Does $a^2+b^2=1$ have infinitely many solutions for $a,b\in\mathbb{Q}$?

I'm fairly certain it does, but I'm hoping to see a rigorous proof of this statement. Thanks.

Here is my motivation. I'm working on a strange little problem. I'm working in a geometry over an ordered field. Suppose I have a circle $\Gamma$ with center $A$ passing through a point $B$. I want to prove that there are infinitely many points on $\Gamma$. Up to a change of variable, I'm considering the unit circle centered on the origin over $\mathbb{Q}$. To show there are an infinite number of points on $\Gamma$, it suffices to show there are an infinite number of solutions to $a^2+b^2=1$ for $a,b\in\mathbb{Q}$. I could then extend this to showing there are infinite number of solutions to $a^2+b^2=r^2$ for some $r$, which proves that any circle over $\mathbb{Q}$ has an infinite number of points. Then since any ordered field has a subfield isomorphic to $\mathbb{Q}$, I would be finished.


Just to make explicit something which was left implicit (or hyperlinked) in the other answers: for any rational $t$ we have

$\left(\frac{1-t^2}{1+t^2}\right)^2 + \left(\frac{2t}{1+t^2}\right)^2 = 1$.

This is just what you get by projecting the $y$ axis through the point $(-1,0)$ on the circle $x^2+y^2=1$.

  • $\begingroup$ Thanks, doubt I would have seen that any time soon. $\endgroup$ – Dani Hobbes Apr 4 '11 at 5:26
  • $\begingroup$ This is also what you get from the substitution $t=\tan(\phi/2)$, which might be a familiar trick to integrate trig functions. (See the picture at the bottom of this page: en.wikipedia.org/wiki/Weierstrass_substitution; the angle at $(-1,0)$ is $\phi/2$ because of the inscribed angle theorem: en.wikipedia.org/wiki/Inscribed_angle#Theorem.) $\endgroup$ – Hans Lundmark Apr 4 '11 at 6:41
  • $\begingroup$ this is the parameterization you get by taking the line through the point $(-1,0)$ with slope $t$ and finding the point of intersection with the circle. taking $t$ rational gives rational points. $\endgroup$ – yoyo Apr 4 '11 at 15:33

Generally, if a conic curve with rational coefficients has one rational point $\rm\:P\:$ then it has infinitely many, since any rational line through $\rm\:P\:$ will intersect the curve in another point, necessarily rational, since if one root of a rational quadratic is rational then so is the other. Thus by sweeping lines of varying rational slopes through $\rm\:P\:$ we obtain infinitely many rational points on the conic. Further, projecting these points onto a line leads to a rational parametrization of the conic. For a very nice exposition see Chapter 1 of Silverman and Tate: Rational Points on Elliptic Curves.

For Pythagorean Triples there is even more structure. For example one may employ ascent in the Ternary Tree of Pythagorean Triples to simply and beautifully generate them all. Picking one simple ascending path yields this formula:

$\rm\quad\ (x,\:x+1,\:z) \to (X,\:X+1,\:Z),\ \ \ X\ =\ 3\:x+2\:z+1,\ \ \ Z\ =\ 4\:x+3\:z+2\:.\ \ $ For example

$\rm\quad (3,4,5)\to (20, 21, 29)\to (119, 120, 169)\to (696, 697, 985)\to (4059, 4060, 5741)\to\cdots$

Since $\rm\ a^2 + b^2 =\: c^2\ \Rightarrow\ (a/c)^2 + (b/c)^2 =\: 1\ $ this yields infinitely many solutions to your equation.


There are infinitely many (primitive) solutions to $$a^2 + b^2 = c^2$$ with $a$, $b$, and $c$ positive integers. Given any such solution, dividing through by $c^2$ gives $$\left(\frac{a}{c}\right)^2 + \left(\frac{b}{c}\right)^2 = 1.$$

Since $\gcd(a,c) = \gcd(b,c)=1$, this is already expressed in least terms, so distinct primitive pythagorean triples give distinct rational solutions.

  • $\begingroup$ Thanks, nice to have a basic number theoretic look at it. $\endgroup$ – Dani Hobbes Apr 4 '11 at 5:23
  • $\begingroup$ Good to know is that if $(a, b, c)$ and $(m, n, p)$ are two Pythagorean triples, then also $(am-bn, an+bm, cp)$ is also a Pythagorean triple. $\endgroup$ – md2perpe May 24 '17 at 8:04

The answer is yes. This is most easily seen via stereographic projection, as described here.

  • $\begingroup$ Thanks, the picture gives it a nice geometric way to see it. Quite fitting. $\endgroup$ – Dani Hobbes Apr 4 '11 at 5:22
  • $\begingroup$ If you want a pointset topology perspective, QxQ is a dense subset of the plane, so that its restriction (to the unit circle here) is also dense, and so infinite; if it was not infinite, there would be a point p in the circle, and an e>0 so that a ball (an arc, actually) centered at p, with radius e, would contain no point of QxQ. One can generalize to saying that dense subsets of the plane with standard topology must be infinite. $\endgroup$ – gary Apr 4 '11 at 7:53
  • $\begingroup$ @gary There are many dense subsets of the plane whose restriction to the unit circle are not dense. This is because the unit circle, as a set, has no topological interior, so its complement is dense. $\endgroup$ – Douglas Zare Apr 4 '11 at 12:16


Alon has used stereographic projection to define pythogarean triples (You can use these in turn to get the half angle formulas for tangent, sine and cosine).

Using stereographic projection if our domain is the points on the circle, then as we move closer to the north pole we will get "further out" in the real line, so we form what's called the extended real axis (or the closure of $\mathbb{R}$) which is $\mathbb{R}\cup \{\infty\}$. Interesting it is to note that here we only talk about 1 infinity and not 2 infinities.




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