Consider the following equations $$ \begin{array}{cccx}\tag{1} z+1&=&\frac{1}{z}&,\\ z+2&=&\frac{1}{z^2}&,\\ \vdots &=& \vdots &,\\ z+k &=&\frac{1}{z^k}&. \end{array} $$ where $k$ is a positive integer number.
Question: How to find all positive real solutions of $(1)$ when $k$ is given.
Example: The only positive real solution of $(1)$ when $k=1$ and $k=2$ is the $z=\frac{1}{\mu}$, where $\mu=\frac{1+\sqrt{5}}{2}$(Golden ratio).
My try: I prove that the system $(1)$ has no positive real solution for $k>2$.
Proof: Consider $(1)$ has a positive real solution such as $z$, then we get $$ z+k=\frac{1}{z^k} \Longleftrightarrow (z+k-1)+1=\frac{1}{z^k} \Longleftrightarrow \frac{1}{z^{k-1}}+1=\frac{1}{z^k} \Longleftrightarrow z^k+z-1=0 $$ but the equation $z^k+z-1=0$ has no positive real solution for $k>2$.
Is my proof correct.
Thanks for any suggestion.
Edit(1): My proof is incorrect since the equation $z^k+z-1=0$ has positive real solution for $k>2$. Is it possible to ask you to improve my proof or make a correct proof for the question. Thanks
The claim that $z^k+z-1=0$ has no positive solutions is false. The left side is $-1$ at $z=0$ and $2$ at $z=1$ so the intermediate value theorem guarantees a solution in $(0,1)$
If you are trying to satisfy the whole system, a much easier approach is to show that $\frac 1\mu$ does not satisfy the third. As you have shown it is the only solution for the first, there is no common solution.