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
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