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 20111010 02:38:30 