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This is famous puzzle that I can came across recently. Place six points on a plane so that distance between any two points is integer such that no three points are collinear.

In 3D it is easy( pyramid) but how to do this in case of 2D?

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Do the distances need to be different? It makes a huge difference. If not, it is easy to find as many as you want. – Ross Millikan Jan 14 '13 at 5:15

Slide 46 of this link shows an answer attributed to Leech. Slide 47 shows $7$ points attributed to Tobias Kreisel and Sascha Kurz. The paper is here

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Or just notice that if $\sin x$ and $\cos x$ are rational, then so is $\sin px$ for all integer $p$, so it is easy to place as many points as you want on a circle. However if you demand that no four lie on one circle, the life gets hard. – fedja Nov 15 '12 at 1:21
@fedja: that doesn't guarantee that the distances are rational. $(1,0)$ and $(0,1)$ are an irrational distance apart. – Ross Millikan Nov 15 '12 at 4:42
Well, I skipped the part that you should go by $2x$, not by $x$ and that you'd better choose $x$ either incommensurable with $\pi$ or just small enough if you want your points to be distinct ;-) – fedja Nov 16 '12 at 3:27
@fedja: I don't understand the last at all. You can find lots of points on the unit circle with rational coordinates, but the distances between them may well not be rational. The angles will be incommensurable with $\pi$ but that still doesn't help. – Ross Millikan Nov 16 '12 at 3:33
Ross, the distance between the points on the unit circle at angle $2\alpha$ is $2\sin\alpha$. The rational coordinates have nothing to do with it... – fedja Nov 16 '12 at 15:15

This is question 12 at The answer there is brilliant, but actually they provide an answer that has all six points with integer coordinates. The answer already suplied at the slide 46 link above shows seven points, but they are not at integer coordinates.

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