# many rational points on an algebraic curve

Given polynomial $f(x,y)$ with integer coefficients, may be reducible, but without linear factors. For positive integer $n$ denote by $a_n$ the number of points $(x,y)\in \frac1n \mathbb{Z}^2$ on a curve $f(x,y)=0$. May it appear that $a_n$ tends to infinity (when $n$ increases taking all positive integer values), but is always finite? Similar question: if we consider only bounded part of our curve, say $\{ (x,y):f(x,y)=0,x^2+y^2 < R^2 \}$, and define $b_n$ as the cardinality of the intersection of this set with lattice $\frac1n \mathbb{Z}^2$, may $b_n$ tend to infinity?

• Regarding the question about bounded points, if a curve has infinitely many rational points then these points are dense (at least in one connected component of the curve). So the only problem I can think of is the case of a cubic curve with a bounded real component. – François Brunault Feb 14 '12 at 13:38

If your $a_n$'s are unbounded then your curve has infinitely many rational points, so has genus zero or one. In the case of genus one (or genus zero with more than two points at infinity), the $a_n$ will always be finite and, by finiteness of $S$-integral points, the $a_n$ are bounded if e.g. $n$ ranges through integers with a fixed set of prime factors. So $a_n$ does not go to infinity with $n$. But, even in this case, the sequence $a_n$ can be unbounded. E.g. take an elliptic curve with rank one, such as $y^2=x^3+3$.
• For the unit circle $a_p$ equals 4 for prime $p=4k+3$. I think, a priori the problem can not be reduced to components, because different components may have many points with different denominators. – Fedor Petrov Feb 11 '12 at 6:09