I had some free time from my work to do a little exploration regarding the existence (or non existence) of perfect cuboids. A solution is represented by the set of Diophantine equations:

$a^2 + b^2 = x^2$

$a^2 + c^2 = y^2$

$b^2 + c^2 = z^2$

$a^2 + b^2 + c^2 = R^2$

If there exists a perfect cuboid, it implies that there is a consecutive sum of odd numbers from 1 to the $Rth$ odd number representing the space diagonal of a perfect cuboid, and that Six other perfect squares representing the face diagonals and edges also exist with the sum. It also means that each edge and face diagonal can be represented as a sum starting from 1 within $R^2$

I represent this consecutive arithmetic sequence like this:

$1+3+5+...+ R_o = R^2$ Where $R_o$ is the $Rth$ odd number in a consecutive sequence.

The set of edge, face diagonals and the body diagonals $[a,b,c,x,y,z,R]$ can each be represented similarly.

By using the fourth equation in the set, we see that:

$(1+3+5+...+a_o) + (1+3+5+...+b_o) + (1+3+5+...+c_o) = 1+3+5+...+R_o$

Then

$(1+3+5+...+x_o) + (1+3+5+...+c_o) = 1+3+5+...+R_o$

Since by first equation and our definition, $(1+3+5+...+a_o) + (1+3+5+...+b_o) = (1+3+5+...+x_o)$

Which Implies

$1+3+5+...+x_o = (c+1)_o + ... + R_o$

by subtracting $(1+3+5+...+c_o)$

Which implies:

$x^2 = (c+1)_o + ... + R_o$

By using the same process, we can derive this for the others:

$a^2 = (z+1)_o + ... + R_o$

$b^2 = (y+1)_o + ... + R_o$

$c^2 = (x+1)_o + ... + R_o$

$x^2 = (c+1)_o + ... + R_o$

$y^2 = (b+1)_o + ... + R_o$

$z^2 = (a+1)_o + ... + R_o$

Ultimately, the Perfect Cuboid Problem implies that there are two unique ways to represent each side and face diagonal in two different ways. Another way to see this is that it is possible to represent each side and face diagonal between $1+ ... + R_o$ both forwards and backwards. This also demonstrates that $a<b<c<x<y<z<R$

That is about as far as I got. My efforts now are attempts to demonstrate that it is not possible for one of these to be true, and by extension disproving others. I also believe that by extending the set of Diophantine equations, Perfect "Hypercuboids" can be disproved as well.

Suggestions are appreciated.