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Having trouble solving this Math contest problem.

Prove that there are infinitely many points on the unit circle such that the distance between any two of them is a rational number.

It's obvious to see that there are infinitely many points on a unit circle such that the distance between two points on a unit circle is rational and it's also obvious that there are infinitely many rational distances between two points. But how in the world can you construct an infinite set of points such that the distance between any two points is rational?

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    $\begingroup$ I don't know your source on Niven's theorem, but it feels likely that it is only about angles $\theta=\pi q$ with rational $q$. Or some such constraint on the angles. At least then we can relate to a question about real subfields of cyclotomic fields. $\endgroup$
    – Jyrki Lahtonen
    Commented Aug 13, 2023 at 5:14
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    $\begingroup$ See Lulu's answer here: math.stackexchange.com/questions/1978452/… $\endgroup$
    – TheBestMagician
    Commented Aug 13, 2023 at 5:18
  • $\begingroup$ Are there infinitely many rational distances between two points on a unit circle? This question bugs me I don't know where I went wrong with both approaches, yielding opposite answers. $\endgroup$
    – Hammock
    Commented Aug 13, 2023 at 5:56
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    $\begingroup$ I think I found the mistake. Nivne's theorem states that the only rational values of θ in the interval 0° ≤ θ ≤ 90° for which the cosine of θ degrees is also a rational number are 30 deg. The key is that theta is also rational. my bad. I missed that point $\endgroup$
    – Hammock
    Commented Aug 13, 2023 at 7:06

1 Answer 1

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If $z_1$, $z_2 \in \mathbb{C}$ then $|z_1^2 - z_2^2| = |z_1-z_2|\cdot |z_1 + z_2|$. If moreover $|z_1| = |z_2|$ then $|z_1^2 - z_2|^2$ equals twice the area of the rhombus on the vectors $z_1$, $z_2$.

So now consider a set of points $M = \{z_n \ | z_n \in \mathbb{Q}(i), |z_n| = 1\}$. The set $\{z_n^2 \ | \ z_n \in M \}$ will work ( can you see why ? $\ $ $\bf{A:}$ The area of the said rhombus is rational)

Note that $M = \{-1\} \cup \{ \frac{1-t^2}{1+t^2} + i \frac{2 t}{1+t^2} \ |\ t \in \mathbb{Q} \}$

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