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  "body": "\n## Proof\n\nWe argue by [[induction|0ASJVEFA9J]] on $n$.\n\n1. Base case $n = 0$. We have $F_0 = 0$ and $F_1 = 1$. The only positive divisor of $1$ is $1$, so $\\gcd(F_0, F_1) = 1$.\n2. Induction step. Let $n \\ge 0$ and assume $\\gcd(F_n, F_{n+1}) = 1$.\n3. Let $e$ be a positive integer that divides $F_{n+1}$ and $F_{n+2}$.\n4. By the recurrence, $F_n = F_{n+2} - F_{n+1}$. A common divisor of two integers divides their difference, so $e \\mid F_n$.\n5. So $e$ is a positive common divisor of $F_n$ and $F_{n+1}$. By the induction hypothesis, the greatest such divisor is $1$. So $e \\le 1$, and $e = 1$.\n6. Thus the only positive common divisor of $F_{n+1}$ and $F_{n+2}$ is $1$. So their [[greatest common divisor|02WF878BH1]] is $1$.\n7. By induction, $F_n$ and $F_{n+1}$ are [[coprime|0GFKP8F8AG]] for each $n \\ge 0$.\n\n## Idea\n\nA common divisor of two neighbours in the sequence also divides the term before them, because that term is their difference.\nWalking down the sequence, it divides $F_1 = 1$, so it is $1$.\n",
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