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.

Solution 1:

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.

Solution 2:

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 $\bmod 4$.

Solution 3:

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 $3$.