infinite subset in real plane and straight line Let $A\subset\mathbb{R}^2$ be an infinite set such that the distance between any points $a,b \in A$ is an integer. Prove that A is a subset of a straight line.
For any finite n, give an example of a set A containing precisely n points such that A is not a subset of a straight line but the distance between any two points $a,b \in A$ is an integer.  
My attempt: for the first part, i tried cosine law. However, the angle can be quite arbitrary for the distance between 2 points to be integer.
Second part: i can find right triangle with side 3-4-5 for 3 points or up to 5 points. Not sure for arbitrary n though.
 A: Your first statement is the Erdős-Anning Theorem. You can look it up. The proof is not easy to find on your own, but not hard to understand when it's written out for you. 
For the second part, choose $n-1$ numbers of the form $a_i=(u^2-v^2)/2uv$. Then show that the $n$ points $(0,1)$ and $(a_i,0)$, $i=1,2,\dots,n-1$, are such that the distance between any two is rational. Then show that you can pick an integer $N$ such that the points $(0,N)$ and $(Na_i,0)$, $i=1,2,\dots,n-1$, give you what you want. 
A: Here’s an extended hint for the first part, if you’d like to try it on your own. Suppose that $A$ and $B$ are two of the points and that they are $d$ units apart. Show that every other point $P$ in the set must lie on one of the $2d+1$ (possibly degnerate) hyperbolae having $A$ and $B$ as foci and satisfying $$|PA|-|PB|=n$$ for some integer $n$ such that $-d\le n\le d$. If $C$ is a point in the set that is not on the line through $A$ and $B$, every point $P$ in the set other than $B$ and $C$ must also lie on one of $2d\,'+1$ hyperbolae having $B$ and $C$ as foci, where $d\,'=|BC|$. What’s the largest possible number of points of intersection of these two families of hyperbolae?
A: I totally forgot about this question, sorry and here is my attempt after reading the hints:  
An example of n points on the plane (not all co-linear) such that distance between any 2 points is an integer:
Choose  $ (N,0); (0, a_1); (0, a_2)...(0, a_{n-1} )$ with $ N > 0;  a_i \neq a_j$ , all are integers where $a_i=N\frac{u_i^2-v_i^2}{2u_iv_i} $; $u_i , v_i \in \mathbb{N}$ and N is the least common multiple of $2u_iv_i$.
n-1 points $(0, a_i)$are on the same line Oy and their inter-distance are all integers. Also distance $d_i = \sqrt{N^2+a_i^2}=N\frac{u_i^2+v_i^2}{2u_iv_i}$ between $(N,0)$ and n-1 other points is also an integer.  
Another example, consider n points $\exp(2ni\theta)$ in the complex plane where $cos \theta $ and $sin \theta $ are both rational, and no two points coincide (such choice is always possible since $x^2+y^2 =1$ has infinite number of rational solutions). The distance between any 2 points j,k is $d_{j-k}= 2\sin (j-k)\theta $ which is just a polynomial of $\sin \theta $ and $ \cos \theta$ since j-k is an integer. Then $d_{j-k} $ is rational for all pairs j,k. Thus n points $ N exp(2ni\theta) $ where N is the least common multiple of the denominators of $d_{j-k} $ are the points we need.       
Now, if there exists 3 non-co-linear points A,B,C, we prove that there are only finite points that have integer distance to all these 3 points (and therefore any infinite set of points that have all integer distances must lie in a single line).
Assume P on the plane s.t PA, PB, PC are all integers. Triangle inequality implies $|PA-PB| \leq AB $, and since $PA - PB$ is a positive integer, it has a finite set of values: $|PA-PB| = i$ means P is on a hyperbola with axis AB, and there are finitely many such curves. Similarly, P also belongs to some hyperbolas with axis BC,CA. Since A,B,C are non-co-linear, these curves are pairwise different. P is intersection of 3 curves, with axis AB, BC and CA respectively. There are only finite number of curve, thus there are finite number of intersections.
Thanks everyone ~
