Is it possible to find 8 non collinear points with their mutual distances being rational? Is it possible to find 8 non collinear points with their mutual distances being rational?
This came to my mind after reading this article (http://mathworld.wolfram.com/RationalDistances.html)
Note that in my question 4 points can lie on a circle.
But no three points are collinear.
Answers referencing to a article/proof are appreciated.
 A: On a unit circle, you can produce countably many points so that the mutual distances are rational. That is why they exclude the circle points in their problem.
To explain what happens here: say $z_l= e^{i\alpha_k}$, $k = 1,2$. Then 
$$|z_1 - z_2| = 2 |\sin(\frac{\alpha_1 - \alpha_2}{2})|$$
Let now $\alpha_k = 2 \beta_k$. Then $|z_1 - z_2| = 2 |\sin(\beta_1 - \beta_2)| = 2 | \sin \beta_1 \cos \beta_2 - \sin \beta_2 \cos \beta_1|$. 
Take now $\beta_k$ so that $\tan \frac{\beta_k}{2}$ is rational. Then both 
$\sin \beta_k$, and $\cos \beta_k$ will be rational.  Moreover, the distances between all these points will be rational. 
Therefore, for every $s\in \mathbb{Q}$ consider an angle $\beta_s$ with $\tan \frac{\beta_s}{2}= s$. Then the corresponding point on the circle will be
$$P_s= e^{i 2 \beta_s} = 
\left(\frac{1 - 6 s^2 + s^4}{(1 + s^2)^2}, \frac{4 s (1 -s^2)}{(1 + s^2)^2}\right)$$
One can check that
$$d(P_s ,P_t) = \frac{4 |(s-t)(1+s t)|}{(1+s^2)(1+t^2)}$$
We are taking squares of rational points on the circle. Their pairwise distances are also rational. 
$\bf{Added:}$ Check a similar problem on this site at the link.
