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/
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