cube of an integer

Theorem.  Any cube of integer is a difference of two squares, which in the case of a positive cube are the squares of two successive triangular numbers.

For proving the assertion, one needs only to check the identity

$$a^3\equiv {\left(\frac{(a+1)a}{2}\right)}^2-{\left(\frac{(a-1)a}{2}\right)}^2.$$

For example we have  ${(-2)}^3=1^2-3^2$  and  $4^3=64={10}^2-6^2$.

Summing the first $n$ positive cubes, the identity allows http://planetmath.org/encyclopedia/TelescopingSum.htmltelescoping between consecutive brackets,

$1^3+2^3+3^3+4^3+\mathrm{\dots }+n^3$ $\mathrm{\hspace{0.33em}}=[1^2-0^2]+[3^2-1^2]+[6^2-3^2]+[{10}^2-6^2]+\mathrm{\dots }+\left[{\left({\displaystyle \frac{(n+1)n}{2}}\right)}^2-{\left({\displaystyle \frac{(n-1)n}{2}}\right)}^2\right]$

$\mathrm{\hspace{0.33em}}={\left({\displaystyle \frac{n^2+n}{2}}\right)}^2,$

saving only the square ${\left(\frac{(n+1)n}{2}\right)}^2$.  Thus we have this expression presenting the sum of the first $n$ positive cubes (cf. the Nicomachus theorem).

Title	cube of an integer
Canonical name	CubeOfAnInteger
Date of creation	2013-03-22 19:34:33
Last modified on	2013-03-22 19:34:33
Owner	pahio (2872)
Last modified by	pahio (2872)
Numerical id	11
Author	pahio (2872)
Entry type	Theorem
Classification	msc 11B37
Classification	msc 11A25
Related topic	NicomachusTheorem
Related topic	TriangularNumbers
Related topic	DifferenceOfSquares