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May 27th, 2014, 03:01 PM   #1
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cube

You know the integer triangles? I used those to form a cube.


$\displaystyle (2mn)^2+(m^2-n^2)^2+(((m^2+n^2)^2-1)/2)^2=(((m^2+n^2)^2+1)/2)^2$

Think of a 3 4 5 triangle combined with a 5 12 13 triangle, using the hypotenuse of the first triangle to generate the second triangle. Remove the 5 and the 13.

So a 3w x 4l x 12h cube has a 13 distance
$\displaystyle 3^2+4^2+12^2=13^2$

When done with consecutive pell numbers the product is a perfect square. Kind of a nice coincidence which ends up being the volume.
$\displaystyle 3*4*12=12^2$

20 21 29, 29 420 421
$\displaystyle 20*21*420=420^2$

Last edited by tejolson; May 27th, 2014 at 03:35 PM.
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May 27th, 2014, 08:49 PM   #2
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So your question is...?
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May 27th, 2014, 10:23 PM   #3
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Math Focus: Geometry
It is not a cube.

And it is more like an algebra problem.

$\displaystyle a^2 + b^2 + c^2 = d^2$

Has infinitely many solutions.

Assume, $\displaystyle a^2 + b^2 = u^2$ and $\displaystyle u^2 + c^2 = d^2$

$\displaystyle a,b,c,d,u \in \mathbb{N}$
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