TITLE: Is there a bijective seacucumber?
QUESTION [32 upvotes]: A friend defined a seacucumber as a continuous function $f:\mathbb{C}\to\mathbb{C}$ such that $f(z+1)+f(z+i)+f(z-1)+f(z-i)=0$ for all $z\in\mathbb{C}$.
He wanted to know if there exists a bijective seacucumber.
An example of a non-trivial seacucumber is $f(x+yi):=e^{i\pi(x+y)/2}$.
Assume there exists a bijective seacucumber $f$.
If we observe $f^{-1}(D)$ for some large enough disc $D$, then this is bounded, so it is contained in a large enough disk $E$.
Then in any direction we can get outside $E$ and find $4$ points very close together where the function values add up to $0$.
As $f$ is continuous and bijective, any path through these $4$ points must go all the way around $D$ in order for this sum to be $0$.
If you draw some of these paths, you quickly notice that $f$ must have some very spirally properties.
You see what we described as tentacles if you let a path pass through multiple such sets of $4$ points all the way around $E$.
Also, if you go further from the origin, you can find more of these sets of $4$ points, which means the amount of tentacles should keep increasing.
This is all very heuristic unfortunately, and we could not come to a contradiction or come up with an example.
We also tried to look whether a function from $\mathbb{Z}+i\mathbb{Z}$ to $\mathbb{C}$ satisfying the seacucumber equation can even be injective, but we expect this is possible.
I would love to see how far we can come with this problem.
Any progress would be greatly appreciated.

REPLY [3 votes]: This is not an answer, but it is too long for a comment.
Let's define two functions:

$g:\mathbb{R}\rightarrow\mathbb{C}$ will be defined as $g(x)=xe^{i\pi x}$,
$f:\mathbb{C}\rightarrow\mathbb{C}$ will be defined as $f(x+yi)=g(x)+g(y)i$

The function $g$ has some nice properties:
$$g(x)+g(x-1)=xe^{i\pi x}+(x-1)e^{i\pi (x-1)}=e^{i\pi x}\left(x-(x-1)\right)=e^{i\pi x}$$
Applying this identity to $x$ and $(x+1)$ yields
$$g(x)+g(x-1)+g(x+1)+g(x)=e^{i\pi x}+e^{i\pi(x+1)}=e^{i\pi x}(1-1)=0$$
Now we can conclude that $f$ is a sea-cucumber (where $z=x+yi$):
$$\sum_{d\in\{\pm 1, \pm i\}} f(z+d) = \left[g(x+1)+g(x-1)+2g(x)\right] + \left[2g(y) + g(y+1)+g(y-1)\right]i=0$$
Furthermore, $f$ restricted to the domain of Gaussian integers reduces to a very simple expression (note, both $x$ and $y$ are integers): $$f(x+yi)=(-1)^xx + (-1)^yyi$$
which allows us to see that it is actually its own inverse and thus a bijection on the set of Gaussian integers.
Unfortunately, it is far from being a bijection on the whole $\mathbb{C}$. For example, consider just numbers of the form $a(1+i)+\frac{1}{2}$ and define a function $h:\mathbb{R}\rightarrow\mathbb{C}$ as $$h(a)=f\left(a(1+i)+\frac{1}{2}i\right)$$
This simplifies to $h(a)=g(a)+g(a+\frac{1}{2})i=ae^{i\pi a}+(a+\frac{1}{2})e^{i\pi (a+1/2)}i=-\frac{1}{2}e^{i\pi a}$, a periodic function with period $2$. This alone implies existence of infinitely many different complex values, each of which is attained by $f$ infinitely often (in fact, for each such value $z$, the set $f^{-1}(z)$ is unbounded).