The intersection of a circle and a rank 3 subgroup of the plane Let $A$ be a rank 3 subgroup of the Euclidean plane, i.e. $A = \mathbb{Z} v_1 + \mathbb{Z} v_2 + \mathbb{Z} v_3$, where $v_1, v_2, v_3 \in \mathbb{R}^2$ are three $\mathbb{Q}$-linearly independent vectors.
Let $S^1 = \{v \in \mathbb{R}^2 ; \, |v|=1\}$ denote the unit circle in the plane.
Is it possible that $|A \cap S^1| = \infty$?
Remark: I could find rank 4 subgroups intersecting $S^1$ in infinitely many points.
The idea was to take a complex algebraic integer $\alpha$ such that $|\alpha| = 1$ and $\alpha$ is not a root of unity. Then the additive group generated by $1, \alpha, \alpha^2, \dots$ has finite rank.
 A: Here's at least a method that probably solves the problem. I say "probably" because I'm not entirely confident in my treatment of degenerate cases. Anyway, the main idea is to try to construct
a second quadratic function that vanishes at all the points of your $S^1 \cap A$, 
thus reducing the problem to finiteness of integral points on a certain curve. 
Let $q(x,y,z) = \|x v_1 + y v_2 +z v_3\|^2$, a quadratic form in $x,y,z$.
Note that $q(x,y,z)$ has a one-dimensional radical.
We assume that the set $S = \{(x,y,z) \in \mathbb{Z}^3: q(x,y,z) =1\}$ 
is infinite and (try to) derive a contradiction. 
For $\lambda \in \mathbb{Q}$, consider the set $\mathcal{Q}_{\lambda}$ of all quadratic forms $Q$ in $x,y,z$ which have the property that $Q = \lambda$
on $S$. This is an affine subspace of the $5$-dimensional real vector space
of all quadratic forms in $x,y,z$.  Moreover, it is defined over $\mathbb{Q}$, since it is defined by affine equations with rational coefficents.
$\mathcal{Q}_{1}$ cannot consist of the single point $\{q\}$, because otherwise
$q$ would have rational coefficients, and then the radical of $q$
would contain a rational vector $(x,y,z)$, contradicting the assumed
independence of $v_1, v_2, v_3$.
Now $S$ is contained in the intersection of the affine quadric surfaces
$Q=1$ for $Q \in \mathcal{Q}_1$. Call $X$ that intersection.  Because $\mathcal{Q}_1 \neq \{q\}$ we see that  $X$, considered
as an algebraic variety, is
of dimension $\leq 1$. If $X$ is of dimension $0$ there is nothing to show;
we will show that if $X$ has dimension $1$ it is (an open subset of) an elliptic curve. 
Then Siegel's theorem on integral points yields a contradiction.  
$\mathcal{Q}_0$ contains a nondegenerate form $R$, for
otherwise we may pick degenerate $Q \in \mathcal{Q}_0$ with rational coefficients.
In that case, the set of zeroes of $Q$ is contained in the union of two $\mathbb{Q}$-rational proper subspaces of $(x,y,z) \in \mathbb{Q}^3$.  So $S$ is contained
in the union of two sublattices of $\mathbb{Z}^3$ of rank $2$. That's easily seen to be impossible.
We have $q \in \mathcal{Q}_1$ and $R \in \mathcal{Q}_0$. 
If $X$ has dimension $1$,
it must be the intersection of $R=0$ and $q=1$ (these equations are not defined over $\mathbb{Q}$, but $X$ is anyway...)   Now the intersection of two   quadrics
is an elliptic curve under a certain nonsingularity condition.  I THINK that, in our situation, the nonsingularity condition amounts to asking that no linear combination $aR+bq$ of $R,q$ is a square of a linear form  $\ell$ (could certainly be wrong about this!). But if that were so,  $S$ would be contained in the union
of two affine planes $\ell =  \pm \sqrt{b}$, and that is again easily seen to be impossible. 
