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I'm trying to understand the following proof of why

No two Fermat numbers have a common divisor greater than $1$.

Hardy and Wright show the following proofs:

If $x = 2^{2^n}$, we have

$$\frac{F_{n+k} - 2}{F_n}(\circ) = \frac{2^{2^{n+k}}-1}{2^{2^n}+1} = \frac{x^{2^k}-1}{x+1} = x^{2^k-1} - x^{2^k-2} + \ldots -1$$

The main issue I'm having is the $\circ$ part. I don't understand how they get that. The rest of the proof makes sense but the first part is really confusing me. Can someone please explain where they get that fraction?

Thanks a bunch!

Jeel Shah
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1 Answers1

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The point is: if we multiply that equation by $F_n\,$ we 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\ \ \ mod\,\ \color{#c00}{c\!+\!1}\!:\ \color{#c00}c^{2J}\!+1\: \equiv\ (\color{#c00}{-1})^{2J}\!+1\:\equiv\ 2,\ $ $ {\rm by}\,\ \color{#c00}{c\equiv -1},\ {\rm by}\,\ \color{#c00}{c\!+\!1\equiv 0_{\phantom{|_|}}}\ \ $ QED

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

Remark $ $ Aternatively, we 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 $\,\gcd,\,$ i.e. $\rm\ \gcd(a,b) = \gcd(a,\:b\ mod\ a),\,$ 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\!-\!n,\, f(c))\, =\, \gcd(c\!-\!n,\, f(n))\ $ for any polynomial $\rm\,f(x)\,$ with integer coeff's, because $\rm\ mod\,\ c\!-\!n\!:\ c\equiv n\,\Rightarrow\,f(c)\equiv f(n).\,$ Above is special case $\rm\, n = -1\,$ and $\rm\, f(x) = x^{2J}\!+1.$

Bill Dubuque
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