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How many rational points (a point $(a, b)$ is called rational if $a$ and $b$ are rational numbers) can exist on the circumference of a circle having centre $(\pi, e)$?

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marked as duplicate by José Carlos Santos, Isaac Browne, max_zorn, Taroccoesbrocco, Jyrki Lahtonen Jul 11 '18 at 6:00

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Either 1 or 2 (but probably just 1).

First, note that it's easy to find a circle centered at this point (let's call it $P$) whose circumference contains 1 rational point; just pick your favorite rational point, and set the radius of this circle to be the distance from this point to $P$.

On the other hand, any circle with 3 (or more) rational points on its circumference must be centered at a rational point; to see this, note we can construct the center by finding the intersection of two of the perpendicular bisectors of the sides of the triangle formed by these 3 points. If all of these 3 points are rational, the perpendicular bisectors will have rational coefficients, so their intersection will be a rational point. Since $\pi$ and $e$ are irrational, any circle centered at $(\pi, e)$ therefore contains at most 2 points on its circumference.

Can you find a circle centered at $P$ with 2 rational points on its circumference? It turns out you can iff $\pi$ and $e$ are linearly dependent over $\mathbb{Q}$; that is, iff there exist rational numbers $p$, $q$, and $r$ such that $p\pi + qe = r$. To see this, first assume such rational numbers exist. Then, note that $(\pi, e)$ is equidistant from the two points $(q, p+\frac{r}{q})$ and $(-q, -p+\frac{r}{q})$; indeed, the perpendicular bisector of these two points is simply the line $px + qy = r$, which $P$ lies on. Conversely, assume two rational points $Q$ and $R$ lie on a circle centered at $P$. Then the perpendicular bisector of the segment $QR$ passes through $P$. But this perpendicular bisector must have rational coefficients (since $Q$ and $R$ are both rational), which implies that some relation of the form $p\pi + qe = r$ exists.

Unfortunately, whether $\pi$ and $e$ are linearly dependent over $\mathbb{Q}$ is an open problem: see http://en.wikipedia.org/wiki/Algebraic_independence. It's generally believed, however, that they are independent over $\mathbb{Q}$, and therefore that at most one rational point should exist on this circle.

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    $\begingroup$ Also, by a cardinality argument, for most radius there are no rational points on the circle. $\endgroup$ – mercio Jun 13 '14 at 8:30

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