Suppose that $S\subseteq\mathbb{Z}[i]$ has the following properties:


For convenience, let $A\_n = \{z : z\in\mathbb{Z}[i], \text{Nm}(z)\le n\}$


$$\limsup\_{n\rightarrow\infty} \frac{|S\cap A\_n|}{|A\_n|} > 0$$


Then is it the case that $S$ contains arbitrarily long arithmetic progressions.



Szemeredi's theorem in the Gaussian integers
This is true, and follows from one-dimensional Szemeredi.


Fix $\delta > 0$ such that $|S\cap A\_n| \geq \delta |A\_n|$ infinitely often.
Let $r = \lfloor \sqrt n \rfloor$, so $A\_n$ is contained in
the square $S\_r: \{x+iy \in {\bf Z}[i] \colon |x|,|y| < r\}$.
Define the additive homomorphism $h\_r: {\bf Z}[i] \rightarrow {\bf Z}$
by $h\_r(x+iy) = 4rx+y$. Then $h\_r$ is injective on $S\_r$,
and any arithmetic progression in $h\_r(S\_r)$ lifts to $S\_r$.
[The point is that $h\_r$ regularly lays the horizontal segments
in $S\_r$, each of length $2r-1$, on ${\bf Z}$, but separated by gaps
of length $2r+1$, which is short enough to reduce the density by
only a finite factor but long enough that any identity $2a\_2=a\_1+a\_3$
in the image holds on each coordinate of the $h\_r^{-1}(a\_i)$.]
Then the image $h\_r(S\_r)$ is a subset of $(-4r^2,4r^2)$
of density at least $\frac\pi8 \! \delta - O(1/r)$,
and by Szemeredi is guaranteed to contain arbitrarily long arithmetic
progressions as $r \rightarrow \infty$.


The same technique works with ${\bf Z}[i] \cong {\bf Z}^2$
replaced by ${\bf Z}^k$ for any fixed $k$.


[*Added later*: Naturally this stratagem is far from new.
[This 
2008 entry from Terry Tao's blog](https://terrytao.wordpress.com/tag/freiman-isomorphism/) reminds me that 
it's called the "Ruzsa projection trick", and the map $h\_r$ 
(and more generally the injection from $(-r,r)^k$ to ${\bf Z}$
taking $(x\_1,\ldots,x\_k)$ to $\sum\_{i=1}^k (4r)^{k-i} x\_i$)
is called a "Freiman isomorphism of order $2$" to its image
("order $2$" because coincidences between sums of $2$ elements of $S$
suffice to detect arithmetic progressions of arbitrary length).]


