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  "body": "\n## Proof\n\n1. Let $F_k = 2^{2^k} + 1$ be the [[Fermat numbers|0R03B54V4Z]]. We first show, by [[induction|0ASJVEFA9J]] on $k \\ge 1$, that $F_0 F_1 \\cdots F_{k-1} = F_k - 2$.\n2. For $k = 1$: $F_0 = 3$ and $F_1 - 2 = 5 - 2 = 3$.\n3. Assume $F_0 \\cdots F_{k-1} = F_k - 2$ for some $k \\ge 1$. Then\n$$F_0 \\cdots F_k = (F_k - 2) F_k = (2^{2^k} - 1)(2^{2^k} + 1) = 2^{2^{k+1}} - 1 = F_{k+1} - 2.$$\nThis completes the induction.\n4. Now let $m \\ne n$ be natural numbers. We may assume $m \u003c n$. Then $n \\ge 1$, and $F_m$ is one of the factors $F_0, \\dots, F_{n-1}$.\n5. Let $d$ be a positive common divisor of $F_m$ and $F_n$. Then $d$ [[divides|096PW7Q014]] $F_0 \\cdots F_{n-1} = F_n - 2$, by steps 1 and 4.\n6. So $d$ divides $F_n - (F_n - 2) = 2$. Hence $d = 1$ or $d = 2$.\n7. Every Fermat number is odd, since $2^{2^k}$ is even. So $d \\ne 2$, and $d = 1$.\n8. So the only positive common divisor of $F_m$ and $F_n$ is $1$. That is, $F_m$ and $F_n$ are [[coprime|0GFKP8F8AG]].\n\n## Idea\n\nThe product of all earlier Fermat numbers is $2$ less than the next one. So a common divisor of two Fermat numbers must divide $2$. Fermat numbers are odd, so that divisor is $1$.\n",
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