Does there exist a bi-infinite sequence (for every integer $n$ there exist an $a_n$ item in the sequence) with positive real items, such that for every integer $k$, $a_{k-1}+a_k=a_{k+1}$?
I know it is easy if every item is an integer. But we only know that every item is positive and real. So I believe there doesn't exist. How can I prove it?
On
There is such a sequence: $$a_n=a\,\varphi^n$$ with some real $a>0$ for every integer $n$, where $$\varphi=\frac{1+\sqrt{5}}2,$$ because $\varphi^2=\varphi+1$, and multiplying by $a\,\varphi^{k-1}$, you immediately get $a_{k+1}=a_{k-1}+a_k$.
On
By the general theory of linear recurrences, a sequence satisfies the recurrence $a_{k-1}+a_k=a_{k+1}$ iff it is of the form $a_n=bs^n+ct^n$ for some $b,c\in\mathbb{R}$, where $s=\frac{1+\sqrt{5}}{2}$ and $t=\frac{1-\sqrt{5}}{2}$ are the solutions of the equation $x^2=x+1$. So your question is just whether there exist $b,c\in\mathbb{R}$ such that $bs^n+ct^n>0$ for all $n\in\mathbb{Z}$.
The answer is yes: if $c=0$ and $b>0$, then the sequence $a_n=bs^n$ is always positive, since $s$ is positive.
(On the other hand, these are the only examples. If $c\neq 0$, then when $n$ is very negative, $s^n$ is close to $0$ since $s>1$, while $t^n$ is very large and alternates in sign based on the parity of $n$ since $-1<t<0$. So, choosing $n$ to be sufficiently negative and of the correct parity according to the sign of $c$, $bs^n+ct^n$ will be negative.)
If you choose $x$ as the positive root of $x^2-x-1=0$ then $a_n=x^n$ will do since $x^{n+1}=x^n+x^{n-1}$ and the terms are positive even for negative exponent.