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Multiplying gets a square? (Posted on 2004-06-26) Difficulty: 3 of 5
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.

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  Subject Author Date
SolutionFurther AnalysisK Sengupta2009-01-31 12:33:39
SolutionSolutionDej Mar2008-07-26 06:21:09
SolutionAn Alternative MethodologyK Sengupta2007-06-29 12:19:42
re: quick-oneayala2004-09-30 14:31:04
SolutionThe proof to the solutionMohammad2004-07-17 23:53:32
NoBon2004-07-12 16:10:53
Some Thoughtsre: Solutionvije2004-07-07 13:14:59
Some Thoughtsquick-onevije2004-07-07 13:12:31
Some Thoughtsre: Reply to Ady...FINAL WORDAdy TZIDON2004-07-05 00:49:45
Reply to Ady...Patricia Stamets2004-07-04 20:32:29
Hints/Tipsre: Solution...un-naturalAdy TZIDON2004-07-04 18:08:06
Solution...Patricia Stamets2004-07-04 15:36:11
answerCesar Ali2004-06-30 10:16:09
Some Thoughtsre: NoNick Hobson2004-06-27 04:30:49
re: NoFederico Kereki2004-06-26 22:42:15
NoChristian Hiort2004-06-26 21:04:01
re(2): SolutionNick Hobson2004-06-26 18:53:02
re: SolutionRichard2004-06-26 15:48:42
SolutionSolutionNick Hobson2004-06-26 13:31:01
re(2): SolutionPenny2004-06-26 13:27:49
Some Thoughtsre: SolutionNick Hobson2004-06-26 13:16:42
Hints/Tipsre: Ramblings in wrong directionAdy TZIDON2004-06-26 13:04:19
SolutionPenny2004-06-26 12:47:20
RamblingsVictor Zapana2004-06-26 12:35:53
Three Ways to Approach this Problem (+ more, maybe)Victor Zapana2004-06-26 12:29:10
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