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

12 - 02

2

8

3

1

32 -12

3

27

6

3

62 -32

4

64

10

6

102 -62

5

125

15

10

152 -102

6

216

21

15

212 -152

 

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 Tn represent the nth triangular number.  Might it be that Tn2 – Tn-12 is always a perfect cube?


The conjecture makes the chart look like this:

 

A

A3

x

y

"Solution"

1

1

1

0

12 - 02

2

8

3

1

32 -12

3

27

6

3

62 -32

4

64

10

6

102 -62

5

125

15

10

152 -102

6

216

21

15

212 -152

:::

:::

:::

:::

:::

 

 

n

 

 

n3

 

 


 

???

 

 

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).