A Diophantine graph is a set of vertices in the plane with integer coordinates, all at integer distances from eachother.
An Erdős–Diophantine graph is a maximal Diophantine graph, so that it cannot be extended with any more vertices.
Erdős and Anning proved that any infinite set in the plane with pairwise integer distances must lie on a line, so that any non-collinear Diophantine graph is contained in a finite maximal graph.
What is the largest known Erdős–Diophantine graph? Are there easy examples of non-collinear Diophantine graphs with 8 or more vertices?
This is a sort of a partial (non) answer, we claim: Given any $N$, there exists an Erdős–Diophantine graph $P$ on the plane with more than $N$ points that is not colinear.
Indeed, consider $N$ distinct primitive Pythagorean triples $(a_1,b_1,c_1),\ldots,(a_N,b_N,c_N)$, where $a_i^2+b_i^2=c_i^2$. (This is possible as there are infinitely many such primitive Pythagorean triples).
Then, let $A$, be any positive integral common multiple of $a_1,\ldots,a_N$, and consider on the plane the points $(0,0),(0,A),(Ab_1/a_1,0),\ldots,(Ab_N/a_N,0)$.
Basically, what we have done is put together a bunch of Pythagorean triangles that are scaled to the same height. One can also reflect it to get more points. See illustration, where each 'O' is a point, and that the distance between any two points is integral.
Note also this configuration is not colinear, hence by Erdős–Anning, one can extend it to a maximal Erdős–Diophantine graph, with quite a lot of vertices.
O *** * * * * * * * * * * * * * * * * * * * * * * * * * * O**************O*****O****O****O*****O**************O * * * * * * * * * * * * * * * * * * * * * * * * * * *** ONevertheless, this does not tell us what the actual maximal graph is, just that it has this as a subgraph, which can be made arbitrarily large.
ADDED.
Now say we have such planar set $P$ that is not colinear with all integer distances, where else can we possibly add additional points to $P$ so the mutual distances remain integer? The follows from the proof to Erdős-Anning's:
Lemma.
If $P$ is a planar set with $|p-q|$ integer for all $p,q\in P$, and $a,b,c\in P$ are three points but not colinear in $P$, then we have $$ P\subset \bigcup_{j=0}^{|a-b|}\bigcup_{k=0}^{|a-c|} H(a,b,j)\cap H(a,c,k), $$ where $H(a,b,j)$ is the hyperbola with foci at $a,b$ and shape $j$, namely $H(a,b,j) = \{x: ||x-a|-|x-b||=j\}$.
(In particular $|P|\le 4(|a-b|+1)(|a-c|+1)$, as any two distinct conic sections intersect in at most 4 times, cf. Bezout's theorem)
Proof.
Indeed, take any $p\in P$, then by construction of $P$, we must have $|p-a|$ and $|p-b|$ integers, so their absolute difference $||p-a|-|p-b||=j$ is some integer $j$. But by triangle inequality, we also have $||p-a|-|p-b||\le|a-b|$, so this $j$ is an integer between $0\le j\le |a-b|$.
Notice however, $||x-a|-|x-b||=j$ defines the hyperbola $H(a,b,j)$. Hence every point in $P$ is contained in the finite family of hyperbolas $H(a,b,j)$, for $j=0,1,...,|a-b|$.
Similarly if we use the points $a,c$ instead, that $P$ is contained in the finite faily of hyperbolas $H(a,c,k)$, for $k=0,1,...,|a-c|$. $\blacksquare$
Here is an illustration:
So, to construct an Erdős–Diophantine graph, we can start with any planar set with integer distances and having at least 3 noncolinear points $a,b,c$, and construct the finite familes of hyperbolas $H(a,b,j)$ and $H(a,c,k)$. Then we find their pairwise intersections (finitely many), and then check if these points also have integer distance to a fourth point not $a,b,c$. To reduce the number of hyperbolas, pick noncolinear $a,b,c$ with smallest distances among them. I suspect numerical methods could be sufficient to reject non-integral distances (I do not know the state-of-art on numerical solvers...).