## 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:

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

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

$\displaystyle a^2 = 1 + 3 + 5 + ... + a$_o$$Where \displaystyle a_o$$ is the $\displaystyle a$_t_h$$odd number in the consecutive sum. With a little work, we can see that: \displaystyle a^2 = (b+1)_o + (b+2)_o + ... + c_o$$
$\displaystyle b^2 = (a+1)$_o$+ (a+2)$_o$+ ... + c$_o$$By solving for \displaystyle a^2 and \displaystyle b^2 respectively. Basically, we see that there are 2 partitions in \displaystyle c^2 that don't start at 1, but end in \displaystyle c_0$$ that equal both $\displaystyle a^2$ and $\displaystyle b^2$

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

$\displaystyle 5^2 = 1 + 3 + 5 + [7 + [9]]$

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

the above partitions only occur in Pythagorean triples.

Now, lets extend this concept to another set of Diophantine equations:

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

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

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

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

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:

$\displaystyle a^2 = (z+1)$_o$+ ... + R$_o$$\displaystyle b^2 = (y+1)_o + ... + R_o$$

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

$\displaystyle y^2 = (b+1)$_o$+ ... + R$_o$$\displaystyle z^2 = (a+1)_o + ... + R_o$$

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

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

Thoughts? Comments on where to go from here?