How do I solve this System of Equations?

Question

How do I begin to solve this system? $$x^2=y+a$$$$y^2=z+a$$$$z^2=x+a$$

Do I take the square roots of $x,y$ and $z$? If so, we get $$x=\pm\sqrt{a+y}$$$$z=\pm\sqrt{a\pm\sqrt{a+y}}$$$$y=\pm\sqrt{a\pm\sqrt{a\pm\sqrt{a+y}}}$$

What do I do now? I'm confused as to what to do.

Answer 1

Proceeding from Sonnhard's solution, besides $z = (1 \pm \sqrt{1+4a})/2$ from the factor $-z^2 + z + a$, we get the two cubics $$z^3 + \frac{1\pm r}{2} z^2 + \frac{\pm r - 2a - 1}{2} z + \frac{a \mp ra -2}{2}$$ where $r = \sqrt{4a-7}$. If you're interested in real solutions, you need $ a \ge -1/4$ for the first pair of solutions, and $a \ge 7/4$ for the cubics.

The solutions for $z = (1 \pm \sqrt{1+4a})/2$ have the same values for $x$ and $y$. The solutions for the cubics have $x, y, z$ the three roots of the cubic, in decreasing order (or cyclically permuted).

Here's a plot: the black curve is $ z = (1 \pm \sqrt{1+4a})/2$, the red and blue curves are the roots of the two cubics.

Answer 2

from the first equation we obtain $$y=x^2-a$$ plugging this in the second one we get $$(x^2-a)^2=z+a$$ with the third equation we get an equation only in $z$ $$((z^2-a)^2-a)^2=z+a$$