Yeah, $A_i(\cos2\alpha_i,\sin2\alpha_i)$,where $\sin\alpha_i=\frac{i^2-1}{i^2+1},\cos\alpha_i=\frac{2i}{i^2+1}$.($i\in\mathbb{N}$) It's obvious that $A_i$ is on the unit circle,so no three is collinear. $A_iA_j=2\sin|\alpha_i-\alpha_j|\in\mathbb{Q}$(consider the angle $A_iOA_j$)

I'm not really sure if this is the same as the previous proof. We will use induction. Consider a circle radius $\frac {\sqrt {3}}{2}$, and an equilateral triangle inscribed in it with vertices A,B,C, side length 1. Clearly no three points on the circle are collinear. These three points obviously work. Now suppose a point D on minor arc AB is a rational distance from A and B. By Ptolemy, it is a rational distance from C. Now if we choose another point F on minor arc AB a rational distance from A and B, also by Ptolemy FC is rational but then we have cyclic quad FDBC, so by Ptolemy FD is rational. It remains to show there are infinitely many points on arc AB a rational distance from A and B. But if N is a point on minor arc AB a distance of x and y from A and B, respectively, then $\angle ANB = 120$, so by the law of cosines, $x^2 + y^2 + xy = 1$. Can anyone show there are infinitely many rationals $(x,y)$ that satisfy this?

($\frac {2n + 1}{n^2 + n + 1}$, $\frac {n^2 - 1}{n^2 + n + 1}$) works for the above proof.

These answers don't seem to have any motivation so let me provide motivation for shobber's proof. Let $S = \{(1, \frac{2s}{s^2 - 1}) : s \in \mathbb{Q}\}$. Consider the inversion centered about the origin $O$, with radius $r \in \mathbb{Q}$. From now on the inverse of a point $X$ will be denoted by $X'$. It suffices to show that for all $P, Q \in S$ that $P'Q' \in \mathbb{Q}$ because then we will have found an infinite set of points, no three collinear (they all lie on a circle), with pairwise rational distances. Now note that $P'Q' = \frac{r^2 \cdot PQ}{OP \cdot OQ}$. It is clear that for all $A, B \in S \cup {(0, 0)}$ we have that $AB \in \mathbb{Q}$ and this implies that $P'Q' \in \mathbb{Q}$ as desired.

dgreenb801 wrote: Can anyone show there are infinitely many rationals $(x,y)$ that satisfy this? Looks like Eisenstein integers?

dgreenb801 wrote: I'm not really sure if this is the same as the previous proof. We will use induction. Consider a circle radius $\frac {\sqrt {3}}{2}$, and an equilateral triangle inscribed in it with vertices A,B,C, side length 1. Clearly no three points on the circle are collinear. These three points obviously work. Now suppose a point D on minor arc AB is a rational distance from A and B. By Ptolemy, it is a rational distance from C. Now if we choose another point F on minor arc AB a rational distance from A and B, also by Ptolemy FC is rational but then we have cyclic quad FDBC, so by Ptolemy FD is rational. It remains to show there are infinitely many points on arc AB a rational distance from A and B. But if N is a point on minor arc AB a distance of x and y from A and B, respectively, then $\angle ANB = 120$, so by the law of cosines, $x^2 + y^2 + xy = 1$. Can anyone show there are infinitely many rationals $(x,y)$ that satisfy this? About this solution, I am not sure if such a point on minor arc $AB$ whose distance from both $A,B$ is rational will always exist. You can guarantee that the point has rational distance from at least one of $A,B$, but not both (since in that equation, if $x$ is rational, it might still happen that $y$ is irrational).

@above, read the next post dgreenb801 wrote: ($\frac {2n + 1}{n^2 + n + 1}$, $\frac {n^2 - 1}{n^2 + n + 1}$) works for the above proof.

Wolstenholme wrote: These answers don't seem to have any motivation so let me provide motivation for shobber's proof. Let $S = \{(1, \frac{2s}{s^2 - 1}) : s \in \mathbb{Q}\}$. Consider the inversion centered about the origin $O$, with radius $r \in \mathbb{Q}$. From now on the inverse of a point $X$ will be denoted by $X'$. It suffices to show that for all $P, Q \in S$ that $P'Q' \in \mathbb{Q}$ because then we will have found an infinite set of points, no three collinear (they all lie on a circle), with pairwise rational distances. Now note that $P'Q' = \frac{r^2 \cdot PQ}{OP \cdot OQ}$. It is clear that for all $A, B \in S \cup {(0, 0)}$ we have that $AB \in \mathbb{Q}$ and this implies that $P'Q' \in \mathbb{Q}$ as desired. Why do $P'$ and $Q'$ lie on the unit circle?