Hi,

Infinite in any number of dimentions (collinear)

If they are non-collinear, I think they are equilateral triangle in 2D and a pyramid in 3D. I don't know how to visualize higher dimensions!

I meant the points to be non-colinear, distances can be any integer. In 2D the best I can find is 4 points in a 3X4 rectangle or any other rectangle based on a pythagorean triple.

In the xy-plane, look at


(0, 8), (-15, 0), (-6, 0), (0, 0), (6, 0), (15, 0), (0, -8)


And look at all of the distances of the 7 C 2 = 21 pairs of these 7 points.


This concept can be extended to greater numbers of Pythagorean triangles
an integer distances using additional points.

Hi;

Muuuwwwwaaaahhahaha (My best Bela Lugosi evil laugh)

Then it looks like I have the only 5 point solution where no 3 points are collinear!

Hi;

hi bobbym and gAr

You just got there before me gAr.  But I make it that the sixth point isn't an integer distance from it's mirror image.  Hence what I typed for bobbym.  Then I'll let you have your four extra points!  (I should have thought of that too.  Brain still half asleep! )

bobbym:  Excellent!  Cannot guess how you got that.

But 6 points.  Yes!  From your solution it's easy.

Multiply all the coordinates by 17, so they're all integers.

Then add the sixth point by symmetry (-4025,6000).

Ok?

Bob

Last edited by bob bundy (2011-08-03 19:37:46)

Hi Bob,

Aargh, yes!

It is not integer for all the points
It got me excited, and posted just after checking 2 or 3 segments!

Hmmm, need to go back searching.

Thanks for telling.

Hi gAr;

Yes, if they do not lie on a circle.

Hi gAr;

Here is how it is phrased:

How large can a set of points be ( that are pairwise integer distances away ) if no 3 points are collinear and no 4 points are on a circle? Answer unknown!

With either of these conditions the answer is infinity.

But our circle does not have even one set of collinear points.

That is true. It is one type of solution. The other type has a bunch of points with one set of three collinear. That is a known solution. For anything else no one knows.

Also for the circle method he is using you need to be able to generate the initial points ( not just reflect those ). He claims he can do that for any number of points on the circle.

Diophantine equations are very hard. But you already somewhat acquainted with them. When you solve a gf you are asking how many solutions to the diophantine equation. Number theory and Diophantine analysis are the other side, they ask what are the solutions.

Hi gAr;

Yes, that makes sense. I am obviously not totally grasping the article. It is very sketchy and has no clear example.

There is also an open problem about putting more than 5 points with pairwise integer distances on the right part of y = x^2. I take it they can find 5 points on a parabola but no one knows if there are more.

I think we have shown that we can generate as many points as needed. That answers the question for 2D.