A little background:

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

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

Where is the odd number in the consecutive sum.

With a little work, we can see that:

By solving for and respectively.

Basically, we see that there are 2 partitions in that don't start at 1, but end in that equal both and

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

The partitions include which is and which is .

the above partitionsonlyoccur in Pythagorean triples.

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

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:

There aresixpartitions in that don't start with 1, but end with

Therefore, if perfect cuboids exist, there must exist a square number such that there are six partitions that also equal perfect squares.

Thoughts? Comments on where to go from here?