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So we know $F_{0} = 0, F_{1} = 1$, and for all $n ≥ 2, F_{n} = F_{n−1} + F_{n−2}$ for the Fibonacci Sequence. How do we prove that for every natural number $n$, $gcd(F_{n}, F_{n+1}) = 1$?

I can make sense that we will prove the contrapositive of the induction step.

And also that only the two following properties will be useful: \begin{align} ∀n ∈ N, gcd(0, n) = n \tag{I} \\ ∀n, a, b, p, q ∈ Z, n | a ∧ n | b ⇒ n | (ap + bq) \tag{II} \end{align}

Jean Marie
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Yahya Farooq
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1 Answers1

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Now, the Fibonacci sequence can be written as $$F_n=F_{n-1}+F_{n-2}$$ and we want to prove $$\forall n\in \mathbb{N},\ gcd(F_n,F_{n+1})=1$$ We will let this statement be $P(n)$.

(Note that the Fibonacci sequence can also be written as $F_{n+2}=F_{n+1}+F_{n}$ and we will use this to our advantage.)

For the base case, $n=0$: \begin{align*} gcd(F_n,F_{n+1})&=gcd(F_0,F_1)\\ &=gcd(0,1) \\ &=1 && \text{(Using Claim 1)} \end{align*} So $P(0)$ holds.

Now, assuming $P(k)$ holds for $n=k$, we let $g=gcd(F_k,F_{k+1})$.

Then $g \mid F_k \land g \mid F_{k+1}$ by the definition of the $gcd$(greatest common divisor).

By Claim 2,\begin{align*} g \mid \overbrace{F_k + F_{k+1}}^\text{their sum} \land \ g \mid \overbrace{F_{k+1}-F_k(=F_{k-1})}^\text{their difference} && \text{(by the definition of the Fibonacci sequence)}\\ \end{align*}

So we know $ gcd(F_{k},F_{k+1})=1$ and $ gcd(F_{k-1},F_{k})=1$ by this and assuming that $P(k)$ holds.

In order to prove $P(k+1)$ for $n=k+1$, observe that: \begin{align*} gcd(F_n,F_{n+1})&=gcd(F_{k+1},F_{k+2})\\ &=gcd(F_{k+1},F_{k+1}+F_k)\\ &=1 && \text{(as we showed earlier using Claim 2 that the sum is also divisible)}\\ \text{Then $g=1$.}\\ \text{This proves $P(k+1)$.} \end{align*}

HKT
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