Partitioning a Sum of Consecutive Odd Numbers and Perfect Cuboids

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?