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December 9, 2016
Posted by **ayan** on Tuesday, April 16, 2013 at 3:38am.

- math -
**Reiny**, Tuesday, April 16, 2013 at 9:38amusing the quadratic equation, I solved for y

where a = 1 , b = -2x and c = 2x^2-225

y = (2x ± √(4x^2 - 4(4x^2 - 225) )/2

= (2x ± √(900 - 20x^2)/2

= x ± √(225 - x^2)

now we want x and y to be integers, so

225 - x^2 must be a perfect square

so possible results for 225- a perfect square giving us a perfect square are

225 - 0 = 225 , good one

225 - 1

225 - 4

225 - 9

225 - 16

225 - 25

225 - 36

225 - 49

225 - 64

225 - 81 = 144 ahhhh

225 - 100

225 -121

225 - 144 = 81 --- another one

225 - 169

225 - 196

225- 225 = 0 --- that one works

so we have x = 0, ±9, ±12, ±15

each one will give an integer for y

(0, ±15) , (9, 21), (9, -3), (-9, 3), (-9, -21) ....

so I count 14 such ordered pairs - math -
**hmmm**, Thursday, April 18, 2013 at 2:51ambrilliant says its wrong... 14 is wrong... it meant 2x^2−2xy+y^2=225