Cubes as Differences of Squares

Date: 07/04/2002 at 10:21:16
From: Steve Lander
Subject: Geometry

Hi, 

I am really stuck here. Can you help me out? It seems to be an 
easy question but I can't see how to approach or attack it. 
Here is the problem: 

  Prove that the cube of any positive integer is equal 
  to the difference of the squares of two integers.

Here is how I think the proof should start:

  Let: a^3 = b^2 - c^2

I know that it is true for example if I let a=2, b=3 and c=1.

However, how would you prove it in a general form?

Date: 07/04/2002 at 10:34:30
From: Doctor Jubal
Subject: Re: Geometry

Hi Steve,

Thanks for writing Dr. Math.

Let's look at the first few examples, and see if we notice any 
patterns.

  1^3 =   1 =   1 -   0 =  1^2 -  0^2
  2^3 =   8 =   9 -   1 =  3^2 -  1^2
  3^3 =  27 =  36 -   9 =  6^2 -  3^2
  4^3 =  64 = 100 -  36 = 10^2 -  6^2
  5^3 = 125 = 225 - 100 = 15^2 - 10^2
  6^3 = 216 = 441 - 225 = 21^2 - 15^2

Now, each of these cubes can be written as the difference of two 
squares.  Moreover, in each case the smaller of the two squares is the 
larger of the two squares for the previous case.  Moreover, the 
squares are all squares of triangular numbers, T(n) = n(n+1)/2

So it looks like

  n^3 = [(n)(n+1)/2]^2 - [(n)(n-1)/2]^2

Can you prove this formula is generally true?  Write back if you'd 
like to talk about this some more, or if you have any other questions.

- Doctor Jubal, The Math Forum
  http://mathforum.org/dr.math/