2

(We are proving the claim for $n \geq 1$, and we have $F_1 = 1, F_2 = 1$ and $F_n = F_{n-1} + F_{n-2}$.)

The proof proceeds by induction on $n$.

Base Case: We have that $F_1 =1$ and $F_2 = 1$, and both clearly only have a factor of $1$. So, this case holds.

Inductive Step: We assume the claim holds for $n=k$ and show this implies the claim holds for $n=k+1$.

For the sake of contradiction, let's assume $F_{k+1}$ and $F_{k+2}$ are not relatively prime. Then, there is some integer $d >1$ such that $d|F_{k+1}$ and $d|F_{k+2}$.

We have that $F_{k+2} = F_{k+1} + F_k$. It follows that since $d|F_{k+2}$ and $d|F_{k+1}$, we must also have $d|F_k$. This implies that $F_{k+1}$ is not relatively prime with $F_k$, since both share a factor $d > 1$. This contradicts the inductive hypothesis, so we conclude $F_{k+1}$ and $F_{k+2}$ must be relatively prime.

Hisoka Moroh
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  • This is not proof by induction, this is a proof by contradiction. – Peter Foreman Mar 31 '19 at 19:36
  • There's several ways of proving this in a previous MSE post : https://math.stackexchange.com/questions/24378/prove-that-any-two-consecutive-terms-of-the-fibonacci-sequence-are-relatively-pr – Martin Hansen Mar 31 '19 at 20:11
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    Contrary to what Peter Foreman said, this is a perfectly good proof by induction. It uses proof by contradiction in the induction step, but that doesn't make it any less a proof by induction. – Andreas Blass Mar 31 '19 at 20:25

1 Answers1

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You can also use the Bezout identity in your induction to have a direct proof.

$\gcd(F_{n+1},F_n)=1\iff \exists (a,b)\mid aF_{n+1}+bF_n=1$

Then $F_{n+2}=F_{n+1}+F_n$ after multiplying by $b$ we get

$\implies bF_{n+2}=bF_{n+1}+1-aF_{n+1}\\\implies bF_{n+2}+(a-b)F_{n+1}=1\\\implies \gcd(F_{n+2},F_{n+1})=1$

zwim
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