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Perfect Square Not (Posted on 20080507) 

The pairs of positive integers (A, B) are such that B divides 2A^{2}.
Prove that A^{2} + B cannot be a perfect square.

Submitted by K Sengupta

Rating: 2.0000 (3 votes)


Solution:

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

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