11.3. Find all real numbers $a$ for which there exist three distinct real numbers $x, y, z$ such that $a=x+\frac{1}{y}=y+\frac{1}{z}=z+\frac{1}{x}$.
Answer. $a= \pm 1$.

Solution. Suppose that $a$ satisfies the condition of the problem. From the first equality, express $x=a-\frac{1}{y}=\frac{a y-1}{y}$ and substitute it into the third: $z=a-\frac{1}{x}=a-\frac{y}{a y-1}=\frac{a^{2} y-y-a}{a y-1}$. Substitute the last expression into the second equality: $a=y+\frac{1}{z}=y+\frac{a y-1}{\left(a^{2}-1\right) y-a}=\frac{\left(a^{2}-1\right) y^{2}-1}{\left(a^{2}-1\right) y-a}$, from which we obtain the relation $\left(a^{2}-1\right)\left(y^{2}-a y+1\right)=0$, which is a quadratic equation in $y$ if $a^{2}-1 \neq 1$. Due to the symmetry of the relations in the condition, the remaining variables $x$ and $z$ also satisfy this equation. All of this together gives three different roots of the quadratic equation $y^{2}-a y+1=0$, which is impossible. Therefore, $a^{2}=1, a= \pm 1$.

If $a^{2}=1$, then for any value of $y$, let $x=a-\frac{1}{y}=\frac{a y-1}{y}, z=\frac{-a}{a y-1}$, then $a=y+\frac{1}{z}=\frac{\left(a^{2}-1\right) y^{2}-1}{\left(a^{2}-1\right) y-a}=\frac{1}{a}$ - the relations in the condition are satisfied. The requirement that $x, y, z$ are distinct is equivalent to $y$ not being a root of the quadratic equation $y^{2}-a y+1=0$. The latter equation has no real roots since its discriminant is $a^{2}-4=-3$ - negative, therefore, for any $y$, the obtained triplet of distinct numbers $x, y, z$ satisfies the condition of the problem.

Grading criteria. It is proven that $a$ can only be 1 or -1: 5 points. It is proven that for $a= \pm 1$ there exist three distinct numbers $x, y, z$: 2 points.