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 partitions only occur 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 are six partitions 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?