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Suppose $f\, : \, \mathbb{Z}\times\mathbb{Z}\to\mathbb{R}\times\mathbb{R}$ is defined by $$f(x,y) = \left(\frac{x}{\sqrt{x^2+y^2}},\frac{y}{\sqrt{x^2+y^2}}\right),$$ and set $P\subset \mathbb{Z}^2$ to be the set $$P = \left\{(x,y) \in \mathbb{N}\times\mathbb{N}\, \left|\, \sqrt{x^2+y^2}\in\mathbb{N}, \gcd\left(x,y,\sqrt{x^2+y^2}\right)=1\right.\right\}$$ (using the convention that $\mathbb{N}$ does not include $0$; i.e., the set of all $(x,y)$, with $x>0$, $y>0$, integers, so that $\left(x,y,\sqrt{x^2+y^2}\right)$ forms a primitive Pythagorean triple).

Define $\Omega_f$ to be the restriction of $f$ to $P$ (or the image of $P$ under $f$).

If $C = \{(x,y)\in\mathbb{R}^2 \mid x> 0,y> 0, x^2+y^2=1\}$ is the segment of the unit circle lying in the first quadrant, and $C_f = \Omega_f \cap C$, then is $C_f \subset C$ dense?

Relevant things I have considered:

  1. There are (countably) infinitely many primitive Pythagorean triples, but this doesn't say anything about their distribution in angle.
  2. Equivalently, set $C = \{(\cos(\theta),\sin(\theta))\in\mathbb{R}^2\mid 0< \theta < \pi/2\}$ (note this is the same set as $C$ above). Is the set $\{(\cos(\theta),\sin(\theta))\in \mathbb{R}^2\mid 0<\theta<\pi/2, \cos(\theta)\in\mathbb{Q},\sin(\theta)\in\mathbb{Q}\}$ dense as a subset of $C$?
  3. Equivalently, let $P$ be as above. If we set $\theta(x,y) = \arctan(y/x)$, is $\text{image}(\theta|_P) \subset (0,\pi/2)$ dense (i.e., is the image of the restriction of $\theta$ to the set $P$ dense as a subset of $(0,\pi/2)$?
  4. I do know that the image of a dense subset under a continuous surjection is itself dense, but I'm not sure how/if that is relevant.

Hope this isn't something that's already been addressed on MSE, or something that is trivially easy, but I have not been able to find a proof (or confirm whether one exists).

Nicholas Stull
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  • $P$ is not dense, so $\Omega_f$ cannot be dense. – Pockets Apr 21 '16 at 16:12
  • @Pockets I assume OP means that $f(P)$ is dense in the unit circle. – Travis Willse Apr 21 '16 at 16:15
  • @Travis, That's exactly what I mean (or at least the portion of the unit circle in the first quadrant). – Nicholas Stull Apr 21 '16 at 16:15
  • Yes, the points with rational coordinates are dense on the unit circle. – André Nicolas Apr 21 '16 at 16:19
  • @AndréNicolas Thank you for the response. Is the proof more or less based on density of $\mathbb{Q}\subset\mathbb{R}^2$? Or is there another proof I should be looking for? – Nicholas Stull Apr 21 '16 at 16:20
  • @NicholasStull: Consider the lines with rational slopes $m$ that pass through $(-1,0)$ (say). It is not hard to show such a line meets the unit circle at another rational point, and it is reasonably clear by continuity that these second meeting points are dense on the unit circle. – André Nicolas Apr 21 '16 at 16:30

2 Answers2

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Let $\ell_m$ be the line with slope $m$ that passes through the point $(-1,0)$.

Let $P_m=(x_m,y_m)$ be the second meeting point of $\ell_m$ with the unit circle. It is reasonably clear by continuity the set of second meeting points as $m$ ranges over the rationals is dense on the unit circle.

The line $\ell_m$ has equation $y=m(x+1)$. Substituting in the equation of the unit circle, we get $x^2+m^2(x+1)^2=1$, which expands to $$(1+m^2)x^2+2m^2 x+m^2-1=0.$$ One root of this quadratic is $-1$, and the product of the roots is $\frac{m^2-1}{m^2+1}$, so the other root is $x_m=\frac{1-m^2}{1+m^2}$. In particular, $x_m$ is rational. Substituting in $y=m(x+1)$ we find that $y_m=\frac{2m}{1+m^2}$, also rational.

For actual Pythagorean triples, we use rational $m$ with $0\lt m\lt \frac{\pi}{4}$, and obtained denseness on the first-quadrant part of the unit circle.

André Nicolas
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  • That's exactly what I was after. Thank you very much. I couldn't think of how one would go about it, but this is rather elegant in its simplicity. – Nicholas Stull Apr 21 '16 at 17:49
  • @NicholasStull: It gives a rational parametrization of the rational points on the unit circle. You can recognize that the ordered pair $(x_m,y_m)$, as described in terms of $m$, is close to the standard description of the (primitive) Pythagorean triples. – André Nicolas Apr 21 '16 at 18:11
  • Does this address the questions about $f(P)$ (the set of primitive Pythagorean triples projected on the unit circle)? – robjohn Apr 22 '16 at 16:47
  • @robjohn: Primitive or not primitive makes no difference. The argument shows we have denseness on the unit quarter-circle. As a consequence, the $x$-coordinates and $y$-coordinates are dense on $[0,1]$. – André Nicolas Apr 22 '16 at 16:59
  • Okay, I am just missing where the Pythagorean triples come in (primitive or not). – robjohn Apr 22 '16 at 18:13
  • @robjohn: Every rational point on (open) quarter-circle comes from a Pythagorean triple (and conversely), a unique one if we specify primitive. – André Nicolas Apr 22 '16 at 18:28
  • Never mind. I see. I missed that the lines were through $(-1,0)$ and so you were using the standard $(m^2-n^2,2mn,m^2+n^2)$ Pythagorean Triangles. (+1) However, this might be confusing to others as well. Furthermore, did you mean $0\lt m\lt 1$ – robjohn Apr 22 '16 at 21:02
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If $p^2+q^2=r^2$ and $\tan(x)=\frac pq$, then $\sin(x)=\frac pr$ and $\cos(x)=\frac qr$. Niven's Theorem, which is proven in this answer, says that if $x\in\mathbb{Q}\pi$ and $\sin(x)\in\mathbb{Q}$, then $\sin(x)\in\left\{0,\pm\frac12,\pm1\right\}$. Since both $\sin(x)\in\mathbb{Q}$ and $\cos(x)\in\mathbb{Q}$, we must have that $\sin(x),\cos(x)\in\left\{0,\pm1\right\}$.

Thus, if $p^2+q^2=r^2$ and $\arg(p+iq)\in\mathbb{Q}\pi$, then $pq=0$. Thus, $$ \arg\left(\frac{4+3i}5\right)\not\in\mathbb{Q}\pi $$ Therefore, we can find an $m,n$ so that $m\arg\left(\frac{4+3i}5\right)+n2\pi$ is arbitrarily close to $0$. This means that $$ \left\{\left(\frac{4+3i}5\right)^n:n\in\mathbb{Z}\right\} $$ is dense (and uniformly distributed) in the unit circle. The same can be done for any other non-trivial Pythagorean Triple.

Therefore, $$ \left\{\frac{p+iq}r:p^2+q^2=r^2\text{ and }p,q,r\in\mathbb{Z}\right\} $$ is dense and uniformly distributed in the unit circle.

In terms of the question, $f(P)$ is dense and uniformly distributed in the unit circle.

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