Determine whether a function exists from the positive integers to the positive integers which satisfies the equation: $$f(n)=f(f(n-1))+f(f(n+1))$$.
My guess is that this function does not exist, as through trial and error I found that $f(n)=\frac{n}{2}$ is a solution to the functional equation. Obviously this is not enough, but I don't know how to tackle this problem.
I found that $f(n+2)-f(n)=f(f(n+3))-f(f(n-1))$, but this doesn't seem very useful to me.
Can anyone help me with this problem?
Here is a sketch of the proof that no such function exists. Say that such a function does exist. Take $n$ be such that $f(n)=m$ is the minimum value that $f$ takes. If $n\neq1$, then then $m$ is expressed as the sum of two outputs of the function $f$ by the given relation. Since the outputs of $f$ are positive integers, this is impossible.
If the minimum is $f(1)$ and is attained only at $1$, then take let $n_1>1$ be such that $f(n_1)$ is the minimum of the values attained by $f$ excluding $f(1)$. Then $f(n_1+1)\geq f(n_1)>f(1)\geq 1$, so by the minimality of $f(n_1)$, we have that $f(f(n_1+1))\geq f(n_1)$. Again this leaves the desired relation impossible, so there is no such function.