It is well known that every planar graph has an embedding such that every edge is drawn as a straight line segment (Fáry's Theorem). Kemnitz and Harborth made the following stronger conjecture


**Conjecture 1.** Every planar graph has a straight line embedding with *integer* edge lengths. 


I was wondering if it is possible to attack this problem with the following approach.


**Conjecture 2.** Let $X:=\{ x\_1, \dots, x\_n \}$ be a finite set of points in the plane such that no three points of $X$ are colinear. For any $\epsilon >0$, there exists $X':=\{x\_1', \dots, x\_n'\}$ such that for all $i, j \in [n]$


1. $d(x\_i, x\_i') < \epsilon$,
2. $d(x\_i', x\_j') \in \mathbb{Q}$, and
3. no three points of $X'$ are colinear.


To prove Conjecture 2, it suffices to prove the following conjecture.


**Conjecture 3.** Let $X:=\{ x\_1, \dots, x\_n \}$ be a finite set of points in the plane such that no three points of $X$ are colinear and all pairwise distances are rational. Then the set of points which are at rational distance from all points in $X$ is a dense subset of the plane.


Conjecture 3 is trivial for $n=1$ and easy for $n=2$. Almering proved it for $n=3$, and I think it is open for $n>3$. Note that Conjecture 3 is (essentially) a weakening of:


**Conjecture 4.** There exists a dense subset of the plane with all pairwise distances rational.


This question was posed by Ulam in 1945 (see [this](https://mathoverflow.net/questions/19127/is-there-a-dense-subset-of-the-real-plane-with-all-pairwise-distances-rational/19129#19129) mathoverflow question for more background). So, the reason I like Conjecture 3 is that it is still strong enough to prove Conjecture 1, but appears much weaker than Conjecture 4.
Unfortunately, Conjecture 3 is beyond my limited area of expertise. Hence:



> 
> **Question.** What are the prospects for proving Conjecture 3? A proof or disproof would be fantastic. However, even arguments suggesting that it is true/false but say beyond current technology would be most welcome. 
> 
> 
> 



Drawing planar graphs with integer edge lengths
I think conjecture 3 is actually stronger then conjecture 4. 


I prove $C\_3\implies C\_4$:


Pick any sequence of integers $a\_n$, which contains all integers infinite times.


Pick any enumeration of all squares $s\_n$ in the plane with corners at rational coordinates.


Then assuming conjecture 3, at step $n$ we can find a point with rational distances in $s\_{a\_n}$ which is not collinear to any previously used rationals, since there are only finitely many straight lines in our set so far.


At step $\omega$, we have $\omega$ many rationals in every square, so a dense set of all rational distances.


