Let $a=2^{2^{35}} +1, b=2^{2^{21}} +1 $. Then $\gcd(a,b)$ is
My little work :
$$a -2 =(2^{2^{22}})^K -1$$ where $K=2^{14}$.
$(2^{2^{22}})^K -1$ is divisible by$(2^{2^{22}}) -1$ $$\implies a-2= (2^{2^{22}} -1)m ,$$ where $m$ is some integer
$$\implies a-2= ((2^{2^{21}})^2 -1 )m $$
$$\implies a-2=((2^{2^{21}}) -1 )bm $$ Help me to figure out things after this.
Let $\ c = 2^{\large 2^{\Large 21}}$ so $\ 2^{\large 2^{\Large 35}}\! = (2^{\large 2^{\Large 21}})^{\large 2^{\Large 14}}\! =\, c^{{\large 2^{\Large 14}}}\!.\ $ By one step of the Euclidean gcd algorithm
our gcd is $\,(\color{#c00}c^{\large 2^{\Large 14}}\!\!+1,c\!+\!1)\, =\, ((\color{#c00}{-1})^{\large 2^{\large 14}}\!\!+1,c\!+\!1) = (2,c\!+\!1) = 1\,$ by $\,c\!+\!1\,$ odd,
by $\,{\rm mod}\ c\!+\!1\!:\,\ \color{#c00}{c\equiv -1}\,\Rightarrow\, \color{#c00}c^N\equiv (\color{#c00}{-1})^N\ $
Remark $\ $ If Euclid's algorithm or mod are unfamiliar you can instead use divisibility
$$ c\!+\!1\mid c^2\!-\!1\mid c^{2N}\!-1\ \ \ {\rm so} \ \ \ c\!+\!1 \mid (c^{2N}\!-\!1)+2\iff \,c\!+\!1\mid 2$$
Or, using orders, if prime $\,p\mid 2^{\large 2^{\Large 35}}\!\!-1\,$ then $\,{\rm mod}\ p\!:\ 2^{\large 2^{\Large 35}}\!\equiv -1\,$ so squaring shows $\,2\,$ has order $\,2^{\large 36}$ by the Order Test. Hence $\,2^{\large 2^{\Large N}}\!\!\not\equiv 1\,$ for all $\,N < 36,\,$ so $\,p\nmid 2^{\large 2^{\Large N}}\!\!-1.\,$ Generally this argument proves that Fermat numbers with distinct indices are coprime.
