Find $ \ f:\mathbb Z \to \mathbb C$ such that $f(n)= -f(-n+m)$ for $m\in \mathbb Z \setminus 3\mathbb Z$ and $n\in \mathbb Z$.

Question

Can we expect to find no-zero function $ \ f:\mathbb Z \to \mathbb C$ such that $f(n)= -f(-n+m)$ for $m\in \mathbb Z \setminus 3\mathbb Z$ and $n\in \mathbb Z$?

Edit: the example in my mind is: $f(n)=(-1)^n= (e^{\pi i})^n$, this satisfy the above situation if we take $m\in \mathbb Z \setminus 2\mathbb Z.$ I'm willing to generalize the property of $f$ for $m\in \mathbb Z \setminus 3\mathbb Z.$

Answer 1

Define $\, q(n, m) := f(n) + f(-n + m). \,$ Now we are given that $\,0 = q(x,y) \,$ for all $\,x,y \in \mathbb Z\,$ where $\,y \ne 3z.\,$ Given $\, n \in \mathbb Z\,,$ $\,n \ne 3z,\,$ let $\,m=2n\,$ implying $\,0 = f(n) + f(n) \,$ and $\, 0 = f(n).\,$ If $\,n = 3z,\,$ let $\,m=2n+1\,$ also implying $\,0 = f(n) + f(n+1)\,$ but since $\, n+1 \ne 3z\,$ we just proved $\, 0 = f(n+1) \,$ implying $\, 0 = f(n).$ There are many other ways to use the given equations.

Answer 2

Apply the condition to $n=0$ and all authorized $m$:

\begin{align*}f(0)&=-f(1)\\&=-f(2)\\&=-f(4)\\&=-f(5)\\&=-f(7)\\&\ldots\end{align*}

So $f$ assumes the same value on all numbers that are not multiples of $3$.

Now do the same with $n=1$. \begin{align*}f(1)&=-f(0)\\&=-f(1)\\&=-f(3)\\&=-f(4)\\&=-f(6)\\&\ldots\end{align*}

Now $f$ assumes the same value on all numbers that are not congruent to $2\bmod 3$.

Therefore $f$ has to be constant. Moreover, the second line above tells you that $f(1)=0$.