Understand the problem

Suppose that real numbers $x,y$ and $z$ satisfy the following equations: \begin{align*} x+\frac{y}{z} &=2,\\ y+\frac{z}{x} &=2,\\ z+\frac{x}{y} &=2. \end{align*}
Show that $s=x+y+z$ must be equal to $3$ or $7$. Note: It is not required to show the existence of such numbers $x,y,z$.
Source of the problem

Germany MO 2019, Problem 6

Topic
Algebra, Simultaneous Equations
Difficulty Level
6/10
Suggested Book
Challenges and Thrills of Pre College Mathematics

Start with hints

Do you really need a hint? Try it first!

Observe that x = y = z = 1 gives a valid solution of the set of equations. In this case s = x+y+z = 3. Now, observe one thing that this set of equations is symmetric in (x,y,z). Observe that we are required to comment on (x+y+z).  Rewriting the equations as: $$xz+y = 2z, \qquad (1)$$
$$xy + z = 2x, \qquad (2)$$
$$yz + x = 2y \qquad (3)$$
and then summing gives us that $x+y+z = xy + yz + zx = s.$ Our aim will be to reduce all the equations into a single variable. ( maybe a polynomial ). Let’s consider the case, where all of x,y,z is not 1.
From now on we consider $x,y,z \neq 1$. This also gives $x \neq y \neq z \neq x$ Solving the first expression  $x=\frac{2z-y}{z}$  then plugging this into the second two gives:
$$y+\frac{z^2}{2z-y}=2 \Rightarrow (2z-y)y+z^2=2(2z-y)$$$$z+\frac{2z-y}{yz}=2 \Rightarrow yz^2+2z-y=2yz \Rightarrow y=-\frac{2z}{z^2-2z-1}$$
as z is not equal to 1. Plugging the latter into the former and simplifying gives:
$$\frac{z^2 (z^4-8z^3+14z^2-7)}{(z^2-2z-1)^2}=0 \Rightarrow z^4-8z^3+14z^2-7=0$$
Plugging the latter into the former and simplifying gives:
$$\frac{z^2 (z^4-8z^3+14z^2-7)}{(z^2-2z-1)^2}=0 \Rightarrow z^4-8z^3+14z^2-7=0$$
Now, observe that we already know z = 1 is a solution. This gives rise to  $$0=z^4-8z^3+14z^2-7=(z-1)(z^3-7z^2+7z+7) \Rightarrow z^3-7z^2+7z+7=0$$  
Observe that the polynomial we have got in terms of z is also satisfied by x,y,z as the equations are symmetric in x,y,z. Hence we can claim that \( t^3 – 7t^2 + 7t + 7 = 0 \) has three solutions x,y,z.  Hence, \( t^3 – 7t^2 + 7t + 7 = (t-x)(t-y)(t-z)\). Therefore, by Vieta’s formula, x+y+z = 7. QED.

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