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

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.

• followup: what are the possible closures of the set of rational points on a circle in the coordinate plane? – Holden Lee Oct 23 '18 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 '18 at 20:06
• @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 '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.

• 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 '18 at 0:38
• I had to read this six times before it hit me. Amazing. – Randall Oct 23 '18 at 0:42
• @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 '18 at 11:47
• That first paragraph brought joy to my morning. Thank you. – John Hughes Oct 23 '18 at 12:29
• 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 '18 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 $$\bmod 4$$.

• what about the case, $a/b$ and $c,d$ case ?, i.e all of them different. – onurcanbektas Oct 23 '18 at 4:43
• @J.G. But that would violate the fact that they have no common factor – onurcanbektas Oct 23 '18 at 5:49
• Even more blatantly, $x^2+y^2=\pi^2$ contains no rational points. – John Dvorak Oct 23 '18 at 9:12
• @JohnDvorak: Even more brazenly, $x^2+y^2 = \sqrt{2}$ contains no rational points. – user21820 Oct 23 '18 at 9:48
• @user21820 Even more braggadociously, $x^2 + y^2 = -1$ contains no rational points. ...wait... – 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 $$3$$.

• Thanks, that addition was really necessary for me. Without it, @OscarLanzi's answer didn't make sense to me. – 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:

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

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 $$3.$$