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 October 28th, 2013, 02:57 AM #1 Member   Joined: Nov 2012 Posts: 61 Thanks: 0 maximum value If a,b are natural numbers such that $2013 +a^2= b^2,$ then the maximum possible value of ab is
 October 28th, 2013, 03:31 AM #2 Math Team     Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233 Re: maximum value Here is an 'intuitive' approach. The difference between consecutive integer squares is in arithmetic progression of the odd integers. $1^2 \ - \ 0^2 \ = \ 1 \\ 2^2 \ - \ 1^2 \ = \ 3 \\ 3^2 \ - \ 2^2 \ = \ 5 \\ 4^2 \ - \ 3^2 \ = \ 7$ . . . $(n + 1)^2 \ - \ n^2 \= \ 2n \ + \ 1$ So we want 2n + 1 = 2013 which makes n = 1006 $2013 \ + \ 1006 ^2 \= \ 1007^2$ ab = 1006*1007 = 1013042 This is the maximum since it gives $2013 \= \ b^2 \ - \ a^2$ $2013 \= \ (b \ + \ a)(b \ - \ a)$ $2013 \= \ (1007 \ + \ 1006)(1007 \ - \ 1006)$ $2013 \= \ (2013)(1)$ WE can't do better than that because there are no integer squares between consecutive integer squares.
 November 8th, 2013, 02:29 AM #3 Member   Joined: Nov 2012 Posts: 61 Thanks: 0 Re: maximum value Can you tell what will be the minimum value?

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