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  "body": "\n## Proof\n\nWe show that $\\gcd(F_a, F_b)$ obeys the subtraction step of the [[Euclidean algorithm|0TRNT4X7J0]], and argue by [[strong induction|0KMVFSHFGZ]].\n\n1. Let $F_k$ be the [[Fibonacci numbers|0GA9K57A39]]. They are natural numbers. Two pairs of integers with the same common divisors have the same [[greatest common divisor|02WF878BH1]].\n2. For integers $x, y, t$, the pairs $(x, y)$ and $(x, y + t x)$ have the same common divisors. Indeed, $d \\mid x$ and $d \\mid y$ imply $d \\mid y + tx$, and conversely $d \\mid x$ and $d \\mid y + tx$ imply $d \\mid y$.\n3. Let $x, y, c$ be integers with $\\gcd(x, y) = 1$. We claim that $(x, yc)$ and $(x, c)$ have the same common divisors.\n4. A common divisor of $x$ and $c$ divides $yc$. Conversely, let $d \\mid x$ and $d \\mid yc$. A common divisor of $d$ and $y$ divides $x$ and $y$, so it is $\\pm 1$. Also $d$ and $y$ are not both $0$, since otherwise $x = 0 = y$ and $\\gcd(x, y) = 0$. Thus $d$ and $y$ are [[coprime|0GFKP8F8AG]].\n5. By the [[generalized Euclid's lemma|03BAREW1Z5]], $d \\mid c$. This proves the claim of step 3.\n6. Let $a \\ge 1$ and $c \\ge 0$. The [[addition formula|00CN2F997Q]] with $m = a - 1$ and $n = c$ gives $F_{a+c} = F_{a-1} F_c + F_a F_{c+1}$.\n7. By step 2 with $t = -F_{c+1}$, $\\gcd(F_a, F_{a+c}) = \\gcd(F_a, F_{a-1} F_c)$.\n8. By [[consecutive Fibonacci numbers are coprime|0D3PFXEAKF]], $\\gcd(F_a, F_{a-1}) = 1$. By step 3, $\\gcd(F_a, F_{a-1} F_c) = \\gcd(F_a, F_c)$.\n9. So $\\gcd(F_a, F_b) = \\gcd(F_a, F_{b-a})$ whenever $1 \\le a \\le b$, and also $\\gcd(a, b) = \\gcd(a, b - a)$ by step 2.\n10. For a natural number $x$, $\\gcd(x, 0) = x$. For $x \u003e 0$ the largest divisor of $x$ is $x$, and $\\gcd(0, 0) = 0$ by convention.\n11. Let $Q(s)$ be the claim that $\\gcd(F_a, F_b) = F_{\\gcd(a, b)}$ for all natural numbers $a, b$ with $a + b = s$. We prove $Q(s)$ for all $s$ by strong induction.\n12. Fix $s$ and assume $Q(s')$ for all $s' \u003c s$. Let $a + b = s$.\n13. If $a = 0$, then by step 10, $\\gcd(F_0, F_b) = \\gcd(0, F_b) = F_b = F_{\\gcd(0, b)}$. The case $b = 0$ is the same with the roles exchanged.\n14. If $1 \\le a \\le b$, then $a + (b - a) = b \u003c s$. By step 9 and $Q(b)$, $\\gcd(F_a, F_b) = \\gcd(F_a, F_{b-a}) = F_{\\gcd(a, b-a)} = F_{\\gcd(a, b)}$.\n15. If $1 \\le b \u003c a$, the same argument with $a$ and $b$ exchanged applies, since $\\gcd$ is symmetric in its two arguments.\n16. So $Q(s)$ holds. By strong induction, $\\gcd(F_m, F_n) = F_{\\gcd(m, n)}$ for all natural numbers $m$ and $n$.\n\n## Idea\n\nThe addition formula writes $F_{a+c}$ as $F_{a-1} F_c$ plus a multiple of $F_a$. Since $F_{a-1}$ and $F_a$ are coprime, $\\gcd(F_a, F_{a+c}) = \\gcd(F_a, F_c)$. So the indices follow the Euclidean algorithm, which ends at $\\gcd(m, n)$.\n",
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