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Can we find a circle in $\mathbb{R}^2$ with exactly 5 points with rational coordinates?

What is obvious is that a circle with a rational center and a rational radius has infinitely many rational points. And that a circle with a radius whose square is irrational and a rational center has no rational points.

I tried finding a circle with 1 rational point but to no avail.

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    $\begingroup$ What have you tried? Can you, for example, prove that the center must be rational? $\endgroup$
    – lulu
    Jun 24 '21 at 17:46
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    $\begingroup$ "What is obvious is [...] that a circle with an irrational radius and a rational center has no rational points." ... Um ... $x^2+y^2=2$? $\endgroup$
    – Blue
    Jun 24 '21 at 17:49
  • $\begingroup$ What are the rational points you imply? $\endgroup$ Jun 24 '21 at 17:50
  • $\begingroup$ Hint: If $(x,y)\in\mathbb{Q}^2$ are rational points on the circle, then all $(\pm x, \pm y)$ are rational and on the same circle. So, you cannot have odd number of rational points. $\endgroup$
    – Bumblebee
    Jun 24 '21 at 17:50
  • $\begingroup$ @Bumblebee Are you assuming that the circle must be centered at the origin (or at least a rational point)? $\endgroup$
    – Calvin Lin
    Jun 24 '21 at 17:53
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Claim: If the circle has (at least) 3 rational points, then

  1. the center is rational (so we can translate to the origin), and
  2. the set of rational points is infinite, actually dense.

So the answer is no.

If you're stuck proving these statements, show what you've tried.

1) (If you don't have a slick argument,) You can find the coordinates of the center by taking the intersection of the perpendicular bisectors.

2) From a rational point on the circle, take a line with rational slope. Show that it intersects the circle again at a rational point.


Notes:

  • I know of a circle with exactly 0 rational points.
  • I know of a circle with exactly 2 rational points -> This is similar to (arguably identical to) 2009 Putnam B1, which is why the above approach was so familiar to me.
  • Lulu provided an example of a circle with exactly 1 rational points in the comments.
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  • $\begingroup$ To make an example with exactly $2$ rational points: Let $P,Q$ be your two (distinct) rational points and let $C$ be an irrational point on the perpendicular bisector of $P,Q$. Taking $C$ as the center, gives you an example. And, for exactly $1$, use $x^2+(\pi-y)^2=\pi^2$. $\endgroup$
    – lulu
    Jun 24 '21 at 18:11
  • $\begingroup$ @lulu Thanks for that example :) $\endgroup$
    – Calvin Lin
    Jun 24 '21 at 18:20
  • $\begingroup$ I proved the first point. Which means proving the property for a circle centered at the origin is sufficent. By simply considering $(x,y)$ and $(-x,y)$ etc... one can prove that the number of solutions is even. But how can I prove that it's infinite? $\endgroup$ Jun 24 '21 at 18:30
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    $\begingroup$ @AymaneMaaitat Hint: From a rational point on the circle, take a line with rational slope. Show that it intersects the circle again at a rational point. $\endgroup$
    – Calvin Lin
    Jun 24 '21 at 22:20

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