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I did a proof to show that:

If $m|F_n$ and $m|F_{n+1}$ then $m=1$

My approach was to show that if $m|F_n$ and $m|F_{n+1}$ then $m|F_{n-1}$ which means that $m$ divides every Fibonacci number which means it divides $F_1$ which means it divides 1.

Is there an easier way of doing this?

Bill Dubuque
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Mathematical Dummy
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    This seems like a pretty easy approach. What about about this proof seemed too complicated, so that it should be eliminated in another approach? – Elchanan Solomon Aug 29 '22 at 14:16
  • This argument shows that the claim is almost trivial. Which further simplification do you expect ? You could use that $\frac{F_{n+1}}{F_n}$ are the convergents of $[1,1,1,\cdots]$ and they are always in lowest terms , a very elegant approach , but to prove this all rigorously is surely not easier. – Peter Aug 29 '22 at 15:40
  • Another observation (clearly related): Euclid's algorithm directly shows the GCD is $1$. – Brian Tung Aug 29 '22 at 16:56

2 Answers2

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I think your approach is probably the easiest one. However, if you are familiar with the fact that $$F_{n+1}\cdot F_{n-1} - F_n^2 = \pm 1$$ then you can immediately conclude that $\gcd(F_n,F_{n+1}) = 1$ (recall that $\gcd(a,b)=1$ iff there exists $x,y \in \mathbb{Z}$ s.t $xa+yb=1$).

Bill Dubuque
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Giorgos Giapitzakis
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That's likely simplest. Generally we can show that the gcd of $k$ consecutive elements (satisfying a $k$'th order linear recurrence) is constant if the recurrence is bi-monic (first & last coef's are $\color{#c00}{\bf \pm1}$), hence all such gcds are equal to the gcd of the initial $k$ elements (so, e.g. for the Fibonacci sequence this yields $\,\gcd(F_{n+1},F_{n+2}) = \gcd(F_1,F_2) = \gcd(1,1) = 1)$

Theorem $ $ If the integer sequence $\,a_i$ satisfies the below $\color{#c00}{\text{bi-monic}}$ recurrence for all $n\ge 1$ $$ \color{#c00}{\bf 1}\:\! a_{n+k} = c_{n+k-1} a_{n+k-1} + \cdots + c_1 a_{n+1} \color{#c00}{\bf \pm 1}\:\! a_n,\ \ c_i\in\Bbb Z\quad$$

then this implies: $\ \gcd(a_{n+1},\ldots,a_{n+k}) = \gcd(a_1,a_2,\ldots,a_k)\ $ for all $\,n\ge 0$

Proof $\ $ We induct on $n$. Base case $\,n=0\,$ is clear. Else by by $\rm\color{#0af}{GMR}$ = gcd mod reduction

$\qquad\begin{align} &\gcd(\ \ \ \ \ \ a_{n+1},a_{n+2},\ldots,a_{n+k-1},\,a_{n+k})\\[.2em] =\, &\gcd(\ \ \ \ \ \ \color{#90f}{a_{n+1}},\color{#0a0}{a_{n+2}},\ldots,\color{#c00}{a_{n+k-1}},\,\pm a_n + c_1 \color{#90f}{a_{n+1}} + c_{2} \color{#0a0}{a_{n+2}} + \cdots + c_{n+k-1} \color{#c00}{a_{n+k-1}})\\[.2em] =\, &\gcd(\ \ \ \ \ \ a_{n+1},a_{n+2},\ldots,a_{n+k-1},\,\pm \color{darkorange}{a_n})\ \ {\rm by\ \rm\color{#0af}{GMR}}\\[.2em] =\, &\gcd(\color{darkorange}{a_n},a_{n+1},a_{n+2},\ldots,a_{n+k-1})\\[.2em] =\, &\gcd(a_1,a_2,\ldots,a_k)\ \ {\rm by\ induction} \end{align}$

Generally if the shift map $S_n$ on such $k$ tuples is linear and unimodular $(\det S_n =\pm1)$

$$S_n \begin{bmatrix} a_{n+k}\\ \vdots\\ a_n \end{bmatrix} = \begin{bmatrix} a_{n+k+1}\\ \vdots\\ a_{n+1} \end{bmatrix},\ \ \det(S_n) = \pm1$$

then again the gcd of $k$ consecutive elements remains constant, with proof a straightforward generalization of the $k=2$ case. Fibonacci's are the special case $S_n = \begin{bmatrix}1 \!&\! 1\\ 1 \!&\! 0\end{bmatrix}$ and the proof by Cassini's identity (or addition law) in G.G's answer is a special case of this, e.g. see here.

It can also be viewed as special case $\,j = k\!+\!1\,$ of the strong divisibility property of the Fibonacci's

$$\gcd(F_j,F_k) = F_{\gcd(j,k)}\qquad$$

Although the linked inductive proof of this is not much more difficult, its discovery likely is.

Bill Dubuque
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