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Open problem in Geometry/Number Theory. The real question here is:

Is there an infinite family of points on $y=x^2$, for $x \geq 0$, such that the distance between each pair is rational?

The question of "if not infinite, then how many?" follows if there exists no infinite family of points that satisfies the hypothesis.

We have that there exists a (in fact, infinitely many) three point families that satisfy the hypothesis by the following lemma and proof.

Lemma 1: There are infinitely many rational distance sets of three points on $y=x^2$.

The following proof is by Nate Dean.

Proof. Let $S$ be the set of points on the parabola $y = x^2$ and let $d_1$ and $d_2$ be two fixed rational values. For any point, $P_0(r)=(r, r^2) \in S$, let $C_1(r)$ be the circle of radius $d_1$ centered at $P_0(r)$ and let $C_2(r)$ be the circle of radius $d_2$ centered at $P_0(r)$. Each of these circles must intersect $S$ in at least one point. Let $P_1(r)$ be any point in $C_1(r) \cap S$ and likewise, let $P_2(r)$ be any point in $C_2(r) \cap S$. Now let $dist(r)$ equal the distance between $P_1(r)$ and $P_2(r)$. The function $dist(r)$ is a continuous function of $r$ and hence there are infinitely many values of $r$ such that $P_0(r)$, $P_1(r)$, and $P_2(r)$ are at rational distance. $ \blacksquare $

This basically shows that the collection of families of three points that have pairwise rational distance on the parabola is dense in $S$.

Garikai Campbell has shown that there are infinitely many nonconcyclic rational distance sets of four points on $y = x^2$ in the following paper: http://www.ams.org/journals/mcom/2004-73-248/S0025-5718-03-01606-5/S0025-5718-03-01606-5.pdf

However, to my knowledge, no one has come forward with 5 point solutions, nor has it been proven that 5 point solutions even exist.

But I know that many people have not seen this problem! Does anyone have any ideas on how to approach a proof of either the infinite case or even just a 5 point solution case?

Edit: The above Lemma as well as the paper by Garikai Campbell do not include the half-parabola ($x \geq 0$) restriction. However, I thought that the techniques that he employed could be analogous to techniques that we could use to make progress on the half-parabola version of the problem.

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Any insights/strategies/useful theorems that may be used to make some progress are welcome in the comments, since I know that people seldom submit an answer that they don't consider complete. –  Clark Zinzow May 22 '13 at 19:54
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You add $x\geq0$ in the problem,but it's not appear in Lemma 1,and the paper you refered did not say $x\geq0$,do you need this condition when you try to find a five-point solution? –  Next May 23 '13 at 5:07
    
Ah yes, I apologize, the problem is supposed to be restricted to the half-parabola, so $x \geq 0$ should be included. The paper that I was referencing was only exploring the four point solutions that are nonconcyclic, but I don't think that he included the restriction of $x \geq 0$ in any of his arguments, I should have looked at the paper more thoroughly. However, I think that some of his techniques may still apply in exploring this problem further, even with restricting it to the half-parabola. Thank you for the correction, I'll add that right now and +1 to you for catching that. –  Clark Zinzow May 23 '13 at 5:34
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You have to show that $dist(r)$ is not an irrational constant... –  Georges Elencwajg May 23 '13 at 6:30
    
The idea of Nate Dean's proof was two fix two circles with rational radii $d_1$ and $d_2$ at one point, and since these circles will intersect the parabola at at least one point each, this generates our three points, with $d_1$ being the distance between $P_0(r)$ and $P_1(r)$; $d_2$ being the distance between $P_0(r)$ and $P_2(r)$, each being rational. Since $dist(r)$ is continuous, and since the rationals are dense in the reals, we can shift the three points up and down the parabola, keeping $d_1$, $d_2$ constant and letting $dist(r)$ vary, yielding an infinite number of rational values. –  Clark Zinzow May 23 '13 at 15:43
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1 Answer 1

The answer to the "infinite family" question appears to be, no. Jozsef Solymosi and Frank de Zeeuw, On a question of Erdős and Ulam, Discrete Comput. Geom. 43 (2010), no. 2, 393–401, MR2579704 (2011e:52024), prove (according to the review by Liping Yuan) that no irreducible algebraic curve other than a line or a circle contains an infinite rational set.

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Thank you! I just found the arXiv PDF, here it is if anyone else wants to take a gander: arxiv.org/pdf/0806.3095v2.pdf –  Clark Zinzow Jul 22 '13 at 14:55
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