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**namealreadychosen****Member**- Registered: 2011-07-23
- Posts: 16

How many points can be placed in the plane so that the distance between every pair of points is an integer? What about in higher dimensions?

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**gAr****Member**- Registered: 2011-01-09
- Posts: 3,479

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!

"Believe nothing, no matter where you read it, or who said it, no matter if I have said it, unless it agrees with your own reason and your own common sense" - Buddha?

"Data! Data! Data!" he cried impatiently. "I can't make bricks without clay."

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,426

hi

is an integer?

Do you mean the same, constant, integer? OR may we consider eg. 3, 4, 5 as an acceptable answer?

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**namealreadychosen****Member**- Registered: 2011-07-23
- Posts: 16

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.

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,426

Ok.

I, too, have gone for Pythagorean triples.

How about any rhombus of the type shown below?

Bob

*Last edited by bob bundy (2011-08-01 00:32:00)*

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**reconsideryouranswer****Member**- Registered: 2011-05-11
- Posts: 172

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.

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**gAr****Member**- Registered: 2011-01-09
- Posts: 3,479

He wants the points to be non-collinear.

"Believe nothing, no matter where you read it, or who said it, no matter if I have said it, unless it agrees with your own reason and your own common sense" - Buddha?

"Data! Data! Data!" he cried impatiently. "I can't make bricks without clay."

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,238

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!

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,426

hi bobbym

which is ...?

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,238

Hi;

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**gAr****Member**- Registered: 2011-01-09
- Posts: 3,479

Hi bobbym,

Nice solution!

Yes, you can get 6 point solution just by adding (4025/17 , 6000/17 ).

Did you observe that they all lie on the circumference of a circle?! ( x² + y² = 180625 )

I believe there are more!

For more, Let us reflect those points to the negative y-axis!

So you get 4 more points! Yippee!!

Anyway, how did you find those five points? Trial and error?

*Last edited by gAr (2011-08-02 21:26:40)*

"Believe nothing, no matter where you read it, or who said it, no matter if I have said it, unless it agrees with your own reason and your own common sense" - Buddha?

"Data! Data! Data!" he cried impatiently. "I can't make bricks without clay."

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,426

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 (2011-08-02 21:37:46)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,238

Hi;

According to the author of the idea, which he was not thorough enough to explain so that I could understand it. There are an infinite number of points.

Yes, there are two types of solutions to ths problem, one with all but one point collinear ( 3 in a line ) and points that lie on a circle.

After that the problem is considered an open one.

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**gAr****Member**- Registered: 2011-01-09
- Posts: 3,479

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.

"Data! Data! Data!" he cried impatiently. "I can't make bricks without clay."

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**gAr****Member**- Registered: 2011-01-09
- Posts: 3,479

Hi bobbym,

Let us consider all the points to be non-collinear.

Are you saying this is an open problem?

"Data! Data! Data!" he cried impatiently. "I can't make bricks without clay."

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,238

Hi gAr;

Yes, if they do not lie on a circle.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**gAr****Member**- Registered: 2011-01-09
- Posts: 3,479

Hi bobbym,

But the question only mentions that they need to be non-collinear.

They may lie on the circumference, isn't it? No three points are collinear in what you have done.

"Data! Data! Data!" he cried impatiently. "I can't make bricks without clay."

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,238

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.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**gAr****Member**- Registered: 2011-01-09
- Posts: 3,479

Hi,

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

I was working on a method:

Take a line segment and draw a parallel line through the origin.

Reflect all the points about that line.

Consider the new points it creates, and check those points.

All the distances I have checked are integers till now, I don't know when it'll fail !

If they are not to be on a circle, I think there can be only 4 points.

Anyway, diophantine type equations are so boring - so little data are available.

"Data! Data! Data!" he cried impatiently. "I can't make bricks without clay."

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,238

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.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**gAr****Member**- Registered: 2011-01-09
- Posts: 3,479

Okay.

If we are using a circle, then too the number of points would be finite, since every point is atleast a unit distance from each other.

We may argue to choose an infinitely big circle!

"Data! Data! Data!" he cried impatiently. "I can't make bricks without clay."

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,238

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.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**gAr****Member**- Registered: 2011-01-09
- Posts: 3,479

Okay.

I cannot think of any method other than brute force.

I could verify for about 10 points on that circle. I'll leave it, it is difficult to keep track of all the points!

"Data! Data! Data!" he cried impatiently. "I can't make bricks without clay."

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,238

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

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**gAr****Member**- Registered: 2011-01-09
- Posts: 3,479

Yes, and for 3D, the eqivalent would be a sphere.

"Data! Data! Data!" he cried impatiently. "I can't make bricks without clay."

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