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

Natural numbers are integers starting from 1, 2, etc...

Consider a sequence of numbers n-1, n, n+1, where n>=2

Two cases:
Case 1:  If n is a perfect square itself, then (n-1)(n+1) = nē-1 cannot be a perfect square for n>=2, as no consective squares differ by 1 except n=1.  Therefore the product (n-1)n(n+1) cannot be a perfect square.

Case 2: If n is not a perfect square itself, then when you do the prime factorization, there must be at least one odd power of some prime number p.  (Eg1. 32 = 2^5, so 2 is the prime number and 5 is the odd power; Eg2. 45 = (3^2)*(5^1), so 5 is the prime number and 1 is the odd power)  Now p divides n, so it cannot divide n-1 nor n+1.  Therefore, the prime factorization of the product (n-1)n(n+1) will have an odd power of p.  Thus the product is not a perfect square.

Posted by Bon on 2004-07-12 16:10:53