Prove that $$gcd(2^{2^m}+1, 2^{2^n}+1)=1$$ if $m, n$ are positive integers such that $m \neq n$.
A Hints to solve the problem is also given in the book as follows:
Let $m>n$. Then $2^m=2^{n}2^{m-n}=2^n.2k$ for some integer $k$, (since $m>n$). Let $2^{2^n}=x$. Then $2^{2^m}-1=x^{2k}-1$ and it is divisible by $x+1$.
I am still unable to solve. Please help.
Let $A_n = 2^{2^n}+1$. Then: $$ A_{n+h} = (A_n-1)^{2^h}+1,$$ hence for any $h\geq 1$: $$ \gcd(A_{n+h},A_n)=\gcd((-1)^{2^h}+1,A_n)=\gcd(2,A_n)=1$$ since $A_n$ is odd.
Lemma $\rm\ \gcd(c+1,\ c^{2J}+1)\ =\ gcd(c+1,\:\color{darkorange}2),\, $ by gcd mod reduction,
since: $\,\ \rm\quad\! mod\ c+1\!:\ c^{2J}+1\equiv (-1)^{2J}\!+1\equiv \color{darkorange}2\, $ by $\rm\ c\equiv -1.\,$ $\small \bf QED\ $ In particular:
$\smash[t]{\, \rm gcd\underset{\overbrace{\large \color{#c00}{c+1_{\phantom{|}}\!}}}{({\it F}_N},\,\underset{\overbrace{\large \color{#0a0}{c^{\Large 2J}+1}}}{{\it F}_{N+K}}) = gcd\underset{\overbrace{\large \color{#c00}{c+1}}}{({\it F}_N},\color{darkorange}2) = 1},\,$ for $\rm\,\ \color{#c00}{c=2^{\large 2^{N}}}\!,$ $\ 2J = 2^{K},\,$ so $\,\color{#0a0}{c^{\large 2J}}\! = (2^{\large 2^{N}})^{\large 2^K}\!\! = \color{#0a0}{2^{\large 2^{N+K}}}$
Remark $\ $ Aternatively, we can employ that $\rm\:c^{2J}+1 = (c^{2J}-1) + 2\equiv 2\pmod{\!c+1}\ $
by $\rm\ c+1\ |\ c^2-1\ |\ c^{2J}-1.\,$ But this requires some ingenuity, whereas the above proof does not, being just a trivial congruence calculation using said modular reduction property of the $\rm\,gcd,\:$ namely $\rm\, \gcd(a,b) = \gcd(a,\:b\ mod\ a),\,$ a reduction which applies much more generally. Said equivalently, the result follows immediately by applying a single step of the Euclidean algorithm. Notice how abstracting the problem a little served to greatly elucidate the innate structure.
Generally $\,\rm \gcd(c\!-\!a, f(a))\, =\, \gcd(c\!-\!a, f(a))\ $ for any polynomial $\rm\,f(x)\,$ with integer coeff's, since $\rm\ mod\,\ c\!-\!a\!:\ c\equiv a\,\Rightarrow\,f(c)\equiv f(a)\,$ by the Polynomial Congruence Rule. Above is simply the special case $\rm\, a = -1\,$ and $\rm\, f(x) = x^{2J}\!+1.$