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Someone posted a question on the notice board at my University's library. I've been thinking about it for a while, but fail to see how it is possible. Could someone verify that this is a valid question and point me in the right direction?

'Given an infinite set of points in a plane, if the distance between any two points is an integer, prove that all these points lie on a straight line.'

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Can we turn your question around? What is (or might be) invalid about the above problem? I am having trouble imagining where your doubts might be coming from. – Alex B. Apr 19 '11 at 0:34
How what is possible? – Qiaochu Yuan Apr 19 '11 at 0:35
Please use more descriptive titles. The title is meant for people to be able to see what questions are about from looking at the titles. – joriki Apr 19 '11 at 0:37
You can't have a circle of infinite radius in euclidean space. You can though have an infinite line. – JSchlather Apr 19 '11 at 0:46
@xbonez: what reasonable interpretation of "circle of infinite radius" does not evaluate to "line"? – Qiaochu Yuan Apr 19 '11 at 1:01
up vote 10 down vote accepted

MR0013511 (7,164a) Anning, Norman H.; Erdős, Paul Integral distances. Bull. Amer. Math. Soc. 51, (1945). 598–600.

The authors show that for any n there exist noncollinear points $P_1,\dots,P_n$ in the plane such that all distances $P_iP_j$ are integers; but there does not exist an infinite set of non-collinear points with this property. Reviewed by I. Kaplansky

I can add that the first result mentioned requires lots of points to be on a circle. I believe the current record for points in the plane with all distances integers, no three on a line, no 4 on a circle, is 8.

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The article and a follow-up. – J. M. Apr 19 '11 at 1:26
@J.M. It appears both links point the same place, probably the follow-up – Ross Millikan Apr 19 '11 at 1:44
I'm no longer confident that an 8-point configuration is known, but a 7-point configuration is given by Tobias Kreisel and Sascha Kurz, There are integral heptagons, no three points on a line, no four on a circle, Discrete Comput. Geom. 39 (2008), no. 4, 786–790, MR2413160 (2009d:52021). – Gerry Myerson Apr 19 '11 at 1:51
@Ross: yes, I copied wrongly; DJC gives the correct link. – J. M. Apr 19 '11 at 1:55

Say $A$, $B$ and $C$ are three points with pairwise integer distances. Any other point $P$ with integer distances from those three will satisfy $|AP-PB| \leq AB$ and $|AP-PC| \leq AC$, by the triangle inequality. So $P$ lies on an intersection of a hyperbola $|AP-BP| = k$ (for some integer $k \leq AB$) and a hyperbola $|AP-PC| = m$ (for some integer $m \leq AC$). There are finitely many such points $P$ (at most $4\;AB\;AC$).

(I'd be surprised if the argument in the Anning and Erdős paper is very different from this one.)

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I believe this is the proof given in the Yaglom and Yaglom book "Challenging Problems..." – user641 Apr 19 '11 at 6:18

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