x= 4-x/y^2-x
How many ordered pairs of positive integers (x,y) satisfy the equation above.
Multiple Choice answers
a) 0
b) 1
c) 2
d) 3
e) 4
x = (4 - x) / (y^2 - x)
y^2x - x^2 = 4 - x
y^2x = x^2 - x + 4
y^2 = [ x^2 - x + 4 ] / x
y^2 = x - 1 + 4/x
Notice that the first two terms always result in an integer for any integer value of x
But 4/x is only an integer when x = ±1, ±2 or ±4
But we can reject the negative values because they make the right side negative.....so y is not a real number for these values
And when x = 2, y = ±√3 which isn't an integer pair
So.....when x = 1, y = ±2
And when x = 4, y = ±2
So......the ordered pairs of positive integers that make this true are (1,2), (-1,2), (4, 2) and (4 , -2)
So....the correct answer is (e)