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It seems evident from infinitely many primitive pythagorean triples $(a,b,c)$ that there are infinitely many rational points $\left(\frac{a}{c}, \frac{b}{c}\right)$ on the unit circle.

But how would one go about, and show that they are dense, in the sense that for two rational points $x$ and $y$ of angles $α$ and $β$ on the unit circle, if $α<β$ there is a third rational point $z$ of angle $γ$ on the unit circle, such that $α<γ$ and $γ<β$.

Is it to expect that this conjecture holds and that it isn't an unsolved number theory problem?

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    $\begingroup$ For a different argument: $\mathbb{Q}(i) \cap S^1$ is an infinite subgroup of $S^1$, hence $\overline{\mathbb{Q}(i) \cap S^1}$ is an infinite closed subgroup of $S^1$. But the closed subgroups of $S^1$ are 1. $S^1$ itself, and 2. the finite subgroups. Conclusion, $\overline{\mathbb{Q}(i) \cap S^1} = S^1$. $\endgroup$
    – Daniel Fischer
    Commented Jan 14, 2017 at 21:31
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    $\begingroup$ @DanielFischer Nice! Although to be fair, your second-to-last sentence does take a bit of proof. $\endgroup$
    – Noah Schweber
    Commented Jan 14, 2017 at 21:43
  • $\begingroup$ @DanielFischer Nice! Is it easy to show that $\mathbb Q(i) \cap S^1$ is infinite? My first thought was to show that, say, $(3+4i)/5$ has no finite order. I'm not sure how an elementary argument along these lines would go; it's easy with some basic Galois theory. $\endgroup$
    – Dustan Levenstein
    Commented Jan 14, 2017 at 21:44
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    $\begingroup$ @DustanLevenstein Find infinitely many Pythagorean triples which correspond to different rational points. $\endgroup$
    – Noah Schweber
    Commented Jan 14, 2017 at 21:45
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    $\begingroup$ @DustanLevenstein: Alternately, see here for an elementary proof that $(3+4i)/5$ has no finite order. $\endgroup$
    – Micah
    Commented Jan 15, 2017 at 21:57

4 Answers 4

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Consider a nonvertical line through the point $(1, 0)$ of slope $m$. This line meets the circle at exactly one other point, and it's not hard to show that the coordinates of that point are $$P_m=\left({m^2-1\over m^2+1}, {-2m\over m^2+1}\right).$$ As long as $m$ is rational, $P_m$ has rational coordinates; so now think about the lines through $(1, 0)$ of rational slope . . .

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  • $\begingroup$ Can you show that P_m is on the unit circle? $\endgroup$
    – user4414
    Commented Jan 14, 2017 at 21:03
  • $\begingroup$ @j4nbur53 Fixed a (godawful) typo. But yes - $(m^2-1)^2=m^4-2m^2+1$, and $(-2m)^2=4m^2$. So adding the squares of the numerators we get $(m^4+1)^2$. But that's exactly the square of the denominator. $\endgroup$
    – Noah Schweber
    Commented Jan 14, 2017 at 21:05
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    $\begingroup$ I think I did it right. The slope is ${y_2-y_1\over x_2-x_1}$. Let $A=y_2-y_1, B=x_2-x_1$. $y_2={-2m\over m^2+1}$ and $y_1=0$, so $A={-2m\over m^2+1}$. $x_2={m^2-1\over m^2+1}$ and $x_1=1$, so $B={m^2-1\over m^2+1}-1={m^2-1\over m^2+1}-{m^2+1\over m^2+1}={-2\over m^2+1}$. Cancelling denominators, we get ${A\over B}={-2m\over -2}=m$. $\endgroup$
    – Noah Schweber
    Commented Jan 14, 2017 at 21:12
  • $\begingroup$ My bad, was short cutting rational arithmetic. gr8! $\endgroup$
    – user4414
    Commented Jan 14, 2017 at 21:13
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    $\begingroup$ Just for reference this is essentially an inverse stereographic projection. $\endgroup$
    – Derek Elkins left SE
    Commented Jan 15, 2017 at 4:07
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We are interested in angles $\theta$ such that $(\cos \theta, \sin \theta)$ are both rational. Call such an angle a $\mathbb Q-$ angle.

Claim: $\theta$ is a $\mathbb Q-$ angle iff $\tan \frac {\theta}2 \in \mathbb Q$.

Proof:

$\implies$ follows from the half angle formula, $\tan \frac {\theta}2 =\frac {1-\cos \theta}{\sin \theta}$

To see $\impliedby$ it may be easiest to work geometrically. Note that $\frac {\theta}2$ is the angle formed by the line connecting $(-1,0)$ to the point $(\cos \theta, \sin \theta)$. That line has equation $y=h(x+1)$ where $h=\tan \frac {\theta}2$. Solving this and $x^2+y^2=1$ simultaneously we see that we are trying to solve $$x^2+h^2(x^2+2x+1)-1=0\implies x^2+\frac {2h^2}{1+h^2}x+\frac {h^2-1}{1+h^2}=0$$ of course one root is given by $x=-1$ and it follows at once that the other root must also be rational. Thus $\cos \theta \in \mathbb Q$. Now we can use the half-angle formula again to see that this implies that $\sin \theta \in \mathbb Q$.

Note: this is the key point behind the Weierstrass substitution, also known as the $\tan \frac {\theta}2$ substitution.

It is clear that the angles $\theta$ such that $\tan \frac {\theta}2\in \mathbb Q$ are dense so we are done.

