6. Find all pairs of real numbers $a$ and $b$ for which the system of equations

$$
\frac{x+y}{x^{2}+y^{2}}=a, \quad \frac{x^{3}+y^{3}}{x^{2}+y^{2}}=b
$$

with unknowns $x$ and $y$ has a solution in the set of real numbers.

(J. Šimša)
Solution. If the given system has a solution $(x, y)$ for numbers $a=A, b=B$, it clearly also has a solution $(k x, k y)$ for any $k \neq 0$ and for numbers $a=\frac{1}{k} A, b=k B$. From this, we see that the existence of a solution to the given system depends only on the value of the product $a b$.

We will therefore first examine the values of the expression

$$
P(u, v)=\frac{(u+v)\left(u^{3}+v^{3}\right)}{\left(u^{2}+v^{2}\right)^{2}}
$$

where the numbers $u$ and $v$ satisfy the normalization condition $u^{2}+v^{2}=1$. According to this, we have

$$
\begin{aligned}
P(u, v) & =(u+v)\left(u^{3}+v^{3}\right)=(u+v)^{2}\left(u^{2}-u v+v^{2}\right)= \\
& =\left(u^{2}+2 u v+v^{2}\right)(1-u v)=(1+2 u v)(1-u v) .
\end{aligned}
$$

Under the condition $u^{2}+v^{2}=1$, the product $u v$ takes all values from the interval $\left\langle-\frac{1}{2}, \frac{1}{2}\right\rangle$ (if $u=\cos \alpha$ and $v=\sin \alpha$, then $\left.u v=\frac{1}{2} \sin 2 \alpha\right)$. Therefore, it suffices to determine the set of values of the function $f(t)=(1+2 t)(1-t)$ on the interval $t \in\left\langle-\frac{1}{2}, \frac{1}{2}\right\rangle$. From the expression

$$
f(t)=-2 t^{2}+t+1=-2\left(t-\frac{1}{4}\right)^{2}+\frac{9}{8}
$$

it follows that the sought set of values is a closed interval with endpoints $f\left(-\frac{1}{2}\right)=0$ and $f\left(\frac{1}{4}\right)=$ $=\frac{9}{8}$.

This means that if the given system has a solution, the parameters $a$ and $b$ must satisfy $0 \leqq a b \leqq \frac{9}{8}$, with the equality $a b=0$ being possible only if $x+y=0$, in which case, however, $a=b=0$.

On the other hand, if some numbers $a$ and $b$ satisfy the inequalities $0<a b \leqq \frac{9}{8}$, there exist, according to the proven result, numbers $u$ and $v$ such that $u^{2}+v^{2}=1$ and $(u+v)\left(u^{3}+v^{3}\right)=a b$. If we denote $a^{\prime}=u+v$ and $b^{\prime}=u^{3}+v^{3}$, then from the equality $a^{\prime} b^{\prime}=a b \neq 0$ it follows that both ratios $a: a^{\prime}$ and $b^{\prime}: b$ have the same value $k \neq 0$. Then the pair $x=k u$ and $y=k v$ is clearly a solution to the system of equations from the problem statement for the considered values of $a$ and $b$.