Thread: Different of two consecutive triangulars is always a cube

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?

You have stated the theorem incorrectly. The correct statement is:

“The difference between the squares of two consecutive triangular numbers is always a cube.”

Now try proving that.

thanks, I solved the problem now!

Hello, tttcomrader!


Please check the wording of that problem.

The statement is patently impossible.