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 one 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 $\,m=n\!+\!1.\:$

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