Are rational points dense on every circle in the coordinate plane?

First thing first I know that rational points are dense on the unit circle. However, I am not so sure how to show that rational points are not dense on every circle.

How would one come about answering this. Any hits are appreciate it.

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    $\begingroup$ followup: what are the possible closures of the set of rational points on a circle in the coordinate plane? $\endgroup$ – Holden Lee Oct 23 '18 at 19:10
  • $\begingroup$ If the center of the circle is on a rational point it's all of the circle or nothing. Please ask a new question for a fuller answer. $\endgroup$ – Oscar Lanzi Oct 23 '18 at 20:06
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    $\begingroup$ @HoldenLee Once three points on the circle are rational, the centre is also rational, hence there is a rational affine transformation of the circle to the unit circle at the origin, hence rational points are dense on the circle. Thus, the only possibilities are: (1) the whole circle, (2) two rational points, (3) one rational point, (4) the empty set. $\endgroup$ – Emil Jeřábek Oct 24 '18 at 15:20

They're not. No two different circles centered at the origin contain any of the same points. There are uncountably many circles (specifically, one for each real number, corresponding to the radius) , so most circles contain no rational points at all.

We can find some more specific examples. Specifically, any rational point $(a, b)$ on a circle of radius $r$ centered at the origin satisfies $a^2+b^2=r^2$. In particular, $r^2$ must be rational. There are also radii whose squares are rational where there are no rational points. Clearing denominators, say multiplying by some $c^2$ to do so, we have that $c^2r^2$ is a sum of two squares. If $r^2$ is an integer, then $r^2$ must be a sum of two squares, since an integer is a sum of two squares if and only if its prime factorization doesn't contain an odd power of a prime congruent to $3$ mod $4$. $r^2$ was arbitrary, so if we choose it not to be a sum of two squares we get circles with no rational points.

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    $\begingroup$ For example, the circle with center at the origin and radius $\pi$ has no rational points. Replace $\pi$ with any number whose square is irrational. $\endgroup$ – GEdgar Oct 23 '18 at 0:38
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    $\begingroup$ I had to read this six times before it hit me. Amazing. $\endgroup$ – Randall Oct 23 '18 at 0:42
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    $\begingroup$ @Randall For me it would have been slightly more obvious if the answer had pointed out that there are uncountably many different radii. $\endgroup$ – kasperd Oct 23 '18 at 11:47
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    $\begingroup$ I can't follow your argument in the second paragraph. Suppose I claim I have a function that for each rational $r$ will give you a rational point on the circle with radius $\sqrt r$. You then clear the denominators of each of my rational points -- but this gives you new $r$s, and it is not clear to me how you can assume that all integers would be represented in the set of new $r^2$s. $\endgroup$ – hmakholm left over Monica Oct 24 '18 at 13:32
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    $\begingroup$ @MattSamuel: That's for circles with rational centers. If we allow arbitrary centers, then a counting argument similar to your first paragraph shows that there are circles with exactly one rational point, as well as circles with exactly two. (But circles with at least three rational points necessarily have rational centers and therefore their rational points are dense). $\endgroup$ – hmakholm left over Monica Oct 24 '18 at 19:27

In particular, $x^2+y^2=3$ cannot have any rational points. If it had any such points then there would be integers $a,b,c$ having no common factor such that $(a/c)^2+(b/c)^2=3$ therefore $a^2+b^2=3c^2$. But with no common factor at least one of $a,b,c$ must be odd and all possibilities conforming with this requirement fail $\bmod 4$.

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    $\begingroup$ what about the case, $a/b$ and $c,d$ case ?, i.e all of them different. $\endgroup$ – Our Oct 23 '18 at 4:43
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    $\begingroup$ @J.G. But that would violate the fact that they have no common factor $\endgroup$ – Our Oct 23 '18 at 5:49
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    $\begingroup$ Even more blatantly, $x^2+y^2=\pi^2$ contains no rational points. $\endgroup$ – John Dvorak Oct 23 '18 at 9:12
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    $\begingroup$ @JohnDvorak: Even more brazenly, $x^2+y^2 = \sqrt{2}$ contains no rational points. $\endgroup$ – user21820 Oct 23 '18 at 9:48
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    $\begingroup$ @user21820 Even more braggadociously, $x^2 + y^2 = -1$ contains no rational points. ...wait... $\endgroup$ – leftaroundabout Oct 23 '18 at 15:02

For a little more detail to Oscar's answer, the reason we may require that $a,$ $b$, and $c$ are co-prime is that if $$\left(\frac{a}{b}\right)^2 + \left(\frac{c}{d}\right)^2 = 3,$$ we may write $$(ad)^2 + (bc)^2 = 3 (bd)^2.$$ Hence $(ad)^2 = b^2(3 d^2 - c^2),$ so $b^2$ divides $(ad)^2$ and $b$ divides $ad$.

With this, we may write $ad = bk$ and divide the $b^2$ from both sides, getting $$k^2 + c^2 = 3d^2,$$ at which point we may apply Oscar's $\mod 4$ argument.

Also, this example is necessary to flesh out Matt's answer: he ends with

Then $r^2$ must be a sum of two squares, which is not true of all integers.

However, in his setup, we require that $r$ satisfies "$r^2 \in \mathbb{N}$ but $r^2 k^2$ is not a sum of two squares for all $k \in \mathbb{N}$," and it's not clear that such an $r$ exists until you establish an example like $3$.

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    $\begingroup$ Thanks, that addition was really necessary for me. Without it, @OscarLanzi's answer didn't make sense to me. $\endgroup$ – Waggili Oct 24 '18 at 10:09

When will there be a dense subset of rational points in a circle?

If $x^2+y^2=r^2$ has any rational point, then the rational points in it are dense in it.

More generally, the following are equivalent about a (non-degenerate) circle in $\mathbb R^2:$

  1. The set of rational points on a circle are dense in the circle
  2. The circle has a rational center and a rational point
  3. The circle has three rational points.

I'll outline why $2\implies 1.$ We can assume that the center of your circle is $(0,0).$ Your circle has an equation like:


Since it has a rational point, it also means $r^2$ is rational.

Now, if $(x_1,y_1)$ is your rational point, take any line through that point with a rational slope, $m.$ Then the set of pairs $(x_1+t,y_1+mt)$ will (except when $m$ is the tangent of the circle at $(x_1,y_1)$) hit the circle again. But that yields a rational quadratic equation fo $t$ with a known rational root, $t=0.$ So the other root $t$ is also rational, and the other point is rational.

$3\implies 2$ is because finding the circumcenter of three points is a linear process.

And $1\implies 3$ because $(1)$ means there are infinitely many rational points on the circle, so at least $3.$

  • $\begingroup$ Finding the circumcentre is nit a linear process, more like it conserves rationality. Great answer btw! $\endgroup$ – N.S.JOHN Jan 8 at 4:54

First of all, I want to thank @Hidaw for having answered such an interesting question, which was swirling around in my head some days ago. After reading all the answers to this post with great pleasure, I discovered that Humke and Krajewski completely solved the problem in a very simple and beautiful article on the American Mathematical Monthly in 1979 (a few years before I was born!) called A Characterization of Circles Which Contain Rational Points.

In particular they establish the very relevant result that if the radius $r$ is irrational, but $r^2=p/q$, with $p, q \in \mathbb{Q}$, then the circle $C$ of equation $x^2+y^2=r^2$ contains no rational point if $pq$ is not the sum of two square integers, while $\mathbb{Q}$ is dense on $C$ if $pq$ is the sum of two square integers.


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