How many values can the expression

$$
z=u v+(u v)^{-1} \text { take if }
$$

$$
\begin{gathered}
u+u^{-1}=x \text { and } \\
v+v^{-1}=y
\end{gathered}
$$

where $x$ and $y$ are different numbers with absolute values not less than 2? Write an equation in which $u$ and $v$ do not appear, and all possible values of $z$ are simple roots.
Given $x, y$ values, from (2) and (3) we need to calculate the possible values of $u$ and $v$, then by pairing these in every possible way, we need to calculate the value of $z$ based on (1). After this, we can answer how many different $z$ values we obtained. - This examination generally needs to be carried out.

From (2) and (3):

$$
\begin{gathered}
u^{2}-x u+1=0, \quad \text { or } \\
v^{2}-y v+1=0
\end{gathered}
$$

for $u$ and $v$:

$$
u_{1,2}=\frac{x \pm \sqrt{x^{2}-4}}{2}, \quad \text { or } \quad v_{1,2}=\frac{y \pm \sqrt{y^{2}-4}}{2}
$$

These are real due to the assumption $|x| \geq 2,|y| \geq 2$. The number of different roots is two for $|x| \geq 2$, and one for $|x|=2$, and similarly for $|y| \geq 2$ and $|y|=2$. In both equations, the two roots are reciprocals of each other, because from the constant terms of $\left(2^{\prime}\right)$ and ( $3^{\prime}$ ) we have $u_{1} u_{2}=1$, and $v_{1} v_{2}=1$. (From this, we also see that $u_{1}, u_{2}, v_{1}, v_{2}$ are none of them 0.) Thus, the possible values are:

$$
\begin{array}{ll}
|x|>2 \text { case: } u_{1} \text { and } \frac{1}{u_{1}} ; & |y|>2 \text { case: } v_{1} \text { and } \frac{1}{v_{1}} \\
|x|=2 \text { case: } u_{1}=\frac{x}{2} ; & |y|=2 \text { case: } v_{1}=\frac{y}{2}
\end{array}
$$

According to these, the different pairs of $u, v$ values are as follows:

$\alpha)|x|>2$ and $|y|>2$ case: I $\left(u_{1}, v_{1}\right)$, II $\left(u_{1}, \frac{1}{v_{1}}\right)$, II $^{\prime}\left(\frac{1}{u_{1}}, v_{1}\right)$, I' $\left(\frac{1}{u_{1}}, \frac{1}{v_{1}}\right)$;

$\beta)|x|>2$ and $|y|=2$ case: III $\left(u_{1}, \frac{y}{2}\right)$, III' $\left(\frac{1}{u_{1}}, \frac{y}{2}\right)$; where $\frac{y}{2}$ is either +1 or -1;

$\gamma)|x|=2$ and $|y|>2$ case: IV $\left(\frac{x}{2}, v_{1}\right)$, IV' $\left(\frac{x}{2}, \frac{1}{v_{1}}\right)$; where $\frac{x}{2}$ is either +1 or -1;

$\delta) |x|=2$ and $|y|=2$ case: $\left(\frac{x}{2}, \frac{y}{2}\right)$,

where $x \neq y$ so $\frac{x}{2}$ is either +1 or -1, and $\frac{y}{2}$ is -1 or +1.

These pairs do not always give different $z$ values, because, for example, the members of the II' pair are the reciprocals of the members of the II pair, so substituting II' gives the same two terms as from II, just in reverse order, indeed

$$
u_{1} \cdot \frac{1}{v_{1}}+\frac{1}{u_{1} \cdot \frac{1}{v_{1}}}=\frac{1}{u_{1}} \cdot v_{1}+\frac{1}{\frac{1}{u_{1}} \cdot v_{1}}=\frac{u_{1}}{v_{1}}+\frac{v_{1}}{u_{1}}
$$

This relationship holds between I and I', III and III', IV and IV'. Therefore, $z$ can take at most 2 different values in the $\alpha$) case, and only 1 value in the $\beta)-\delta$) cases.

In the $\alpha$) case, from I and II:

$$
z_{1}=u_{1} v_{1}+\frac{1}{u_{1} v_{1}}, \quad z_{2} \frac{u_{1}}{v_{1}}+\frac{v_{1}}{u_{1}}
$$

These are different because their difference can be transformed as follows:

$$
z_{1}-z_{2}=u_{1}\left(v_{1}-\frac{1}{v_{1}}\right)-\frac{1}{u_{1}}\left(v_{1}-\frac{1}{v_{1}}\right)=\left(u_{1}-\frac{1}{u_{1}}\right)\left(v_{1}-\frac{1}{v_{1}}\right),
$$

and this is only 0 if at least one of the equalities $u_{1}=1 / u_{1}$ and $v_{1}=1 / v_{1}$ holds. In this case, however, $\left|u_{1}\right|=1$, or $\left|v_{1}\right|=1$, so from (2) $|x|=2$, and from (3) $|y|=2$, which means we would be in one of the $\beta$) - $\delta$) cases. - For the purpose of the sought quadratic equation, their sum and product can be written as follows:

$$
\begin{gathered}
z_{1}+z_{2}=u_{1}\left(v_{1}+\frac{1}{v_{1}}\right)+\frac{1}{u_{1}}\left(\frac{1}{v_{1}}+v_{1}\right)=\left(u_{1}+\frac{1}{u_{1}}\right)\left(v_{1}+\frac{1}{v_{1}}\right)=x y \\
z_{1} z_{2}=u_{1}^{2}+\frac{1}{v_{1}^{2}}+v_{1}^{2}+\frac{1}{u_{1}^{2}}=\left(u_{1}+\frac{1}{u_{1}}\right)^{2}-2+\left(v_{1}+\frac{1}{v_{1}}\right)^{2}-2=x^{2}+y^{2}-4
\end{gathered}
$$

thus, the sought equation is:

$$
z^{2}-\left(z_{1}+z_{2}\right) z+z_{1} z_{2}=z^{2}-x y z+x^{2}+y^{2}-4=0
$$

In the $\beta$) case:

$$
\begin{aligned}
& y=2 \text { with } z=u_{1}+\frac{1}{u_{1}}=x \\
& y=-2 \text { with } z=-u_{1}-\frac{1}{u_{1}}=-x
\end{aligned}
$$

so the equation is either $z-x=0$, or $z+x=0$.

Similarly, in the $\gamma$) case, we get the equations $z-y=0$, or $z+y=0$.

Finally, in the $\delta$) case, $z=(+1)(-1)+\frac{1}{(+1)(-1)}=-2$, so the equation is $z+2=0$.

The constraint $x \neq y$ only affects the $\delta$) case. If, indeed, $x=y \neq \pm 2$, then the pairs I and II, and the $z$ values derived from them, are different.

Remark. The equations obtained in the $\beta)-\delta$) cases can also be obtained from (4) by substituting +2 or -2 for $y$, or $x$, or both. In each such case (and only in such cases), (4) has a double root because its discriminant is

$$
x^{2} y^{2}-4 x^{2}-4 y^{2}+16=x^{2}\left(y^{2}-4\right)-4\left(y^{2}-4\right)=\left(x^{2}-4\right)\left(y^{2}-4\right)=0
$$

so in such cases, the left side of (4) is the square of the factor corresponding to the single 0-place. Since, however, the sought equation cannot have a double root, we set the square root of the left side of (4) equal to 0. For example, with $y=-2$:

$$
z^{2}+2 x z+x^{2}=(z+x)^{2}=0 \quad \text { instead of: } \quad z+x=0
$$