The scatter graph at the Wikipedia article seems to suggest so. Has anyone attacked this before? Is there a known proof?

  • $\begingroup$ This seems to be related to the paper 70.42 The Density of Pythagorean Rationals, Richard E. Pfiefer, The Mathematical Gazette, Vol. 70, No. 454 (Dec., 1986), pp. 292-294, jstor.org/stable/3616191 $\endgroup$ – Martin Sleziak Jul 7 '11 at 19:14

If you mean a pythagorean triangle having one of the angles arbitrary close to a given acute angle then the answer is yes.

Notice that $\displaystyle (m^2-n^2, 2mn, m^2+n^2)$ is always a pythagorean triangle for any positive integers $\displaystyle m \gt n$.

We have that $\displaystyle \tan \theta = \frac{2mn}{m^2-n^2} = \frac{2t}{1-t^2}$ where $\displaystyle t = \frac{n}{m}$.

By continuity (and surjectivity) of $\displaystyle \frac{2t}{1-t^2}$ we can find an interval of $\displaystyle t$ (which is infact a subinterval of $\displaystyle (0,1)$) for which $\displaystyle \frac{2t}{1-t^2}$ is arbitrarily close to any positive number we want.

Since the rationals are dense, we can find $\displaystyle m,n$ which make the tangent of one the angles arbitrarily close to the tangent of a given acute angle, and hence the angles can be made arbitrarily close.

  • $\begingroup$ looks like we answered at the same time :) $\endgroup$ – Dinesh Oct 18 '10 at 19:18
  • $\begingroup$ @Dinesh: Yes :-) If you incorporate my answer into yours I will delete mine. I think we have to mention that rationals are dense, otherwise we cannot be convinced about arbitrarily close. $\endgroup$ – Aryabhata Oct 18 '10 at 19:22
  • $\begingroup$ I feel your answer is clearer and more complete. You can add the bit about the parametrization of pythagorean triples into your answer and then I will delete mine. $\endgroup$ – Dinesh Oct 18 '10 at 19:29
  • $\begingroup$ @Dinesh: Ok. I have edited it. $\endgroup$ – Aryabhata Oct 18 '10 at 19:32

In fact, more is true -- consider the triples $(3,4,5), (-7,24,25), (-117,44,125), (-527,-336,625), \ldots$, where the $n$th triple is

$$ (a_n, b_n, c_n) := (Re((3+4i)^n), Im((3+4i)^n), 5^n). $$

Since $\tan^{-1} 4/3$ is irrational, the sequence of pairs $(a_n/c_n,b_n/c_n)$

$$ (3/5, 4/5), (-7/25, 24/25), (-117/125, 44/125), (-527/625, -336/625), \ldots $$

is equidistributed on the unit circle. Therefore the sequence $(|a_1/c_1|, |b_1/c_1|), (|a_2/c_2|, |b_2/c_2|)$ obtained by taking absolute values is equidistributed on the arc of the unit circle in the first quadrant.

I'm not sure if we can say that the Pythagorean triples (as opposed to some subsequence of them) are equidistributed in the first quadrant in this sense. Anyone?

  • 1
    $\begingroup$ If planar equidistribution holds I imagine it would be for area measure weighted by 1/r, since this is what is invariant under scaling the triangle. $\endgroup$ – T.. Oct 19 '10 at 4:46
  • 1
    $\begingroup$ Re: equidistribution. I would expect so. The number of lattice points in a circle of radius $R$ is $\pi R^2 + O(R^t)$, where best known bound for $t$ is $131/208$. I would expect (but don't know for sure) that the number of lattice points in the pie-slice $r \in [0,R]$, $\theta \in [\alpha, \beta]$ should be $(\beta - \alpha) R^2/2 + O(R^t)$ as well. Then the ratio of pythagorean triples with norm less than $R$ and angle in $[\alpha, \beta]$ to all pythagorean triples would approach $(\beta - \alpha)/(2 \pi)$, as desired. See en.wikipedia.org/wiki/Gauss_circle_problem $\endgroup$ – David E Speyer Oct 19 '10 at 13:17
  • $\begingroup$ Ah, rereading Wikipedia, I see that bounds with $t=1$ go back to Gauss, and that is all we need. I now feel much more confident that we should be able to get bounds of that quality for the pie-slice problem. $\endgroup$ – David E Speyer Oct 19 '10 at 13:22
  • $\begingroup$ Is the equidistribution question about triples in the plane (see picture at wikipedia, linked in question) or their rescaling onto the unit circle? $\endgroup$ – T.. Oct 19 '10 at 20:00
  • $\begingroup$ @David: for $t=1$ I think you just need that the sector of circle is convex (or something weaker like rectifiable boundary with bounded curvature). Drawing a tube of radius 1 around the perimeter of the figure, or covering it with unit squares, should be enough. $\endgroup$ – T.. Oct 19 '10 at 20:07

