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 Quadratic and Perfect Square Query (Posted on 2016-04-30)
Determine all possible pairs (X,Y) of positive integers such that each of
X2 +Y and Y2 + 17*X is a perfect square.

 No Solution Yet Submitted by K Sengupta No Rating

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 Solution Comment 2 of 2 |
Let A^2 = X^2 + Y and B^2 = Y^2 + 17X

A must be at least X+1 and B must be at least Y+1.  Substituting these inequalities yields Y >= 2X+1 and (17X-1)/2 >= Y.

Substituting the latter inequality into the first square then implies X^2 < A^2 <= X^2 + (17X-1)/2 < (X+5)^2.  Then A is equal to one of X+1, X+2, X+3, or X+4.

Case 1: A=X+1
Then Y=2X+1, which implies (2X+1)^2 < B^2 = 4x^2 + 21x + 1 < (2X+6)^2.  B is then one of 2X+2, 2X+3, 2X+4, or 2X+5.  Two of these yield integer values: B=2X+4 yields X=3 and B=2X+5 yields X=24.  Then (X,Y) = (3,7) or (24,49) are solutions.

Case 2: A=X+2
Then Y=4X+4, which implies (4X+4)^2 < B^2 = 16x^2 + 49x + 16 < (4X+7)^2.  B is then one of 4X+5 or 4X+6.  Both of these yield integer values: B=4X+5 yields X=1 and B=4X+6 yields X=20.  Then (X,Y) = (1,8) or (20,84) are solutions.

Case 3: A=X+3
Then Y=6X+9, which implies (6X+9)^2 < B^2 = 36x^2 + 125x + 81 < (6X+11)^2.  B is then 6X+10.  This does not yield an integer value.

Case 4: A=X+4
Then Y=8X+16, which implies (8X+16)^2 < B^2 = 64x^2 + 273x + 256 < (8X+18)^2.  B is then 6X+10.  This yields an integer value X=33.  Then (X,Y) = (33,280) is a solution.

Then the set of all positive integer solutions is (X,Y) = (1,8), (3,7), (20,84), (24,49), or (33,280).

 Posted by Brian Smith on 2016-07-22 23:53:36

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