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 March 16th, 2013, 07:08 PM #1 Senior Member   Joined: Feb 2013 Posts: 114 Thanks: 0 theory of numbers Find n, if $2^{200}-2^{192}.31+2^n$ is a perfect square. $2^{200}-2^{192}.31+2^n= 2^{192}(2^8-31)+2^n = 2^{192}(256-31)+2^n = 2^{192}.225+2^n$ For, some $m \in N$ $2^n= m^2-2^{192}.225 = (m-2^{96}.15)(m+2^{96}.15)$ Please guide further how to get the result...
 March 17th, 2013, 02:37 AM #2 Global Moderator   Joined: Dec 2006 Posts: 21,015 Thanks: 2250 It's easier to complete the solution if you notice that n = 198 works.

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