Why the $GCD$ of any two consecutive Fibonacci numbers is $1$?

Question

Note: I've noticed that this answer was given in another question, but I merely want to know if the way I'm using could also give me a proof.

I did the following:

$$F_n=F_{n-1}+F_{n-2} \\ F_n=[F_{n-2}+F_{n-3}]+[F_{n-3}+F_{n-4}]\\F_n=F_{n-2}+2F_{n-3}+F_{n-4}\\ F_n=[F_{n-3}+F_{n-4}]+2[F_{n-4}+F_{n-5}]+[F_{n-5}+F_{n-6}]\\F_n=F_{n-3}+3F_{n-4}+3F_{n-5}+F_{n-6}\\ \dots \tag{1}$$

I guess that the coefficients of the $F_n$'s might indicate something that could prove it. But I'm not sure if it's possible. Perhaps the impossibility of writing the expression as:

$$n(b_1a_1+b_2a_2+\dots +b_n a_n)$$

With $n,b_n\in\mathbb{N}$ would show that. But I'm not sure on how to proceed. This should be true because if $a$ and $b$ have a common divisor $d$, then:

$$a+b=a'd+b'd=d(a'+b')$$

It is possible to extend this:

$$a+b+c=da'+db'+dc'=d(a'+b'+c')$$

I have noticed that the numbers that appear in the expansion I've shown in $(1)$ seems to be the Pascal's triangle. So perhaps these numbers as coefficients of the $F_n$'s might indicate that it's not possible to write them as:

$$d(a'+b'+c')$$

Answer 1

If $\,f_n = a f_{n-1}\! + f_{n-2}\,$ then induction shows $\,(f_n,f_{n-1}) = (f_1,f_0)\,$ since $\, (f_n,f_{n-1}) = (a f_{n-1}\! + f_{n-2},\,f_{n-1}) = (f_{n-2},f_{n-1}) = (f_1,f_0)\,$ by induction

Remark $\ $ Similarly we can prove much more generally that the Fibonacci numbers $\,f_n\:$ comprise a strong divisibility sequence: $\,(f_m,f_n) = f_{(m,n)},\:$ i.e. $\,\gcd(f_m,f_n) = f_{\gcd(m,n)}.\:$ Then the above is just the special case $\,m=n\!+\!1.\:$

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.

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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.

Answer 2

Let us take $3$ consecutive terms, call them $a, b, a+b$, with $a<b$. As $\gcd(b,a+b)=\gcd(a,b)$, we can by iterating see that the gcd is the same for any pair of consecutive Fibonacci numbers and we can conveniently calculate it for the first two numbers which is $1$.

Answer 3

This can be done by a simple induction argument:

Let $F_1=1$ and $F_2=1$. Clearly, $(F_1,F_2)=1$.

Suppose, $(F_{n+1},F_n)=1$. Then let $(F_{n+2},F_{n+1})=(F_{n+1}+F_n,F_{n+1})= d$.

If $m|d$, then $m|(F_n,F_{n+1})$ and so $m=1$. But now the only divisor of $d$ is $1$. Hence $d=1$.

Answer 4

Suppose F_n+2 = F_n+1 + F_n If a is a common divisor Of F_n+1 and F_n , it divides F_n+2. But in the Fibonacci series, (F_n+1)^2 = (F_n)(F_n+2) +1 or -1. Hence a^2 divides +1 or -1 , a contradiction. Edwin Gray