The Proof of Fibonacci GCD

\[\Large \text{The Proof of Fibonacci GCD} \]

\[\text{By Sangber} \]

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\(\text{Fibonacci Sequence}\)

\(\quad \quad \text{If we record sequence}\ \{F_n\}\ \text{as Fibonacci Sequence, then }\{F_n\} \text{ will have these properties:}\)

\[F_i = \begin{cases}1, & i\leq 2 \\ F_{i-1} + F_{i-2}, & \mathrm{otherwise}\end{cases} \]

\(\quad \quad \text{So we can see that } \{F_n\} \text{ is like “1,1,2,3,5,8,13,21...”.}\)

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\(\text{GCD}\)

\(\quad \quad \text{“GCD” is the abbreviation of “Greatest Common Divisor”.}\)
\(\quad \quad \text{And we often record } (a, b) \text{ as the GCD of } a \text{ and } b.\)
\(\quad \quad \text{For example, } (12, 15) = 3, (48, 36) = 12.\)

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\(\text{Two Theorems about GCD}\)

\(\text{Euclidean theorem}\)

\[(a, b) = (b, a \bmod b), a, b \in N_{+} \]

\(\text{The proof: }\)

• \(\text{If } a = b, (a, b) = a = b, \text{we can surely tell that the theorem is established.}\)
• \(\text{If } a < b, (b, a \bmod b) = (b, a) = (a, b), \text{the theorem is established.}\)
• \(\text{If } a > b, \text{we can assume that } a = k \times b + r, (k \in N_{+}, 0 \le r < b), \text{ so } a \bmod b = r. \\ \text{As for } \forall d, \text{meeting that } d|a \land d|b, d|(a - k \times b), \text{ actually it means } d | r. \\ \text{So we can tell that } (a, b) = (b, r), \text{ that is, } (a, b) = (b, a \bmod b).\)

\(\text{Stein's Algorithm}\)

\[(a, b) = (a, b - a),b > a \]

\(\text{The proof:}\)
\(\text{Actually, the proof of it is similar to that of Euclidean theorem, you can finish it yourselves.}\)

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\(\text{One Lemma}\)

\[(F_n,F_{n - 1})=1, n \in N_{+} \]

\(\text{The Proof}\):

• \(\text{In the case of } n = 1 \text{ and } n = 2,\text{ the lemma is established obviously.}\)
• \(\text{In the case of } n \ge 2 :\)
  \(\text{Obviously:}\)

\[F_n > F_{n - 1} > F_{n - 2}, F_n = F_{n - 1} + F_{n - 2} \]

\[\therefore F_n\ \bmod F_{n - 1} = F_{n - 2} \]

\(\quad \ \text{According to Euclidean theorem and the lemma above:}\)

\[(F_n,F_{n-1})=(F_{n-1},F_n\ \bmod\ F_{n-1})=(F_{n-1},F_{n-2}) \]

\[\therefore (F_n,F_{n-1})=(F_{n-1},F_{n-2})=(F_{n-2},F_{n-3})= \cdots =(F_1,F_2)=1 \]

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\(\text{Fibonacci GCD}\)

\(\quad \quad \text{Here is the expression of the Fibonacci GCD:}\)

\[\forall n,m \in \text{Z}^{+},(F_n, F_m) = F_{(n, m)} \]

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\(\text{The Proof of Fibonacci GCD}\)

\(\text{We assume that } n<m.\)
\(\text{Then we use } F_n \text{ and }F_{n+1} \text{ to express } F_{n + 2},F_{n + 3},F_{n + 4},F_{n + 5}\cdots\)

\[\begin{aligned} & F_{n + 2} = 1 F_n + 1 F_{n + 1} \\ & F_{n + 3} = 1 F_n + 2 F_{n + 1} \\ & F_{n + 4} = 2 F_n + 3 F_{n + 1} \\ & F_{n + 5} = 3 F_n + 5 F_{n + 1} \end{aligned} \\ \cdots \cdots \]

\(\text{We can see that in the expressions above, the coefficients of } F_n \text{ and } F_{n + 1} \text{ meet the properties of the Fibonacci Sequence.}\)

\[\therefore F_m=F_{m - n - 1} \times F_n + F_{m - n} \times F_{n + 1} \]

\[\therefore (F_n, F_m) = (F_n, F_{m - n - 1} \times F_n + F_{m - n} \times F_{n + 1}) \]

\(\text{And obviously we can see that :}\)

\[F_n|F_{m-n-1}\times F_n \]

\[\therefore (F_n,F_m)=(F_n,F_{m-n}\times F_{n+1}) \]

\(\text{According to the lemma above:}\)

\[(F_n, F_{n + 1}) = 1 \]

\[\therefore (F_n, F_m) = (F_n, F_{m - n}) \]

\(\text{All in all:}\)

\[\text{If }n<m,(F_n,F_m)=(F_n,F_{m-n}) \]

\(\text{We can see that }, \text{the change rule of } n \text{ and } m \text{ meet Stein's Algorithm,so we finally discovered that }\)

\[(F_n,F_m)=F_{(n,m)} \]

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\[\Large \text{That's all, thanks!} \]

\[\Huge Q\omega Q \]