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.



The intersection of a circle and a rank 3 subgroup of the plane
I am grateful to Edgardo whose answer supplied me with the Siegel theorem on integral points (I did not know it), and I am happy to see that my approach coincides with Edgardo's in many points.


The answer is negative: there is no abelian subgroup $A\subset{\mathbb R}^2$ of rank $3$ such that its intersection $I\_A:={\mathbb S}^1\cap A$ with the unit circle ${\mathbb S}^1:=\big\{v\in{\mathbb R}^2\mid|v|=1\big\}$ is infinite.


In a few words: we show that such $I\_A$ is a part of the intersection of two quadrics in ${\mathbb P}\_{\mathbb C}^3$ defined over ${\mathbb Q}$, observe
that this intersection in interesting cases is generic, hence, is an elliptic curve, and, finally, apply the Siegel theorem <http://en.wikipedia.org/wiki/Siegel_theorem> to an affine part of this curve. 


Let $L\_{\mathbb Q}:={\mathbb Q}A$ denote the ${\mathbb Q}$-linear space spanned by $A$. Besides the intersection $I\_A$, we will also deal with the intersection $I\_{\mathbb Q}:={\mathbb S}^1\cap L\_{\mathbb Q}$. One can choose another ${\mathbb Q}$-basis $B'$ in $L\_{\mathbb Q}$ and generate another $A'$. It is immediate that $A\subset\frac1nA'$ and $A'\subset\frac1nA$ for a suitable $n\in{\mathbb N}$. We conclude that $I\_A$ is infinite for some $A$ iff, for any ${\mathbb Q}$-basis $B$ in $L\_{\mathbb Q}$, there are infinitely many elements in $I\_{\mathbb Q}$ whose denominators in terms of $B$ are limited.


Assuming that some $I\_A$ is infinite, we can choose a ${\mathbb Q}$-basis of $L\_{\mathbb Q}$ inside $I\_{\mathbb Q}$. Indeed, it is easy to pick a couple of ${\mathbb R}$-linearly independent $b\_1,b\_2\in I\_{\mathbb Q}$. They generate a discrete ${\mathbb Z}$-lattice $A\_0\subset{\mathbb R}^2$; so, $\frac1nA\_0$ is also discrete for any $n\in{\mathbb N}$. Consequently, the intersection ${\mathbb S}^1\cap\frac1nA\_0$ is finite. Including $b\_1,b\_2$ in some basis of $L\_{\mathbb Q}$, we can see that $I\_{\mathbb Q}\subset{\mathbb Q}A\_0$ would imply $I\_A\subset\frac1nA\_0$ for a suitable $n\in{\mathbb N}$. A contradiction. For a similar reason, any $2$-dimensional ${\mathbb Q}$-linear subspace in $L\_{\mathbb Q}$ contains finitely many rational points from $I\_{\mathbb Q}$ with limited denominators.


Let $b\_1,b\_2,b\_3\in I\_{\mathbb Q}$ be a ${\mathbb Q}$-basis in $L\_{\mathbb Q}$. Its Gram matrix has the form $G:=\left[\begin{smallmatrix}1&c\_3&c\_2\\c\_3&1&c\_1\\c\_2&c\_1&1\end{smallmatrix}
\right]$
with $-1<c\_i<1$. Since the $b\_i$'s are ${\mathbb R}$-linearly dependent,
$\det G=0$. So, if $c\_1,c\_2,c\_3\in{\mathbb Q}$, then, by means of the Gram-Schmidt orthogonalization in $L\_{\mathbb Q}$ and by the Sylvester criterion, we could find, in view of $\det G=0$, an isotropic element
$0\ne v\in L\_{\mathbb Q}$, i.e., such that $\langle v,v\rangle=0$, which is impossible. Hence, $c\_i\notin{\mathbb Q}$ for some $i$. 


Actually, we study rational solutions of
$$x\_1^2+x\_2^2+x\_3^2+2c\_1x\_2x\_3+2c\_2x\_3x\_1+2c\_3x\_1x\_2=1.\qquad(1)$$


As $c\_3^2<1$, there are finitely many rational solutions of the form $[x\_1,x\_2,0]$ with limited denominators, because $(1)$ becomes
$(x\_1+c\_3x\_2)^2+(1-c\_3^2)x\_2^2=1$. In other words, we can assume that $x\_1x\_2x\_3\ne0$ and rewrite $(1)$ in terms of $y\_i:=x\_i^{-1}$ as
$$\frac{y\_1y\_2}{y\_3}+\frac{y\_2y\_3}{y\_1}+\frac{y\_3y\_1}{y\_2}+2c\_1y\_1+2c\_2y\_2+2c\_3y\_3=y\_1y\_2y\_3.\qquad(2)$$


