Understanding the proof for $\gcd (F_n, F_{n+k}) = 1$ where $F_n$ is a Fermat number

Solution 1:

The point is: if you multiply that equation by $\,F_n\,$ you get $\,F_{n+k}-\color{#c00}2\, =\, m F_n,\,$ for some integer $\,m,\,$ which implies that any common divisor of $\,F_{n+k}\,$ and $\,F_n\,$ must divide $\,\color{#c00}2.\,$ As the Remark below explains, this is essentially a divisibility-based method of showing that $\,\rm c^{2J}\!+1\equiv 2\pmod{c+1}.\,$ Simpler is to use modular arithmetic and a single step of the Euclidean algorithm.

Lemma $\rm\ \gcd(c+1,\ c^{2J}+1)\ =\ gcd(c+1,\:\color{#c00}2)$

Proof $\rm\quad mod\ c+1\!:\ c^{2J}+1\: \equiv\ (-1)^{2J}+1\:\equiv\ 2\,\ \ $ QED

For $\rm\,\ \color{#c00}{c=2^{\large 2^{N}}}\!,\ \, 2J\, =\, 2^{K} \Rightarrow\ \color{#0a0}{c^{\large 2J}} = (2^{\large 2^{N}})^{\large 2^K}\!\! = \color{#0a0}{2^{\large 2^{N+K}}}\! $ $\Rightarrow$ $\smash[t]{\, \rm gcd\overset{\underbrace{\large \color{#c00}{c+1}}}{({\it F}_N},\,\overset{\underbrace{\large \color{#0a0}{c^{\large 2J}+1}}}{{\it F}_{N+K}}) = gcd\overset{\underbrace{\large \color{#c00}{c+1}}}{({\it F}_N},\color{}2) = 1.}$

Remark $\ $ Aternatively, one could 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 the 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.

More generally $\,\rm \gcd(c-n, f(c))\, =\, \gcd(c-n, f(n))\ $ for any polynomial $\rm\,f(x)\,$ with integer coeff's, since $\rm\, mod\,\ c\!-\!n\!:\ c\equiv n\,\Rightarrow\,f(c)\equiv f(n).\,$ Above is the special case $\rm\, n = -1\,$ and $\rm\, f(x) = x^{2J}\!+1.$