This is just a summer-time curiosity arisen after a recent question by Dominic van der Zypen.
For a finite subset $S$ of $\mathbb{Z}\times\mathbb{Z}$ and a function $f$ on $\mathbb{Z}\times\mathbb{Z}$ let $f_S$ consider the discrete convolution $$f_S(x,y)=(f*\chi_S)(x,y)=\sum_{(a,b)\in S} f(x-a,y-b),\;\;\;\;\;\;(x,y)\in\mathbb{Z}\times\mathbb{Z}.$$ For some simple shapes $S$ there are bijections $f:\mathbb{Z}\times\mathbb{Z}\to\mathbb{Z}$ with $f_S$ vanishing identically. For instance, a square $S:=\{0,1\}\times\{0,1\}$ works, because one can take an $f$ of the form $$f(x,y)=(-1)^{x+y}\big(\alpha(x)+\beta(y)\big),$$ that has $f_S=0$, and that is a bijection for a suitable choice of $\alpha$ and $\beta$. Variation of this construction show that other simple shapes allow such a bijection with $f_S=0$: a diamond $ \{(0,1) \,,(0,-1) \,,(1,0)\,,(-1,0)\}$, a duck $ \{(0,0) \,,(0,1)\,,(1,1)\,,(1,2)\}$ &c. On the other hand, if $S$ is a finite set of points of $\mathbb{Z}\times\mathbb{Z}$ in arithmetic progression, that implies that the symmetric difference of $S$ and some traslate of it, $S\Delta(S+v)$, contains only two points, then, clearly, no injective $f$ can have $f_S=0$.
Question: Is there a bijection $f:\mathbb{Z}\times\mathbb{Z}\to\mathbb{Z}$ such that $f_S=0$ w.r.to the three points set $$S:=\{(0,0) \,,(0,1)\,,(1,0)\}$$ $$\bf{?}$$
