May 27th, 2014, 03:01 PM  #1 
Member Joined: Oct 2012 Posts: 56 Thanks: 0  cube
You know the integer triangles? I used those to form a cube. $\displaystyle (2mn)^2+(m^2n^2)^2+(((m^2+n^2)^21)/2)^2=(((m^2+n^2)^2+1)/2)^2$ 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. 
May 27th, 2014, 08:49 PM  #2 
Senior Member Joined: Sep 2012 From: British Columbia, Canada Posts: 764 Thanks: 53 
So your question is...?

May 27th, 2014, 10:23 PM  #3 
Senior Member Joined: Nov 2013 From: Baku Posts: 502 Thanks: 56 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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