A little background:

Say you have a pythagorean triple, which is a diophantine solution to the pythagorean theorem:

a^2 + b^2 = c^2

Each of these square numbers can be represented as a sum of odd consecutive integers starting at 1. For example;

a^2 = 1 + 3 + 5 + ... + a$_o$

Where a$_o$ is the a$_t_h$ odd number in the consecutive sum.

With a little work, we can see that:

a^2 = (b+1)$_o$ + (b+2)$_o$ + ... + c$_o$
b^2 = (a+1)$_o$ + (a+2)$_o$ + ... + c$_o$

By solving for a^2 and b^2 respectively.

Basically, we see that there are 2 partitions in c^2 that don't start at 1, but end in c$_0$ that equal both a^2 and b^2

A simple example would be the Pythagorean Triple {3, 4, 5},

5^2 = 1 + 3 + 5 + [7 + [9]]

The partitions include [7+9] which is 4^2 and [9] which is 3^2.

the above partitions only occur in Pythagorean triples.

Now, lets extend this concept to another 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

A solution to this set of equations would represent the sides and diagonals of a perfect cuboid.

Using the same method as before, we easily see that:

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$_o$

There are six partitions in R$_o$ that don't start with 1, but end with R$_o$

Therefore, if perfect cuboids exist, there must exist a square number R^2 = 1 + 3 + ... + R$_o$ such that there are six partitions that also equal perfect squares.

Thoughts? Comments on where to go from here?