Let $h(x,y)$ be a polynomial with real coefficients. Suppose there are infinitely many integer solutions to $|h(x,y)|<1$. What can I say about $h$?

When $h$ itself has integer coefficients, a famous theorem of Siegel tells me that the curve $h(x,y)$ has geometric genus zero and either $1$ or $2$ points at infinity. The main reason I can make no progress on this question is that I know no analogous result for $h \in \mathbb{R}[x,y]$.

All I need is something very crude. Basically, if $(x_n,y_n)$ is the sequence of solutions ordered by $x_n^2+y_n^2$, and you can give me any reasonable upper bound on the growth rate of $x_n^2+y_n^2$, that is good enough to solve the linked problem. (When $h$ has integer coefficients, it follows from Siegel's theorem that $(x_n, y_n)$ are more or less the images of $(a_n, b_n)$ under a polynomial map, where $(a_n, b_n)$ are the solutions to a Pell equation, or else are of the form $(f(n), g(n))$ for some polynomials $f$ and $g$. So $x_n^2+y_n^2$ can grow at worst exponentially.)

Adding more details here: If $h$ has integer coefficients, then $|h(x,y)| < 1$ is the same as $h(x,y)=0$. Curves of genus $\geq 2$ have only finitely many rational points (Faltings). Affine curves of genus $1$ have only finitely many integer points (Siegel). $\mathbb{P}^1 \setminus \{ 0,1, \infty \}$ has only finitely many $\mathcal{O}_{K,S}$ points for any number field $K$ and any finite $S$ (this is the $S$-unit equation, I think finiteness was also proved by Siegel.) If the normalization of our curve is isomorphic over $\bar{\mathbb{Q}}$ to $\mathbb{A}^1 \setminus \{ z_1, z_2, \ldots, z_s \}$ for some $s \geq 2$ and some $z_i \in \bar{\mathbb{Q}}$ then, after extending the ground field and inverting finitely many primes, we can apply a linear change of variables making $z_1=0$ and $z_2=1$. So we can embed integer solutions into the $S$-unit equation for some $(K,S)$.

Thus, the only remaining options are a genus $0$ curve with one puncture or two punctures.

A genus zero curve with one puncture is rational over $\mathbb{Q}$, since it has a point (the puncture). So there is a parametrization $(f(t), g(t))$ of $h(x,y)=0$ where $f$ and $g$ are polynomials with rational coefficients. I haven't quite been careful with the details here, but the integer points should wind up being the image of some finite collection of arithmetic progressions under $(f(t), g(t))$.

A genus zero curve with two punctures is either $uv=1$ or $u^2-D v^2 = C$ for some nonsquare $D$. Once again, we get a parametrization $(u,v) \to (f(u,v), g(u,v))$ for some polynomials $(f,g)$ with rational coefficients. Again, we need to be careful with denominators from the coefficients of $(f,g)$, but we should get more or less the image of a pell sequence under a polynomial map in the second case, and only finitely many integer points in the first case.

I'm sure that I have seen (Faltings)+(Siegel)+($S$-unit) all stated together as "only finitely many integer points on a curve with $3g+n \geq 3$" and treated as Siegel's result; but I couldn't quickly find a reference that puts it that way.

  • $\begingroup$ Where might I look for this particular Siegel theorem? Also, with integral coefficients your condition on $h$ would be $h=0.$ That does not appear to fit the Pell thing in your last sentence. Anyway, pending replies on those, I am looking in Cusick and Flahive, The Markoff and Lagrange Spectra, where real coefficient indefinite binary quadratic forms are discussed. $\endgroup$ – Will Jagy Aug 7 '13 at 20:43
  • $\begingroup$ just riffing here, what about $x^2 - L y^2,$ where $L$ is a Liouville transcendental number? $\endgroup$ – Will Jagy Aug 7 '13 at 20:53

Nice question! And it is also not hard at all if the degree is not too small. The only downside is that the answer is negative: the growth can be as fast as one wishes.

