32. [16] Over all pairs of complex numbers $(x, y)$ satisfying the equations
$$
x+2 y^{2}=x^{4} \quad \text { and } \quad y+2 x^{2}=y^{4},
$$
compute the minimum possible real part of $x$.
Answer: $\sqrt[3]{\frac{1-\sqrt{33}}{2}}$
Solution 1: Note the following observations:
(a) if $(x, y)$ is a solution then $\left(\omega x, \omega^{2} y\right)$ is also a solution if $\omega^{3}=1$ and $\omega \neq 1$.
(b) we have some solutions $(x, x)$ where $x$ is a solution of $x^{4}-2 x^{2}-x=0$.

These are really the only necessary observations and the first does not need to be noticed immediately. Indeed, we can try to solve this directly as follows: first, from the first equation, we get $y^{2}=\frac{1}{2}\left(x^{4}-x\right)$, so inserting this into the second equation gives
$$
\begin{aligned}
\frac{1}{4}\left(x^{4}-x\right)^{2}-2 x^{2} & =y \\
\left(\left(x^{4}-x\right)^{2}-8 x^{2}\right)^{2}-8 x^{4}+8 x & =0 \\
\left(x^{8}-2 x^{5}-7 x^{2}\right)^{2}-8 x^{4}+8 x & =0 \\
\underbrace{x^{16}+\cdots+41 x^{4}+8 x}_{P(x)} & =0
\end{aligned}
$$

By the second observation, we have that $x\left(x^{3}-2 x-1\right)$ should be a factor of $P$. The first observation gives that $\left(x^{3}-2 \omega x-1\right)\left(x^{3}-2 \omega^{2} x-1\right)$ should therefore also be factor. Now $\left(x^{3}-2 \omega x-1\right)\left(x^{3}-\right.$ $\left.2 \omega^{2} x-1\right)=x^{6}+2 x^{4}-2 x^{3}+4 x^{2}-2 x+1$ since $\omega$ and $\omega^{2}$ are roots of $x^{2}+x+1$. So now we see that the last two terms of the product of all of these is $-5 x^{4}-x$. Hence the last two terms of the polynomial we get after dividing out should be $-x^{3}-8$, and given what we know about the degree and the fact that everything is monic, the quotient must be exactly $x^{6}-x^{3}-8$ which has roots being the cube roots of the roots to $x^{2}-x-8$, which are $\sqrt[3]{\frac{1 \pm \sqrt{33}}{2}}$. Now $x^{3}-2 x-1$ is further factorable as $(x-1)\left(x^{2}-x-1\right)$ with roots $1, \frac{1 \pm \sqrt{5}}{2}$ so it is not difficult to compare the real parts of all roots of $P$, especially since 5 are real and non-zero, and we have that $\operatorname{Re}(\omega x)=-\frac{1}{2} x$ if $x \in \mathbb{R}$. We conclude that the smallest is $\sqrt[3]{\frac{1-\sqrt{33}}{2}}$.

Solution 2: Subtracting the second equation from the first, we get:
$$
\begin{array}{c}
\left(y+2 x^{2}\right)-\left(x+2 y^{2}\right)=y^{4}-x^{4} \Longrightarrow \\
(x-y)+2\left(x^{2}-y^{2}\right)=\left(x^{2}-y^{2}\right)\left(x^{2}+y^{2}\right) \Longrightarrow
\end{array}
$$
$$
(x-y)\left(1-(x+y)\left(x^{2}+y^{2}+2\right)\right)=0
$$

Subtracting $y$ times the first equation from $x$ times the second, we get:
$$
\begin{array}{c}
\left(x y+2 y^{3}\right)-\left(x y+2 x^{3}\right)=x^{4} y-x y^{4} \Longrightarrow \\
2\left(y^{3}-x^{3}\right)=x y\left(x^{3}-y^{3}\right) \Longrightarrow \\
\left(x^{3}-y^{3}\right)(2+x y)=0
\end{array}
$$

Subtracting $y^{2}$ times the second equation from $x^{2}$ times the first, we get:
$$
\begin{array}{c}
\left(x^{3}+2 x^{2} y^{2}\right)-\left(y^{3}+2 x^{2} y^{2}\right)=x^{6}-y^{6} \Longrightarrow \\
x^{3}-y^{3}=\left(x^{3}+y^{3}\right)\left(x^{3}-y^{3}\right) \Longrightarrow \\
\left(x^{3}-y^{3}\right)\left(1-x^{3}-y^{3}\right)=0
\end{array}
$$

We have three cases.
Case 0. $x=0$ Thus, $(x, y)=(0,0)$ is the only valid solution.
Case 1. $x=\omega y$ for some third root of unity $\omega$. Thus, $y^{2}=\omega^{4} x^{2}=\omega x^{2}$
$$
\begin{array}{c}
x+2 y^{2}=x^{4} \Longrightarrow \\
x+2 \omega x^{2}=x^{4} \Longrightarrow \\
x(1+\omega)\left(2-\omega x^{2}\right)=1
\end{array}
$$

Note that $x=-\omega$ is always a solution to the above, and so we can factor as:
$$
\begin{array}{c}
x^{3}+2(1+\omega) x-1=0 \\
(x+\omega)\left(x^{2}-\omega x-\omega^{2}\right)=0
\end{array}
$$
and so the other solutions are of the form:
$$
x=\frac{1 \pm \sqrt{5}}{2} \cdot \omega
$$
for the third root of unity. The minimum real part in this case is $-\frac{1+\sqrt{5}}{2}$ when $\omega=1$.
Case 2. Since $x^{3}-y^{3} \neq 0$, we have $x y=-2$ and $x^{3}+y^{3}=1$.
Thus, $x^{3}-y^{3}=\sqrt{\left(x^{3}+y^{3}\right)^{2}-4(x y)^{2}}= \pm \sqrt{33} \Longrightarrow x^{3}=\left(\frac{1 \pm \sqrt{33}}{2}\right)$
This yields the minimum solution of $x=\left(\frac{1-\sqrt{33}}{2}\right)^{1 / 3}$ as desired. This is satisfied by letting $y=\left(\frac{1+\sqrt{33}}{2}\right)^{1 / 3}$.