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For how many ordered pairs (x , y) that are solutions of the equation
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16 Oct 2020, 02:44
Given: (x-y)^2 +2y^2 = 18
To Find: No. of pairs of (x,y) for which x and y will be integers.
(x-y)^2 +2y^2 = 18 can be written as (x-y)^2 = 18 - 2y^2
=> (x-y)^2 = 2(9-y^2)
=> (x-y)^2 = 2 (3+y)(3-y)
Now, since L.H.S is a perfect square, y can take values between [-3,3] only.
(Any value greater than 3 or less than -3 will make R.H.S negative and L.H.S can't be negative as it's a perfect square.)
i.e. -3,-2,-1,0,1,2,3
For y = +3 or -3, R.H.S will be 0
=> x = +3 or -3
=> 2 pairs (3,3) and (-3,-3)
For y = +2 or -2,R.H.S will be 10
Now, 10 is not a perfect square so we will not get integral solutions for x.
For y = +1 or -1,R.H.S will be 16
Now, 16 is a perfect square
=> (x+1)^2 = 16 and (x-1)^2 = 16
=> x+1 = +-4 and x-1 = +-4
we will get 4 pairs by the above 2 quadratics i.e. (5,1),(-3,1),(3,-1),(-5,-1)
For y = 0,R.H.S will be 18
Now, 18 is not a perfect square so we will not get integral solutions for x.
=> total 6 integral solutions i.e. (3,3),(-3,-3),(5,1),(-3,1),(3,-1),(-5,-1)
Option C.