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 Perfect Square Not (Posted on 2008-05-07)
The pairs of positive integers (A, B) are such that B divides 2A2.

Prove that A2 + B cannot be a perfect square.

 Submitted by K Sengupta Rating: 2.0000 (3 votes) Solution: (Hide) At the outset, we substitute 2A2 = c*B. Since B divides 2A2, it follows that c must be a positive integer. If possible, let us suppose that: A2 + B = G2, for some positive integer G Then, we have: A2*c2 + B*c2 = G2*c2 or, A2*c2 + 2A2*c = G2*c2 or, A2(c2 + 2c) = (G*c)2......(*) But, c2 < c2 + 2c < (c+1)2, and accordingly, the lhs of (*) cannot be the square of a positive integer. This leads to a contradiction. Consequently, A2 + B cannot be a perfect square. *************************************** For an alternative methodology, refer to the solution submitted by Praneeth in this location.

 Subject Author Date re: Solution K Sengupta 2008-05-26 14:13:59 Another solution Jonathan Lindgren 2008-05-24 15:52:32 Solution Praneeth 2008-05-09 04:00:07 re: To B not a square >>>. Revisited Ady TZIDON 2008-05-07 19:15:48 To B not a square Gamer 2008-05-07 17:22:09 NO WAY Ady TZIDON 2008-05-07 11:14:10

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