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 Multiplying gets a square? (Posted on 2004-06-26)
Is it possible to get a perfect square if you multiply three consecutive natural numbers?

 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

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