The other answers here explain that it is possible to find arbitrarily good approximations to any angle using Pythagorean triples. Here's a footnote on how to turn this into a constructive procedure.

Moron above says to pick a rational number $n/m$ near $t$ where $t$ satisfies $\tan \theta = 2t/(1 - t^2)$. You could pick $n$ and $m$ using a Farey algorithm to find the best rational approximation to $t = (1 - \cos \theta)/\sin\theta$ with a given denominator.

Start with $a = 0$, $b = c = d = 1$. The number $t$ is between $a/b$ and $c/d$. Next compute the "mediant" $(a+c)/(b+d)$. If $t$ is bigger than the mediant, update $a$ and $b$ with $a+c$ and $b+d$. Otherwise update $c$ with $a+c$ and $d$ with $b+d$. Each iteration of this process produces the best rational approximation to $t$ for a given denominator size.


Although you already have a couple of answers proving the result you ask for, I think it is worth adding a further answer using a completely different approach. There is a simple way of parameterizing all points on an algebraic curve of degree two which, once you know, the answer to your question just drops out really easily.

[Edit: Actually, I didn't realize how close this was to Aryabhata's answer when I posted. Still, my parameterization is a bit more general, so I'll leave it up.]

First, the parameterization: For any Pythagorean triple a2 + b2 = c2 you can divide through by c to get the equation a2 + b2 = 1 in rational pairs (a,b) ∈ Q2. The set of all such rational (a,b) is nothing other than the rational points in the unit circle S1. Conversely, if you have rational points (a,b) ∈ S1 then you obtain a Pythagorean triple by multiplying through by a common denominator c (which does not affect the angle).

Now pick any rational point on the unit circle (a0,b0) and a nonzero vector (u,v), and consider the line t → (a0+tu,b0+tv). This must intersect the circle at exactly one other point (a1,b1) -- unless (u,v) is tangent to.the circle (in which case we can say that it intersects it twice at (a0,b0) ). In fact, the equation a2 + b2 - 1 = 0 is a quadratic in t with zero constant term (as it is solved by t = 0), so extracting out the factor of t gives a linear expression. This means that it intersects the curve at a rational point. To be explicit, the equation you get for t is

$$ (u^2+v^2)t^2+2(a_0u+b_0v)t=0\ \Rightarrow\ t=-2(a_0u+b_0v)/(u^2+v^2). $$

Conversely, given any rational point (a1,b1) on the circle distinct from (a0,b0) you can always take (u,v) = (a1 - a0,b1 - b1) to see that every such point is reached in this way. This parameterizes the rational points on the circle by the nonzero rational vectors (u,v) ∈ Q2 (up to recaling). In fact, dividing through by u (if it is nonzero), this parameterizes the rational points on the circle by v ∈ Q and the additional point parameterized by (0,1) curresponding to u = 0. This can be seen to be nothing other than stereographic projection.

Now to your question:

Pick any point (x,y) on the unit circle and approximate the vector (x - a0,y - b0) as closely as you like by rational (u,v). The curve t → (a0+tu,b0+tv) intersects the circle at a rational point, which can be made as close as you like to (x,y). That is, the rational points are dense in the unit circle. So, the angles of pythagorean triples will also be dense in the first quadrant.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.