6. Solve the system of equations in the domain of real numbers

$$
\begin{aligned}
& k-x^{2}=y \\
& k-y^{2}=z \\
& k-z^{2}=u \\
& k-u^{2}=x
\end{aligned}
$$

with the real parameter $k$ from the interval $\langle 0,1\rangle$.
SOLUTION. By subtracting the third equation of the given system from the first equation, we get

$$
z^{2}-x^{2}=(z-x)(z+x)=y-u .
$$

Similarly, from the second and fourth equations of the given system, we obtain

$$
y^{2}-u^{2}=(y-u)(y+u)=x-z .
$$

From relations (1) and (2), it follows that $x=z$ if and only if $y=u$. Therefore, we further distinguish two cases denoted as a) and b).

a) First, assume that $x=z$ and $y=u$, i.e., we will look for solutions of the given system of equations in the form of ordered quadruples $(x, y, z, u)=(x, y, x, y)$ with unknowns $x$ and $y$. In this case, we can reduce the original system of equations to a system of two equations with unknowns $x, y$ in the form

$$
\begin{aligned}
k-x^{2} & =y \\
k-y^{2} & =x
\end{aligned}
$$

By subtracting one equation from the other, we arrive after a simple rearrangement at the following equation in product form

$$
(y-x)(y+x-1)=0 \text {, }
$$

so one of two (as it turns out, not entirely disjoint) possibilities will occur.

In the first possibility $y-x=0$, the reduced system, after substituting $y=x$, turns into a single quadratic equation

$$
x^{2}+x-k=0 \text {. }
$$

This equation has for any value of the parameter $k \in\langle 0,1\rangle$ (as specified in the problem) two distinct real roots, namely

$$
x_{1,2}=\frac{-1 \pm \sqrt{4 k+1}}{2} .
$$

Therefore, the given problem has at least two solutions

$$
x_{1}=y_{1}=z_{1}=u_{1}=\frac{-1+\sqrt{4 k+1}}{2}, \quad x_{2}=y_{2}=z_{2}=u_{2}=\frac{-1-\sqrt{4 k+1}}{2} .
$$

In the second possibility $x+y-1=0$, by substituting $y=1-x$ we get from the reduced system to the quadratic equation

$$
x^{2}-x+(1-k)=0 \text {. }
$$

Since its discriminant is equal to $4 k-3$, the given equation has real roots only in the case when $k \geqq 3 / 4$. These roots are real numbers

$$
x_{3}=\frac{1+\sqrt{4 k-3}}{2} \quad \text { and } \quad x_{4}=\frac{1-\sqrt{4 k-3}}{2} \text {. }
$$

The corresponding values of $y=1-x$ are then

$$
y_{3}=\frac{1-\sqrt{4 k-3}}{2} \quad \text { and } \quad y_{4}=\frac{1+\sqrt{4 k-3}}{2} .
$$

For the smallest value $k=3 / 4$, however, it holds that $x_{3}=y_{3}=x_{4}=y_{4}=1 / 2$, so we do not get any new solutions not included in the analysis of the first possibility $x=y$ - we come again to the first solution from (3), which of course corresponds to the parameter $k=3 / 4$. For other possible values of the parameter $k$ given by the derived condition $k>3 / 4$ and the restriction $k \leqq 1$, the given problem has two additional solutions

$$
(x, y, z, u)=\left(x_{3}, y_{3}, x_{3}, y_{3}\right) \quad \text { and } \quad(x, y, z, u)=\left(x_{4}, y_{4}, x_{4}, y_{4}\right)
$$

These two solutions are, thanks to the inequalities $x_{3} \neq x_{4}, x_{3} \neq y_{3}$ and $x_{4} \neq y_{4}$, indeed distinct from each other and different from both solutions from (3).

b) Now assume that $x \neq z$ and $y \neq u$. In such a case, we can substitute $x-z$ from relation (2) into the left side of equation (1) and then divide both sides of the equation by the non-zero number $y-u$ to obtain the equation

$$
(x+z)(y+u)=-1
$$

We will use it further to the only obvious conclusion: all four numbers $x, y, z, u$ cannot be negative (otherwise the left side of (5) would be positive). Therefore, some of the numbers $x, y, z, u$ is non-negative. If, however, we show that due to the given assumption $k \in\langle 0,1\rangle$ it easily follows from the equations of the original system the implications

$$
x \geqq 0 \Rightarrow y \geqq 0 \Rightarrow z \geqq 0 \Rightarrow u \geqq 0 \Rightarrow x \geqq 0
$$

then the derived conclusion about the non-negativity of some of the numbers $x, y, z, u$ will mean the non-negativity of all four, and thus we will arrive at a contradiction with the already mentioned consequence of equation (5). In case b), the given problem will have no solution.

It remains to prove, for example, the first of the implications (6), as the proofs of the other three will be analogous. Let then $x \geqq 0$. From the fourth equation of the original system, it follows that $x \leqq k$, so with regard to $k \leqq 1$ we have $0 \leqq x \leqq k \leqq 1$. This, however, means that $x^{2} \leqq k$, since $t^{2} \leqq t$ for every $t \in\langle 0,1\rangle$. From the derived inequality $x^{2} \leqq k$, it follows according to the first equation of the system that $y \geqq 0$, as we wanted to prove. The analysis of case b) is thus complete (with a negative conclusion).

Conclusion. In the case $0 \leqq k \leqq 3 / 4$, the given system has exactly two solutions, which are given by formulas (3), and in the case $3 / 4<k \leqq 1$ it has exactly four solutions, which are determined by formulas (3) and (4).