, 2024

Solution:
Note the following observations:
(a) if $(x, y)$ is a solution then $(\omega x, \omega^{2} y)$ 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 $x^{4}-2x^{2}-x=0$.

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 $y^{2}=\frac{1}{2}(x^{4}-x)$, so inserting this into the second equation gives
$$\begin{aligned} \frac{1}{4}(x^{4}-x)^{2}-2x^{2} & = y \\ \left((x^{4}-x)^{2}-8x^{2}\right)^{2}-8x^{4}+8x & = 0 \\ \left(x^{8}-2x^{5}-7x^{2}\right)^{2}-8x^{4}+8x & = 0 \\ \underbrace{x^{16}+\cdots+41x^{4}+8x}_{P(x)} & = 0 \end{aligned}$$
By the second observation, we have that $x(x^{3}-2x-1)$ should be a factor of $P$. The first observation gives that $(x^{3}-2\omega x-1)(x^{3}-2\omega^{2} x-1)$ should therefore also be a factor. Now $(x^{3}-2\omega x-1)(x^{3}-2\omega^{2} x-1)=x^{6}+2x^{4}-2x^{3}+4x^{2}-2x+1$ 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 $-5x^{4}-x$. 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 $x^{3}-2x-1$ is further factorable as $(x-1)(x^{2}-x-1)$ 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:
Subtracting the second equation from the first, we get:
$$\begin{gathered} \left(y+2x^{2}\right)-\left(x+2y^{2}\right)=y^{4}-x^{4} \Longrightarrow \\ (x-y)+2\left(x^{2}-y^{2}\right)=\left(x^{2}-y^{2}\right)\left(x^{2}+y^{2}\right) \Longrightarrow \end{gathered}$$
$$(x-y)\left(1-(x+y)(x^{2}+y^{2}+2)\right)=0$$
Subtracting $y$ times the first equation from $x$ times the second, we get:
$$\begin{gathered} \left(xy+2y^{3}\right)-\left(xy+2x^{3}\right)=x^{4}y-xy^{4} \Longrightarrow \\ 2\left(y^{3}-x^{3}\right)=xy\left(x^{3}-y^{3}\right) \Longrightarrow \\ \left(x^{3}-y^{3}\right)(2+xy)=0 \end{gathered}$$
Subtracting $y^{2}$ times the second equation from $x^{2}$ times the first, we get:
$$\begin{gathered} \left(x^{3}+2x^{2}y^{2}\right)-\left(y^{3}+2x^{2}y^{2}\right)=x^{6}-y^{6} \Longrightarrow \\ x^{3}-y^{3}=\left(x^{3}+y^{3}\right)\left(x^{3}-y^{3}\right) \Longrightarrow \\ \left(x^{3}-y^{3}\right)\left(1-x^{3}-y^{3}\right)=0 \end{gathered}$$
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, $y^{2}=\omega^{4} x^{2}=\omega x^{2}$
$$\begin{gathered} x+2y^{2}=x^{4} \Longrightarrow \\ x+2\omega x^{2}=x^{4} \Longrightarrow \\ x(1+\omega)\left(2-\omega x^{2}\right)=1 \end{gathered}$$
Note that $x=-\omega$ is always a solution to the above, and so we can factor as:
$$\begin{gathered} x^{3}+2(1+\omega)x-1=0 \\ (x+\omega)\left(x^{2}-\omega x-\omega^{2}\right)=0 \end{gathered}$$
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 $xy=-2$ and $x^{3}+y^{3}=1$.
Thus, $x^{3}-y^{3}=\sqrt{\left(x^{3}+y^{3}\right)^{2}-4(xy)^{2}}= \pm \sqrt{33} \Longrightarrow x^{3}=\left(\frac{1 \pm \sqrt{33}}{2}\right)$
This yields the minimum solution of $x=\left(\frac{1-\sqrt{33}}{2}\right)^{1 / 3}$ as desired. This is satisfied by letting $y=\left(\frac{1+\sqrt{33}}{2}\right)^{1 / 3}$.