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  • $\begingroup$ The intermediate claim is interesting, but the last "it is clear" part might need some explanation. $\endgroup$
    – user4414
    Commented Jan 14, 2017 at 21:12
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    $\begingroup$ @j4nbur53 The map $f:x\mapsto \tan{x\over 2}$ is continuous and surjective onto $\mathbb{R}$, so the $f$-preimage of any dense subset of $\mathbb{R}$ (e.g. $\mathbb{Q}$) is dense in the domain. $\endgroup$
    – Noah Schweber
    Commented Jan 14, 2017 at 21:15
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    $\begingroup$ May be worth pointing out that, of course, my construction is exactly the same as that of @NoahSchweber . He projects from $(1,0)$ while I project from $(-1,0)$ but any rational point on the circle would do the job. I just added enough details to show that, in fact, every rational point on the circle arises this way. $\endgroup$
    – lulu
    Commented Jan 14, 2017 at 21:23
  • $\begingroup$ Indeed (and incidentally I upvoted this answer and your comment). $\endgroup$
    – Noah Schweber
    Commented Jan 14, 2017 at 21:32
  • $\begingroup$ @NoahSchweber Thanks! Reciprocated. $\endgroup$
    – lulu
    Commented Jan 14, 2017 at 21:36
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If the two given points are in different quadrants, then one of the points $(0, \pm1), (\pm1, 0)$ may fairly be said to be between them. So wlog they are both in the first quadrant. Then for any rational point $P$ on the unit circle, there are $t, u\in \mathbb N$ with $t>u$, where $$P=P(t,u)=\left(\dfrac{2tu}{t^2+u^2},\dfrac{t^2-u^2}{t^2+u^2}\right).$$ If $t+u$ is odd and $t$ and $u$ are coprime, that even yields the fractions in their lowest terms.

So given $X=P(t_1,u_1)$ and $Y=P(t_2,u_2)$, $Z=P(t_1+t_2,u_1+u_2)$ is a rational point between $X$ and $Y$ as required.

For example,

$X=P(2,1)=(\frac45,\frac35)=(.8,.6), Y=P(4,3)=(\frac{24}{25},\frac7{25})=(.96,.28), Z=P(6,4)=(\frac{48}{52},\frac{20}{52})=(\frac{12}{13},\frac5{13})\approx(.92,.38).$

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  • $\begingroup$ It looks like a geometric interpretation of $P(t,u)$ might help clarify the claim that $P(t_1+t_2,u_1+u_2)$ "lies between" $P(t_1,u_1)$ and $P(t_2,u_2)$, given both the latter are in the first quadrant. $\endgroup$
    – hardmath
    Commented Jan 15, 2017 at 13:38
  • $\begingroup$ From t1*u2 < t2*u1 it follows (t1+t2)*u2 < t2*(u1+u2) and t1*(u1+u2) < (t1+t2)*u1. Hence for m1=t1/t2, m2=t2/u2, m=(t1+t2)/(u1+u2) and m1<m2 it follows m1<m & m<m2. This means the same geometry as in the answer by Noah Schweber or lulu can be used. $\endgroup$
    – user4414
    Commented Jan 15, 2017 at 16:39
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It's super easy to prove. For any complex number on the unit circle, there exists a sequence of complex numbers with integer parts such that if you replace each term with itself divided by its magnitude, the new sequence is a sequence of points on the unit circle that approaches the square root of the chosen complex number. If you replace each term of the original sequence with its square, then every term in the sequence will have an integral magnitude. If after that, you replace each term of the new sequence with itself divided by its magnitude, the new sequence is a sequence of complex numbers on the unit circle with rational parts that approaches the chosen complex number.

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  • $\begingroup$ I described a second operation to the sequence that replaces each number with that number divided by its magnitude creating a new sequence which is a sequence of numbers on the unit circle with rational parts. $\endgroup$
    – Timothy
    Commented Jan 15, 2017 at 23:40
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    $\begingroup$ I fixed my answer to give a more direct proof that doesn't have any missing parts that are as hard to figure out what are. $\endgroup$
    – Timothy
    Commented Jan 16, 2017 at 5:38
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    $\begingroup$ I believe I have given a legitimate proof and you just can't understand it. I know it's not a formal proof that completely proves it. $\endgroup$
    – Timothy
    Commented Jan 16, 2017 at 23:44
  • $\begingroup$ @j4nbur53 This argument does indeed work, although it takes a couple steps to make clear. We want to show that, given $z$ on the unit circle there are rational points on the circle arbitrarily close to $z$. It would be enough, then, to show that there's a sequence of rational points on the circle that converge to $z$. Well, our first step is to pick a complex $w$ such that $w^2=z$ (think of $w$ as "halfway around" the circle to $z$). Now, we can find (exercise) a sequence $p_i$ of points in the plane with integer coordinates, such that ${p_i\over\vert p_i\vert}\rightarrow w$. (Cont'd) $\endgroup$
    – Noah Schweber
    Commented Jan 17, 2017 at 1:29
  • $\begingroup$ Let $q_i={p_i\over\vert p_i\vert}$. Good news: $q_i$ is on the unit circle! Bad news: $q_i$ (probably) doesn't have rational coordinates! Moreover, the sequence $q_i$ converges to $w$, not to our desired $z$. But we can fix these two problems at once: let $v_i={(x_i^2, y_i^2)\over\vert p_i\vert ^2}$, where $p_i=(x_i, y_i)$. Since $x_i, y_i$ are integers, $\vert p_i\vert ^2$ is rational (indeed an integer), so $v_i$ has rational coordinates. Moreover, it's not hard to see that the $v_i$s are on the unit circle and approach $z$. So we're done. $\endgroup$
    – Noah Schweber
    Commented Jan 17, 2017 at 1:32

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