# System of equations that is true just for Golden ratio

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

• @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. – orlp Dec 6 '17 at 20:51

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.
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.