Posted by ayan on Tuesday, April 16, 2013 at 3:38am.
using 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
brilliant says its wrong... 14 is wrong... it meant 2x^2−2xy+y^2=225