Clay Kitchings :: EMAT 6600 :: Spring 2007

The following problem was posed on the Kennesaw State University Mathematics Competition from February 7, 2007.

Prove that always has integer solutions (x, y) whenever A is a positive integer.

Let us first consider some concrete examples of this and look for a pattern (if one exists).

I created the following spreadsheet/table to search for solutions for the first A = 1…6.

A	A^3	x	y	"Solution"
1	1	1	0	1^2 - 0^2
2	8	3	1	3^2 -1^2
3	27	6	3	6^2 -3^2
4	64	10	6	10^2 -6^2
5	125	15	10	15^2 -10^2
6	216	21	15	21^2 -15^2

For each A³, I searched for possible solutions and came up with the ones listed in the “Solution” column above.

The pattern so far seems to suggest a familiar sequence of numbers. I notice the formation of the triangular numbers in both the x and y columns. Let T_n represent the nth triangular number. Might it be that T_n² – T_n-1² is always a perfect cube?

The conjecture makes the chart look like this:

A	A^3	x	y	"Solution"
1	1	1	0	1^2 - 0^2
2	8	3	1	3^2 -1^2
3	27	6	3	6^2 -3^2
4	64	10	6	10^2 -6^2
5	125	15	10	15^2 -10^2
6	216	21	15	21^2 -15^2
:::	:::	:::	:::	:::
n	n^3			???

Is there a “Solution” for the above?

Let’s take x and y to be the representatives listed in the above chart and simplify x²-y².

So what just happened? We started with some positive integer n and assumed it could be written in some kind of form – let’s call it, A³. We then found some candidates “x” and “y” such that their difference x²-y² yielded a perfect cube using the convenience of the triangular numbers pattern we found in the first six solutions we found.

Therefore, for any positive integer A, the equation always has integer solutions (x, y).