How many radical solutions has the system:?

Question

x + y= 2
xy-z^2 = 1

Answer 1

Sorry, had to edit a little. All is OK now I hope.

There is only 1 rational solution: x = 1, y = 1, z = 0.

(1) x + y = 2
(2) xy - z^2 =1

From (1), y = 2 - x
Substituting this into (2) gives:
x(2 - x) - z^2 = 1
Expanding:
2x - x^2 - z^2 = 1
Rearranging:
z^2 = -x^2 + 2x - 1
Taking the square root of both sides:
z = sqrt(-x^2 + 2x - 1)

For rational z, the term under the square root sign must be >= 0.
That is: -x^2 + 2x - 1 >= 0
Multiplying through by -1 (don't forget change of sign) gives:
x^2 - 2x + 1 <= 0
Factorising:
(x - 1)^2 <= 0
Taking the square root of both sides:
+(x - 1) <= 0 implies that x <= 1,
or -(x - 1) <= 0 implies that x >= 1.

Now x and y are symmetrical in the original equations,
so if x <= 1 then also, y <= 1, and if x >= 1 then also, y >= 1.

If both x and y are < 1 or both > 1, then they will
never add up to 2 in the equation, x + y = 2.

Therefore, x and y must both be equal to 1,
and in that case, z = 0.

Answer 2

basically you have a system with two equations and 3 variables. Therefore you should have an infinite number of solutions... replacing the first eq. into the second we get y^2 + z^2 = 2x - 1 which the equation of a circle with radius sqrt(2x - 1), square root of 2x - 1. That is, you have infinite circles which satisfy this condition, the only ones excluded are those with x less than 1/2.

Answer 3

you got me this time