No n-th Powers Here! (Posted on 2005-09-05)

	Submitted by McWorter
Rating: 3.4000 (5 votes)
Solution:	(Hide)
(Proof by Claudio Buffara of Brazil) We may write the product as (m-1)m(m+1)=m(m^2-1). Suppose m(m^2-1)=a^n, for some positive integer a and n>0. Since m and m^2-1 are relatively prime, m must be an n-th power too. But then m^2 is also an n-th power. Hence two consecutive integers, m^2-1 and m^2, are n-th powers, with n>1. This contradicts the fact that consecutive n-th powers must differ by at least n ((x+1)^n=x^n+nx^(n-1)+more stuff).

	Subject	Author	Date
	My Solution	Christie	2005-09-20 20:58:36
	solution	goFish	2005-09-14 09:11:17
	re(3): Solution?	McWorter	2005-09-05 22:04:49
	Hmmm...	Dan	2005-09-05 21:37:02
	re(2): Solution?	Tan Kiat Chuan	2005-09-05 17:55:50
	re: Solution?	Old Original Oskar!	2005-09-05 15:03:53
	Solution?	Tan Kiat Chuan	2005-09-05 14:55:00