 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Multiplying gets a square? (Posted on 2004-06-26) Is it possible to get a perfect square if you multiply three consecutive natural numbers?

 See The Solution Submitted by Federico Kereki Rating: 3.0000 (5 votes) Comments: ( Back to comment list | You must be logged in to post comments.) An Alternative Methodology | Comment 23 of 25 | Let the three consecutive natural numbers be denoted by n-1, n and n+1

By the conditions of the problem, we must have:

n-1> 0, so that:
n> 1
Or, n> =2

Now, in accordance with the provisions governing the puzzle, there exists a positive integer p, such that:

n(n-1)(n+1) = p^2
Or, n(n^2 - 1) = p^2

This is possible iff:

n^2 - 1 = n*m^2, for some positive integer m
Or, n(n-m^2) = 1

Now, n is a natural number factor of 1, so that n =1. This violates the restriction n> =2, and accordingly it follows that  it is not possible to get a perfect square if you multiply three consecutive natural numbers.

 Posted by K Sengupta on 2007-06-29 12:19:42 Please log in:

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