Prove that the sum of consecutive perfect cubes (starting with 1) is always a perfect square.

For example:
1=1
1+8=9
1+8+27=36

Not just any old perfect square, but the square of the nth triangular number, whereupon the proof becomes easy:

 

I    SIGMA(x=1 to x) x^3 is 1/4*x^2*(x+1)^2

II   (n(n+1)/2)^2=1/4*n^2*(n+1)^2

 

And the two expressions are clearly equivalent.

 

QED

Edited on October 10, 2011, 2:40 am

Posted by broll on 2011-10-10 02:38:30