The dimension of the ${\mathbb Q}$-linear space spanned by $[x\_2x\_3,x\_3x\_1,x\_1x\_2]$, where $[x\_1,x\_1,x\_3]$ runs over all rational solutions of $(1)$, equals $2$. Indeed, it cannot be $3$ as, otherwise, $c\_1,c\_2,c\_3\in{\mathbb Q}$. It cannot be $0$ because we have at least one rational solution. If the dimension equals $1$, then, for all rational solutions with $x\_1x\_2x\_3\ne0$, the triples $[y\_1,y\_2,y\_3]$ are all ${\mathbb Q}$-proportional. Looking at $(2)$, we see that the coefficient of proportionality should be $\pm1$, and we get only finitely many solutions of $(1)$ with limited denominators.


We conclude that $c\_1=a\_1c+d\_1$, $c\_2=a\_2c+d\_2$, $c\_3=c$ for suitable $a\_1,a\_2,d\_1,d\_2\in{\mathbb Q}$. It follows from $\det G=0$ that $p(c)=0$, where $p(x):=\det\left[\begin{smallmatrix}1&x&d\_2+a\_2x\\x&1&d\_1+a\_1x\\d\_2+a\_2x&d\_1+a\_1x&1\end{smallmatrix}\right]$.


Now, $(1)$ takes the form
$x\_1^2+x\_2^2+x\_3^2+2d\_1x\_2x\_3+2d\_2x\_3x\_1+2(x\_1x\_2+a\_1x\_2x\_3+a\_2x\_3x\_1)c=1$. Since $c\notin{\mathbb Q}$, we deal in fact with the equations
$$x\_1^2+x\_2^2+x\_3^2+2d\_1x\_2x\_3+2d\_2x\_3x\_1=1,\quad x\_1x\_2+a\_1x\_2x\_3+a\_2x\_3x\_1=0.$$


We want to show that, for any integer $0\ne d\in{\mathbb Z}$, the equations
$$x\_1^2+x\_2^2+x\_3^2+2d\_1x\_2x\_3+2d\_2x\_3x\_1=d^2,\quad2x\_1x\_2+2a\_1x\_2x\_3+2a\_2x\_3x\_1=0$$
admit just finitely many integer solution.


The intersection
$$x\_1^2+x\_2^2+x\_3^2+2d\_1x\_2x\_3+d\_2x\_3x\_1-d^2x\_0^2=0,\quad2x\_1x\_2+2a\_1x\_2x\_3+2a\_2x\_3x\_1=0$$
of two quadrics in ${\mathbb P}\_{\mathbb C}^3$ is an elliptic curve if all the $4$ roots of the polynomial
$q(\lambda):=\det(\lambda M\_1+M\_2)$ are pairwise distinct, where $M\_1,M\_2$ stand for the symmetric matrices of the quadrics (see, for example,
<http://archive.numdam.org/ARCHIVE/ASNSP/ASNSP_1980_4_7_2/ASNSP_1980_4_7_2_217_0/ASNSP_1980_4_7_2_217_0.pdf>
especially, from pages 221-222). Since
$q(\lambda)=\det\left[\begin{smallmatrix}-d^2\lambda&0&0&0\\0&\lambda&1&d\_2
\lambda+a\_2\\0&1&\lambda&d\_1\lambda+a\_1\\0&d\_2\lambda+a\_2&d\_1\lambda+
a\_1&\lambda\end{smallmatrix}\right]=-d^2\lambda^4p(\lambda^{-1})$,
in view of the Siegel theorem, it suffices to show that all the $3$ roots of the polynomial $\lambda^3p(\lambda^{-1})=0$ are $(\text{a})$ pairwise distinct and $(\text{b})$ all different from $0$. The polynomial $p(x)$ of degree $\le3$ has rational coefficients, $p(c)=0$, and $c\notin{\mathbb Q}$. Consequently, $p(x)$ cannot have multiple roots thus implying $(\text{a})$. If $(\text{b})$ is not true, the coefficient of degree $3$ in $p(x)$ vanishes, i.e., $2a\_1a\_2=0$. In this case, the second quadric is a couple of rational planes passing through the origin $0\in L\_{\mathbb Q}$, and we already know that any $2$-dimensional ${\mathbb Q}$-linear subspace in $L\_{\mathbb Q}$ contains finitely many rational points from $I\_{\mathbb Q}$ with limited denominators.


