Solve the system ${x}^{2}+ \left( y-1 \right) ^{2}=4,z^{4}+y{z}^{2}+xz+1=0. $

Question

Find all real solutions of the system of equation \begin{cases} {x}^{2}+ \left( y-1 \right) ^{2}=4,\\{z}^{4}+y{z}^{2}+xz+1=0. \end{cases}

Answer 1

From second equation we get $y=-z^2-\frac{x}{z}-\frac{1}{z^2}$ (it's obvious that $z\neq0$).

Thus, $$x^2+\left(z^2+\frac{1}{z^2}+1+\frac{x}{z}\right)^2=4$$ or $$\left(1+\frac{1}{z^2}\right)x^2+\frac{2}{z}\left(z^2+\frac{1}{z^2}+1\right)x+\left(z^2+\frac{1}{z^2}+1\right)^2-4=0,$$ which gives $$\frac{1}{z^2}\left(z^2+\frac{1}{z^2}+1\right)^2-\left(1+\frac{1}{z^2}\right)\left(\left(z^2+\frac{1}{z^2}+1\right)^2-4\right)\geq0$$ or $$(z^4+z^2-1)^2\leq0,$$ which gives $z^2=\frac{\sqrt5-1}{2}$, $x=-\frac{\frac{1}{z}\left(z^2+\frac{1}{z^2}+1\right)}{1+\frac{1}{z^2}}$ and the rest is smooth.

Answer 2

This system of 2 equations has infinite solutions. Let $x=2\cos\theta$ and $y=2\sin\theta+1$ which satisfy the first equation. Then $$z^4+yz^2+xz+1=0$$ $$z^4+(2\sin\theta+1)z^2+(2\cos\theta)z+1=0$$ $$(z^2+\sin\theta)^2+(z+\cos\theta)^2=0$$ $z=-\cos\theta$ and $z^2=-\sin\theta$, this leads us to find solutions of $\theta$.

$z^2=-\sin\theta$ shows $\sin\theta<0$.

$\sin^2\theta-\sin\theta-1=0$ shows $\theta=\arcsin\dfrac{1-\sqrt{5}}{2}\sim-38.17^\circ$. Only two solution $(x,y,z)$ exist. one for $\theta=(180+38.17)^\circ$ and other for $\theta=(360-38.17)^\circ$ and then $$(x,y,z)=(2\cos\theta,2\sin\theta+1,-\cos\theta)$$

Answer 3

let $t = y-1$

$x^2+t^2 = 4$

$tz^2 + xz = -z^4 - z^2 - 1$

Solutions (x,t) exist if and only if $\frac{z^4 +z^2 +1}{\sqrt{z^4 +z^2}} \le 2$ (geometrically)

Let $k = z^4 + z^2$

$\frac{k +1}{sqrt(k)} \le 2$ equals $k^2 +2k +1 \le 4k$ equals $k^2 -2k +1 \le 0$ equals k = 1;

So, $z^4 + z^2 = 1$