66th Czech and Slovak Mathematical Olympiad

Subtracting the third equation from the first one we get $$z^2-x^2=(z-x)(z+x)=y-u.\text{\hspace{1em}(1)}$$ Similarly, the second and the fourth equation imply $$y^2-u^2=(y-u)(y+u)=x-z.\text{\hspace{1em}(2)}$$ Relations (1) and (2) then imply that $x=z$ holds if and only if $y=u$ holds. We distinguish two cases, denoted a), b).

a) First, let us assume that $x=z$ and $y=u$, that is we are looking for quadruplets of the form $(x,y,z,u)=(x,y,x,y)$ with unknown $x$ and $y$. The original system reduces to $$\begin{array}{rl}k-x^2& =y,\\ k-y^2& =x.\end{array}$$ Subtracting the equations and rewriting we obtain $$(y-x)(y+x-1)=0.$$ We distinguish two (not quite disjoint) subcases.

If $y-x=0$, the reduced system further reduces to a single quadratic equation $$x^2+x-k=0.$$ For any $k\in (0,1]$, this equation has two real solutions. $$x_{1,2}=\frac{-1\pm \sqrt{4k+1}}{2}.$$ The original system therefore has at least two solutions $$x_1=y_1=z_1=u_1=\frac{-1+\sqrt{4k+1}}{2},x_2=y_2=z_2=u_2=\frac{-1-\sqrt{4k+1}}{2}.\text{\hspace{1em}(3)}$$ If $x+y-1=0$, the reduced system further reduces to quadratic equation $$x^2-x+(1-k)=0.$$ Since its discriminant equals $4k-3$, this quadratic equation has solution if and only if $k\ge \frac{3}{4}$. The solutions are $$x_3=\frac{1+\sqrt{4k-3}}{2}\text{and}{x}_4=\frac{1-\sqrt{4k-3}}{2},$$ and the corresponding values of $y=1-x$ are then $$y_3=\frac{1-\sqrt{4k-3}}{2}\text{and}{y}_4=\frac{1+\sqrt{4k-3}}{2}.$$ If $k=3/4$, these solutions are identical and in fact identical with the solutions already found in (3). On the other hand, for $3/4<k\le 1$ we obtain two other distinct solutions $$(x,y,z,u)=(x_3,y_3,x_3,y_3)\text{and}(x,y,z,u)=(x_4,y_4,x_4,y_4).(4)$$

b) Second, let us assume $x\ne z$ and $y\ne u$. In this case, plugging $x-z$ from (2) into the left-hand side of (1) and dividing by nonzero $y-u$ we arrive at $$(x+z)(y+u)=-1.(5)$$ Since the right-hand side of (5) is negative, at least one of the numbers $x,y,z,u$ is positive and at least one of them is negative. However, this contradicts a sequence of implications $$x\ge 0\Rightarrow y\ge 0\Rightarrow z\ge 0\Rightarrow u\ge 0\Rightarrow x\ge 0,(6)$$ that we prove now: It suffices to prove the first implication (the proofs of the others are analogous). Assume $x>0$. The fourth equation of the original system implies $x\le k$ and since $k\le 1$, we have $0\le x\le k\le 1$. This implies $x^2\le k$ (as $t^2\le t$ for any $t\in (0,1)$). The first equation of the original system now implies $y\ge 0$ as desired. Therefore there are no solutions in case b).