Let $F_n$ be the Farey sequence of order n. For $S\subseteq F_n$ we let $\mathcal{Q}(S)=\{x/y:x,y\in S,x\le y\text{ and }y\ne 0\}$. We show that if $\mathcal{Q}(S)\subseteq F_n$, then $|S|\le n+1$. Moreover, we prove that in any of the following cases: (1) $\mathcal{Q}(S)=F_n$; (2) $\mathcal{Q}(S)\subseteq F_n$ and $|S|=n+1$, we must have $S=\{0,1,\frac{1}{2},\dots ,\frac{1}{n}\}$ or $S=\{0,1,\frac{1}{n},\dots ,\frac{n-1}{n}\}$ except for $n=4$, where we have an additional set $\{0,1,\frac{1}{2},\frac{1}{3},\frac{2}{3}\}$ for the second case. Our results are based on Graham's GCD conjectures, which have been proved by Balasubramanian and Soundararajan.

中文翻译：

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法里数列和格雷厄姆猜想

让 $F_n$是n阶 Farey 序列。为了$秒\subseteq F_n$ 我们让 $\mathcal{问}(秒)=\{X/是：X,是\in 秒,X\le 是\text{ 和 }是\ne 0\}$. 我们证明如果$\mathcal{问}(秒)\subseteq F_n$， 然后 $|秒|\le n+1$. 此外，我们证明在以下任何一种情况下：（1）$\mathcal{问}(秒)=F_n$; (2)$\mathcal{问}(秒)\subseteq F_n$ 和 $|秒|=n+1$，我们必须有 $秒=\{0,1,\frac{1}{2},\dots ,\frac{1}{n}\}$ 要么 $秒=\{0,1,\frac{1}{n},\dots ,\frac{n-1}{n}\}$ 除了 $n=4$，我们有一个额外的集合 $\{0,1,\frac{1}{2},\frac{1}{3},\frac{2}{3}\}$对于第二种情况。我们的结果基于 Graham 的 GCD 猜想，该猜想已被 Balasubramanian 和 Soundararajan 证明。