Question

X^2 = Y + a
Y^2 = Z + a
Z^2 = X + a - need help ??

Answer 1

I need to find X,Y,Z

Answer 2

By substitution:

y = x^2-a <<<<this what I got
(x^2-a)^2 = z+a

so
z = (x^2-a)^2-a

then using z^2 = x+a we get
((x^2-a)^2-a)^2 = x+a

Then working this out

x^8-4x^6a+2x^4(3a^2-a)-4x^2(a^3-a^2)-x+a^4-2a^3+a^2-a=0

Also:

x^2- x - a = 0

because
x^2 = a + y
so
x^2 = a + sqrt(z + a)
which exands to
x^2 = a + sqrt(a + sqrt(a + x))

Answer 3

Can someone check it if its right ??