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Topic: almost-square Pythagorean rectangles (Read 1055 times) |
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JocK
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almost-square Pythagorean rectangles
« on: Jun 12th, 2005, 12:32pm » |
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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?
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| « Last Edit: Jun 12th, 2005, 3:29pm by JocK » |
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solving abstract problems is like sex: it may occasionally have some practical use, but that is not why we do it.
xy - y = x5 - y4 - y3 = 20; x>0, y>0.
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Barukh
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Re: almost-square Pythagorean rectangles
« Reply #1 on: Jun 13th, 2005, 6:00am » |
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on Jun 12th, 2005, 12:32pm, 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. |
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It seems this boils down to solving the Pell equation x2 - 2y2 = +-1, and this was discussed several times at this site.
Quote:| Can you generalise the result to higher dimensions? |
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What parameters should be integers in n-dimensional cuboid?
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JocK
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Re: almost-square Pythagorean rectangles
« Reply #2 on: Jun 13th, 2005, 10:03am » |
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on Jun 13th, 2005, 6:00am, Barukh wrote:
It seems this boils down to solving the Pell equation x2 - 2y2 = +-1, and this was discussed several times at this site. |
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True. The 2D problem is merely intended as a stepping stone to the more general problem in higher dimensions.
on Jun 13th, 2005, 6:00am, Barukh wrote:
What parameters should be integers in n-dimensional cuboid? |
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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.
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solving abstract problems is like sex: it may occasionally have some practical use, but that is not why we do it.
xy - y = x5 - y4 - y3 = 20; x>0, y>0.
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Deedlit
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Re: almost-square Pythagorean rectangles
« Reply #3 on: Jun 13th, 2005, 4:49pm » |
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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.
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| « Last Edit: Jun 13th, 2005, 4:52pm by Deedlit » |
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JocK
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Re: almost-square Pythagorean rectangles
« Reply #4 on: Jun 14th, 2005, 10:18am » |
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on Jun 13th, 2005, 4:49pm, Deedlit wrote:
We can take all but one dimension of the brick to be x, and the remaining one x + 1. |
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Can you do this WLOG?
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| « Last Edit: Jun 14th, 2005, 10:22am by JocK » |
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solving abstract problems is like sex: it may occasionally have some practical use, but that is not why we do it.
xy - y = x5 - y4 - y3 = 20; x>0, y>0.
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ThudnBlunder
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Re: almost-square Pythagorean rectangles
« Reply #5 on: Jun 14th, 2005, 12:44pm » |
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Quote:An almost-cubic Pythagorean brick satisfies the additional constraint that no two edges differ in length by more than one unit.
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on Jun 14th, 2005, 10:18am, JocK 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.
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| « Last Edit: Jun 14th, 2005, 10:41pm by ThudnBlunder » |
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JocK
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Re: almost-square Pythagorean rectangles
« Reply #6 on: Jun 14th, 2005, 3:49pm » |
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on Jun 14th, 2005, 12:44pm, 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.
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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.
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| « Last Edit: Jun 14th, 2005, 3:51pm by JocK » |
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solving abstract problems is like sex: it may occasionally have some practical use, but that is not why we do it.
xy - y = x5 - y4 - y3 = 20; x>0, y>0.
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Deedlit
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Posts: 476
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Re: almost-square Pythagorean rectangles
« Reply #7 on: Jun 14th, 2005, 4:33pm » |
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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.)
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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.
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