Show that the difference between two consecutive triangular numbres is always a cube.
Proof. Let x be a triangular number, then x = k(k+1)/2 for some integers k.
Let y be a consecutive triangular number, then y = (k+1)(k+2)/2.
y-x = (k+2-k)(k+1)/2 = k+1
But then there isn't any promise that it is a cube?