Problem
ALG-B1-M01-P005 A Cube as a Difference of Squares
#5
★★☆☆☆ Level 2 of 5
Prove that the cube of any positive integer can be written as the difference of squares of two integers.
Try to write \(n^3\) as \((A-B)(A+B)\).
If \(n\) is odd, then \(n^3\) is odd, and \[ n^3=\left(\frac{n^3+1}{2}\right)^2-\left(\frac{n^3-1}{2}\right)^2. \]
If \(n\) is even, write \(n=2m\). Then \(n^3=8m^3\). Take \(A=2m^3+1\), \(B=2m^3-1\). Then \(A^2-B^2=(A-B)(A+B)=2\cdot4m^3=8m^3=n^3\).
The method is to unfold the problem through factors \(A-B\) and \(A+B\).