Is there a bijective seacucumber? 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.
 A: Note that, for every $z_{0} \in \mathbb{C}$, the condition
 $$ f(z + 1) + f(z - 1) + f(z + i) + f(z - i) = 0$$
gives a constraint on the values of $f$ taken on the lattice
 $$ \{ z_{0} + n + m i : n, m \in \mathbb{Z} \}.$$
In particular, the constraint is that for every point in the lattice, the sum over the neighboring points of $f(z)$ is zero. In fact, this is equivalent to the original condition. More generally, if we have any basis $\beta = \{ u_{1}, u_{2} \}$ of $\mathbb{C}$ over the reals, then the statement that for all $z \in \mathbb{C}$
$$ \sum_{u \in \beta} [f(z + u) + f(z - u)] = 0$$
is equivalent to the statement that for all $z_{0} \in \mathbb{C}$, the values of $f$ on the lattice
  $$ \{ z_{0} + n u_{1} + m u_{2} : n, m \in \mathbb{Z} \} $$
are constrained such that for any lattice point, the sum of $f$ over the neighboring points is zero.
Why go to such extent for generalities? Well, if I remember correctly, such constraints (on complex lattices) often give rather strict constraints on the original function, at least for analytic functions. One particular example (which is admittedly only of indirect relevance) is that if an analytic function is constant on such a lattice (called a multi-periodic function) it is necessarily constant over the whole plane (for example, see this question).
Another possibility is to consider the continuous analogue. The original condition could be generalized so that for some finite set $\{ \alpha_{i} : 1 \leq i \leq n \}$ satisfying $\sum_{i} \alpha_{i} = 0$ we have
 $$ \sum_{i} f(z + \alpha_{i}) = 0. $$
Then we extend to a continuous path $\alpha: [0, 1] \to \mathbb{C}$ with
 $\int_{0}^{1} \alpha(t) \ \mathrm{dt} = 0$
and require
  $$ g(z) = \int_{0}^{1} f(z + \alpha(t)) \ \mathrm{dt} = 0 $$
for every $z \in \mathbb{C}$.
This is tantalizingly similar to the integral in the residue theorem (or perhaps the Cauchy integral formula?) but it many not be similar enough to be helpful.
A: 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).
