Problem 1898

At the beginning we can see that property 3) implies that in any bounded subset of a plane there is only a finite number of points from $M$. (*)

Next, we can see that if for some points $A, B \in M$ we define by $\phi$ a translation by vector $\overrightarrow{AB}$, then for any point $C \in M$ we also have $\phi(C) \in M$. (**)

Indeed: thanks to property 1) we know that there is a point $P \in M$ that does not belong to line $AB$. Therefore, from 2) we can imply that there is a point $R \in M$, such that $ABRP$ is a parallelogram, and therefore $\overrightarrow{PR} = \overrightarrow{AB}$. Now it is sufficient to see that point $C$ does not belong to line $AB$ or does not belong to line $PR$, so $\phi(C) \in M$ by property 2), as the fourth vertex of parallelogram $BAC\phi(C)$ or $RPC\phi(C)$, which concludes the proof of (**). It is worth mentioning that we can say the same about $\phi^{-1}$ (translation by vector $\overrightarrow{BA}$). It shows that $\phi$ is a one-to-one mapping of set $M$ on itself.

We will show now that we can choose such a parallelogram (we will call it “basic”) with vertices in $M$, which does not contain any other points from $M$ (except vertices). Indeed: thanks to (*) we can pick line segment $AB$ with ends in $M$, which won't contain any other points from $M$. Thanks to properties 1) and 2) we know that we can find parallelogram $ABCD$ with vertices in $M$. If $ABCD$ is not basic, by (*), from the finitely many points from $M$ contained inside $ABCD$ we can pick point $E$ that lies closest to the line $AB$. Then, as we know from 2) we can get parallelogram $ABEF$ with vertices in $M$, which either is basic or it contains point $Q \in M$ belonging to $ABEF$ but not on segment $EF$ (then translation of $Q$ by vector $\overrightarrow{AB}$ belongs to $ABCD$ and lies closer to line $AB$ than $E$, a contradiction) or it contains point $Q \in M$ on segment $EF$ (in this case the translation of $A$ by vector $\pm\overrightarrow{EQ}$ is a point of $M$ lying inside segment $AB$, a contradiction).

To sum things up, we have to see that if we get basic parallelogram $ABCD$ with vertices in $M$, and define $\phi$ as a translation by vector $\overrightarrow{AB}$, and define $\psi$ as a translation by vector $\overrightarrow{AD}$, then we can define a set $Z = \{\phi^k \circ \psi^l(A) : k, l \in \mathbb{Z}\}$, which now is easy to see, is equal to $M$. It is obvious that $Z = M$ fulfills the thesis.