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        riddles >> hard >> almost-square Pythagorean rectangles
        
(Message started by: JocK on Jun 12th, 2005, 12:32pm)
Title: almost-square Pythagorean rectangles
Post by JocK on Jun 12th, 2005, 12:32pm
A Pythagorean rectangle is a rectangle with integer edges and integer diagonal. We define an almost-square Pythagorean rectangle as a Pythagorean rectangle whose edges differ by one unit.

Rectangles with sides (3,4), (20,21), (119,120), (696,697) and (4059,4060) are all almost-square Pythagorean rectangles.

Prove that there are infinitely many almost-square Pythagorean rectangles, and give a closed expression for the n-th pair.

Can you generalise the result to higher dimensions?
Title: Re: almost-square Pythagorean rectangles
Post by Barukh on Jun 13th, 2005, 6:00am

on 06/12/05 at 12:32:29, JocK wrote:
A Pythagorean rectangle is a rectangle with integer edges and integer diagonal. We define an almost-square Pythagorean rectangle as a Pythagorean rectangle whose edges differ by one unit.

Rectangles with sides (3,4), (20,21), (119,120), (696,697) and (4059,4060) are all almost-square Pythagorean rectangles.

Prove that there are infinitely many almost-square Pythagorean rectangles, and give a closed expression for the n-th pair.


It seems this boils down to solving [hide]the Pell equation x2 - 2y2 = +-1[/hide], and this was discussed several times at this site.


Quote:
Can you generalise the result to higher dimensions?

What parameters should be integers in n-dimensional cuboid?
Title: Re: almost-square Pythagorean rectangles
Post by JocK on Jun 13th, 2005, 10:03am

on 06/13/05 at 06:00:57, Barukh wrote:
It seems this boils down to solving [hide]the Pell equation x2 - 2y2 = +-1[/hide], and this was discussed several times at this site.


True. The 2D problem is merely intended as a stepping stone to the more general problem in higher dimensions.


on 06/13/05 at 06:00:57, Barukh wrote:
What parameters should be integers in n-dimensional cuboid?


In D dimensions we define a Pythagorean (hyper)brick as a rectangular D-dimensional block with integer edges and integer volume diagonal.

An almost-cubic Pythagorean brick satisfies the additional constraint that no two edges differ in length by more than one unit.

Demonstrate that for any non-square dimension D = {2, 3, 5, 6, 7, 8, 10, 11, ...} infinitely many almost-cubic Pythagorean bricks exist, and give a closed expression for their sizes.
Title: Re: almost-square Pythagorean rectangles
Post by Deedlit on Jun 13th, 2005, 4:49pm
[hideb]
We can take all but one dimension of the brick to be x, and the remaining one x + 1.  Then, letting y be the volume diagonal, we have

D x2 + 2x + 1 = y2
D2 x2 + 2D x + D = D y2
(D x + 1)2 + D - 1 = D y2
(D x + 1)2 - D y2 = 1 - D. (1)

A particular solution to the above is x = 0, y = 1.  We can then generate infinitely many positive solutions by using solving the Pell equation

z2 - D y2 = 1. (2)

For a solution (r, s) to the above equation, we get

(1 - D) (r2 - D s2) = (r + D s)2 - D (r + s)2,

so D x + 1 = r + D s, y = r + s is a solution to (1).  x will be an integer when r = 1 mod D.

Let (r, s) be the smallest positive solution to (2).  (This can be found via the continued fraction expansion of sqrt(d). ) Then the set of all positive solutions is generated by

p  = [(r + s sqrt (D) )n + (r - s sqrt (D) )n] / 2
q  = [(r + s sqrt (D) )n - (r - s sqrt (D) )n] / (2 sqrt (D))

for nonnegative n.  When we expand the expression for p, all rational terms will have a factor of D in them except rn. So p = rn (mod D), and rn = 1 (mod D) infinitely often. (this requires (r, D) = 1, but that is clearly true.)

This is certainly not what is generally considered a closed form expression, but I'm rather doubtful that there is one.

[/hideb]
Title: Re: almost-square Pythagorean rectangles
Post by JocK on Jun 14th, 2005, 10:18am

on 06/13/05 at 16:49:48, Deedlit wrote:
We can take all but one dimension of the brick to be x, and the remaining one x + 1.  


Can you do this WLOG?
Title: Re: almost-square Pythagorean rectangles
Post by THUDandBLUNDER on Jun 14th, 2005, 12:44pm

Quote:
An almost-cubic Pythagorean brick satisfies the additional constraint that no two edges differ in length by more than one unit.



on 06/14/05 at 10:18:43, JocK wrote:
Can you do this WLOG?

As the proof (of an infinite number of solutions) involves a stronger constraint than the one given above, I don't see how any loss of generality is important in this case.
Title: Re: almost-square Pythagorean rectangles
Post by JocK on Jun 14th, 2005, 3:49pm

on 06/14/05 at 12:44:31, THUDandBLUNDER wrote:
As the proof (of an infinite number of solutions) involves a stronger constraint than the one given above, I don't see how any 'loss of generality' is important in this case.


Of course, there is no LOG if the intention was merely to demonstrate there are infinitely many solutions.

However, that is not what I read in Deedlit's post.
Title: Re: almost-square Pythagorean rectangles
Post by Deedlit on Jun 14th, 2005, 4:33pm
THUDandBLUNDER had it right - I was just looking for an infinite family of solutions.  (and I don't think I exhausted that case either - a specific solution to the 1-D case and a general solution to the 1 case doesn't necessarily generate all solutions to the 1-D case.)  

[hideb]
Having a different number of x+1's will certainly change things.  You wind up with

(D x + c)2 - D y2 = - c(D-c)

which only has an obvious solution when c is a perfect square.
[/hideb]


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