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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    followup: what are the possible closures of the set of rational points on a circle in the coordinate plane? – Holden Lee Oct 23 at 19:10
  • 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. – Oscar Lanzi Oct 23 at 20:06
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    @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. – Emil Jeřábek Oct 24 at 15:20
up vote 104 down vote accepted

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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    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. – GEdgar Oct 23 at 0:38
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    I had to read this six times before it hit me. Amazing. – Randall Oct 23 at 0:42
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    @Randall For me it would have been slightly more obvious if the answer had pointed out that there are uncountably many different radii. – kasperd Oct 23 at 11:47
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    That first paragraph brought joy to my morning. Thank you. – John Hughes Oct 23 at 12:29
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    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. – Henning Makholm Oct 24 at 13:32

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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    what about the case, $a/b$ and $c,d$ case ?, i.e all of them different. – onurcanbektas Oct 23 at 4:43
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    @J.G. But that would violate the fact that they have no common factor – onurcanbektas Oct 23 at 5:49
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    Even more blatantly, $x^2+y^2=\pi^2$ contains no rational points. – John Dvorak Oct 23 at 9:12
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    @JohnDvorak: Even more brazenly, $x^2+y^2 = \sqrt{2}$ contains no rational points. – user21820 Oct 23 at 9:48
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    @user21820 Even more braggadociously, $x^2 + y^2 = -1$ contains no rational points. ...wait... – leftaroundabout Oct 23 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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    Thanks, that addition was really necessary for me. Without it, @OscarLanzi's answer didn't make sense to me. – Waggili Oct 24 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:

$$x^2+y^2=r^2$$

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 $3.$

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