For a counterexample, we'll just use $P(x,y)=a^px^p-y^p$ with $a\in(0.4,0,6)$ and $p$ to be chosen later. Note that, given an integer $x>0$, $P(x,y)$ is less than $1$ in absolute value for some $y>0$ if and only if $a\in E_x=(\cup_ y I_{x,y})$ where $I_{x,y}$ is an interval almost centered at $\frac yx$ of length $\ell_{x,y}\approx x^{-p}$.

Now we just make a sequence of simple claims:

Claim 1: Assume that $x,y\in\mathbb N$ and $x$ is prime and not ridiculously small. Then $|I_{x,y}\setminus(\cup_{X>x}E_X)|>\frac 18\ell_{x,y}$.

Indeed, if $X>x$ and $I_{X,Y}\cap I_{x,y}\ne\varnothing$, then either $x\vert X$ and $\frac YX=\frac yx$, so $I_{X,Y}$ is contained in the middle half of $I_{x,y}$, or $\frac 1{xX}\le |\frac YX-\frac yx|\le \ell_{x,y}+\ell_{X,y}\le Cx^{-p}$, so $X\ge cx^{p-1}$, in which case the total measure of $E_X$ is about $X^{-(p-1)}$, and the whole union of such pieces can occupy only the length $\sum_{X> cx^{p-1}}X^{-(p-1)}\approx Cx^{-(p-1)(p-2)}\ll x^{-p}\approx \ell_{x,y}$, provided that $p\ge 5$. A lot is, obviously, left.

Claim 2: Under the same assumptions, there exists an arbitrarily large prime $X>x$ and $Y\in\mathbb N$ such that $\operatorname{Clos}I_{X,Y}\subset I_{x,y}$ and $I_{X,Y}$ is disjoint with $E_z$ for $x<z<X$.

Indeed, we have already seen that the set of points in $I_{x,y}$ not covered by any $E_z$, $z>x$ has measure about $\ell_{x,y}$. Choose large $N$ and notice that the difference $I_{x,y}\setminus(\cup_{x<z\le N})$ is a union of at most $N^2$ intervals and has measure comparable to $\ell_{x,y}$. Thus, there will be an open interval $J$ of length $N^{-2}\ell_{x,y}$ somewhere in $I_{x,y}$ free from every $E_z$ with $x<z<N$. Choose a prime $X\approx 100N^2\ell_{x,y}^{-1}$ (note that we can choose $X$ first and $N$ afterwards, if we want, so we do not need Bertrand here, just Euclid!) and consider an interval $I_{X,Y}$ contained in the middle half of $J$. The only problematic $z$ now are between $N$ and $X$. However, for every such $z$, we have $|\frac tz-\frac YX|>\frac 1{Xz}>\frac {\ell_{x,y}}{z^3}\gg z^{-p}$ if $N$ is large enough and $p\ge 4$, so they just stand outside as poor beggars with hands too short to reach and steal a point from $I_{X,Y}$.

Claim 3: The rest is obvious (the straightforward inductive choice with as fast growth as you wish and the nested interval lemma).

  • $\begingroup$ I'm have a little trouble following this. Could you say at the beginning exactly what you are proving? Thanks! $\endgroup$ – DES-SupportsMonicaAndTransfolk Aug 12 '13 at 19:09
  • $\begingroup$ That you can choose a real $a$ so that the integer solutions to $|P(x,y)|<1$ are infinitely many and grow as fast as you wish. $\endgroup$ – fedja Aug 12 '13 at 19:12

I would look at Noam Elkies' paper "Rational points near curves.." which uses a "determinant" method like that of Bombieri-Pila and Heath-Brown to study small values of |x^3- y^2|.


If $h$ is homogeneous, then the question is about rational approximations to the roots of $h(x,1)$. As Will pointed out, if one such root is e.g. a Liouville number, then there will infinitely many solutions. On the other hand, we know (e.g. from the theory of continued fractions) that good rational approximations to a fixed real number are spread out and that should give the very crude statement that you are after. This should also work if $h=h_1+h_2$, $h_1$ homogeneous of the same degree as $h$ and $h_2$ of small degree. In the general case, I don't think this diophantine approximation argument will work and I don't know what to expect.


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