Find z, if it is known that:
a: $$-y^2 + x^2 = 20$$
b: $$y^3 - 2x^2 - 4z \geq -12$$ and
c: x, y and z are all positive integers
Since $$x^2-y^2=20$$ and x,y,z are positive integers,
(x+y)*(x-y) = 20, Hence x-y, x+y are factors of 20.
Since x, y are positive integers, x+y is always positive, and for the product of (x+y)*(x-y) to be positive x-y must be positive.
x, y are positive integers and x-y is positive x must be greater than y.
The possible cases are : (x+y = 10, x-y = 2), (x+y = 5, x-y = 4).
The second case fails because we get x =9/2, y = 1/2 but x, y are integral values
For case one x = 6, y = 4.
$$y^3 - 2x^2 - 4z \geq -12$$
Substituting the values of x and y, we have :
64 - 72 - 4*z $$\ge\ -12$$
-8 - 4*z $$\ge\ -12$$
Since x, y, z are positive integers, the only possible value for z is 1.
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