# Rational points on a circle with centre as $(\pi,e)$ [closed]

What is the number of rational points on a circle having centre as $$(\pi,e)$$?

## closed as off-topic by Saad, Paul Frost, user, Holo, DidJan 1 at 12:45

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• Of what radius? – Richard Martin Nov 9 '18 at 10:29
• @KaviRamaMurthy There may be some: consider the circle with radius $\sqrt{\pi^2+e^2}$, which passes through the origin. – Michael Hoppe Nov 9 '18 at 10:42
• Fix a rationale point, then draw a circle passing for it. This show that there many circle containing at least one rational point. @michael show that tehre are ta most 2. Now assuming there are 2, and write down the equations for both, comapring them (since they have the same radius) you will obtain a rational equation involving $e$ and $\pi$. The problem is that is still an opne problem if $e$ and $\pi$ are algebrically independetly on $\mathbb{Q}$... – ALG Nov 9 '18 at 11:12
• @ALG Both the question of whether $e+\pi$ is rational and the question of whether they are algebraically independent are open. Therefore, any statement of intermediate strength (where the reductions are easy) should also be open. In particular, the question of whether $1,e$ and $\pi$ are linearly independent should also be open. – alphacapture Nov 9 '18 at 11:53
• related (or duplicate) math.stackexchange.com/q/315761/114910 – Burnsba Nov 9 '18 at 14:15

At most $$2$$: if there were three points with rational coordinates, the center would have rational coordinates, too.

Proof: let $$A$$, $$B$$ and $$C$$ be points on a circle with rational coordinates. The center $$M$$ of the circle passing through these points is the intersection of the perpendicular bisectors of $$AB$$ and $$AC$$. Now the equations of those bisectors contain only rational coefficients.

Solving that linear system, Cramer’s rule, for example, reveals that their intersection has rational coefficients, too.

Addendum: There are circles with irrational coordinates of the centre, which have two rational points on it. Take two points with rational coordinates; the equation of their perpendicular bisector has rational coefficients, hence surely there are irrational points on that bisector.

Example: $$(x-1/2-\sqrt2)^2+(y-1/2+\sqrt2)^2=(1/2+\sqrt2)^2 +(1/2-\sqrt2)^2.$$ Obviously $$(0,0)$$ and $$(1,1)$$ belong to that circle.

As comments imply, it is possible to have a point on the circle, and the other answer implies we can't have three. Let me complement this by saying that existence of a circle with this center with two rational points is an open problem.

Indeed, two points $$P,Q$$ lie on some circle centered at $$(e,\pi)$$ iff this point lies on the perpendicular bisector of the interval $$PQ$$. This line has an equation with rational coefficients, and conversely every line with rational coefficients is a bisector of some segment with rational endpoints. Therefore two points can lie on such a circle iff we can have a rational solution to the equation $$ae+b\pi+c=0$$. This, as you can imagine, is widely believed to not be possible, but it's still open - indeed, we don't even know whether $$e/\pi$$ is rational, and if it were, we could even have a relation above with $$c=0$$.

• There should be another reason. It can easily be shown that the only point with rational coordinates on the circle centered in $(\sqrt2,\sqrt2)$ with radius $2$ is the origin. – Michael Hoppe Nov 9 '18 at 12:06
• @MichaelHoppe We know much more about how $\sqrt 2$ behaves algebraically in relation to the rational numbers than we do $\pi$ and $e$. – Arthur Nov 9 '18 at 12:07
• @MichaelHoppe A given point $c$ belongs to a line with rational coefficients if and only if some circle with centre $r$ includes at least two rational points. (Proof of the left-to-right implication: take a circle going through any rational point not on the line; then the mirror image of the point also belongs to the circle.) However, it is crucial here that the radius of the circle is not fixed. Almost all circles with a given centre actually contain no rational points at all. – Emil Jeřábek Nov 9 '18 at 12:25
• @EmilJeřábek Also, I think your comment is the first mention on this page of the fact that there exist circles with center $(\pi,e)$ with no rational points on them. – Jeppe Stig Nielsen Nov 9 '18 at 14:00
• Also, the first statement in my post (a point is a centre of a circle with at least two rational points iff it belongs to a rational line) is not directly related to the second statement (all but countably many circles with any given centre contain no rational points). They together serve to make the point that Michael Hoppe's comment misinterprets Wojowu's answer. – Emil Jeřábek Dec 3 '18 at 12:14

There is at most one rational point on that circle, assuming $$\pi, e$$ and $$1$$ are linearly independent over $$\Bbb Q$$ (which seems to be an open problem).

A rational point $$(p, q)$$ on the circle means that there is a real number $$r$$ such that $$(p-\pi)^2 + (q-e)^2 = r^2$$ If there is another rational point $$(s, t)$$ on that same circle, then we also have $$(s-\pi)^2 + (t-e)^2 = r^2$$ giving us $$(p-\pi)^2 + (q-e)^2 =(s-\pi)^2 + (t-e)^2\\ \underbrace{(2s-2p)\pi}_{\text{rational multiple of }\pi} + \underbrace{(2t-2q)e}_{\text{rational multiple of }e} + \underbrace{p^2 - s^2 + q^2 - t^2}_{\text{rational multiple of }1} = 0$$ That the two points are distinct implies the three braced terms are not all zero, meaning $$\pi, e$$ and $$1$$ are linearly dependent over $$\Bbb Q$$.

• – Burnsba Nov 9 '18 at 14:14