#c00}{a_{\large n+1}}= \ \ f(a_{\large n})\ \ \equiv f(0) \equiv \color{#c00}2$
$\phantom{{\rm mod}\ a_{\large n}\!:}\,\ \ \color{#0a0}{a_{\large n+2}}= f(\color{#c00}{a_{\large n+1}})\equiv f(\color{#c00}2) \equiv\color{#0a0} 2$
$\phantom{{\rm mod}\ a_{\large n}\!:}\,\ \ a_{\large n+3}= f(\color{#0a0}{a_{\large n+2}})\equiv f(\color{#0a0}2) \equiv 2$
$\phantom{{\rm mod}\ a_{\large n}\!:}\qquad\quad\ \vdots$
$\phantom{{\rm mod}\ a_{\large n}} a_{\large n+k+1}= f(a_{\large n+k})\equiv f(2) \equiv 2$ 
$\ \ $ i.e. $\,\quad a_{\large n+k+1}\! =\! f^{\large k}(\color{#c00}{a_{n+1}})\!\equiv\! f^{\large k}(\color{#c00}2)\equiv 2\ $ because $\,2\,$ is a fixed point of $f,\,$ i.e. $\,f(2) = 2$
A: $$a_{n+1}-2=6^{2^{n+1}}-1=(6^{2^n}-1)(6^{2^n}+1)=(6^{2^{n-1}}-1)(6^{2^{n-1}}+1)(6^{2^{n}}+1)=(6^{2^0}-1)(6^{2^{0}}+1)(6^{2^{1}}+1)\cdots(6^{2^{n-1}}+1)(6^{2^n}+1)=5a_1a_2\cdots a_n$$
Hence $\gcd(a_{n+1},a_k)\leq 2$ where $k=1,2,\cdots ,n$ and $\gcd(a_{n+1},a_k)\not= 2$ since non of $a_n$ are even.
