If a and b are relatively prime, and ab is a square, then a and b are squares. Suppose (n-1)n(n+1)= k^2, where n>1. Then n(n^2-1)= k^2. But n and (n^2-1) are relatively prime. Therefore n^2-1 is a perfect square, which only happens for n=1.
Problem taken from http://rec-puzzles.org/sol.pl/arithmetic/consecutive.product |