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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

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  • $\begingroup$ @HagenvonEitzen Look at OPs proof at the bottom. Unless I'm misreading something here he/she does claim that $z^k+z-1=0$ has no positive real solutions for $k > 2$, which isn't true. $\endgroup$ – orlp Dec 6 '17 at 20:51
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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.

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  • $\begingroup$ Your discussion is correct. Thanks for your hint. $\endgroup$ – Amin235 Dec 6 '17 at 21:02
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Your proof is incorrect because it is based on a false promise. $z^k+z-1=0$ most definitely has positive real solutions for $k > 2$. It's hard to solve them exactly, but a plot will quickly show that it has positive real solutions.

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  • $\begingroup$ You right. Do you have any idea about my question. Thank. $\endgroup$ – Amin235 Dec 6 '17 at 21:01

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