Multiplying gets a square? (Posted on 2004-06-26)

	Submitted by Federico Kereki
Rating: 3.0000 (5 votes)
Solution:	(Hide)
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

	Subject	Author	Date
	Further Analysis	K Sengupta	2009-01-31 12:33:39
	Solution	Dej Mar	2008-07-26 06:21:09
	An Alternative Methodology	K Sengupta	2007-06-29 12:19:42
	re: quick-one	ayala	2004-09-30 14:31:04
	The proof to the solution	Mohammad	2004-07-17 23:53:32
	No	Bon	2004-07-12 16:10:53
	re: Solution	vije	2004-07-07 13:14:59
	quick-one	vije	2004-07-07 13:12:31
	re: Reply to Ady...FINAL WORD	Ady TZIDON	2004-07-05 00:49:45
	Reply to Ady...	Patricia Stamets	2004-07-04 20:32:29
	re: Solution...un-natural	Ady TZIDON	2004-07-04 18:08:06
	Solution...	Patricia Stamets	2004-07-04 15:36:11
	answer	Cesar Ali	2004-06-30 10:16:09
	re: No	Nick Hobson	2004-06-27 04:30:49
	re: No	Federico Kereki	2004-06-26 22:42:15
	No	Christian Hiort	2004-06-26 21:04:01
	re(2): Solution	Nick Hobson	2004-06-26 18:53:02
	re: Solution	Richard	2004-06-26 15:48:42
	Solution	Nick Hobson	2004-06-26 13:31:01
	re(2): Solution	Penny	2004-06-26 13:27:49
	re: Solution	Nick Hobson	2004-06-26 13:16:42
	re: Ramblings in wrong direction	Ady TZIDON	2004-06-26 13:04:19
	Solution	Penny	2004-06-26 12:47:20
	Ramblings	Victor Zapana	2004-06-26 12:35:53
	Three Ways to Approach this Problem (+ more, maybe)	Victor Zapana	2004-06-26 12:29:10