Is it possible to get a perfect square if you multiply three consecutive natural numbers?

Let's see if (n)(n+1)(n+2) = k^2.

One possible way to have a perfect square is if 2 of the triplet is equal in value to the third.

(n)(n+1) = n+2
n^2 + n = n+2
n^2 - 2 = 0
IRRATIONAL NUMBER REJECT

(n)(n+2) = n+1
n^2 + 2n = n + 1
n^2 + n + 1 = 0
COMPLEX NUMBER REJECT

(n+1)(n+2) = n
n^2 + 3n + 2 = n
n^2 + 2n + 2 = 0
COMPLEX NUMBER REJECT

So far, it can't be a perfect square... Hm...

Edited on June 26, 2004, 12:40 pm

Edited on June 26, 2004, 12:41 pm

Posted by Victor Zapana on 2004-06-26 12:35:53