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](https://mathoverflow.net/questions/138676) 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.)




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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](http://en.wikipedia.org/wiki/Faltings%27_theorem)). Affine curves of genus $1$ have only finitely many integer points ([Siegel](http://en.wikipedia.org/wiki/Siegel%27s_theorem_on_integral_points)). $\mathbb{P}^1 \setminus \{ 0,1, \infty \}$ has only finitely many $\mathcal{O}\_{K,S}$ points for any number field $K$ and any finite $S$ (this is the [$S$-unit equation](http://en.wikipedia.org/wiki/S-unit#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.



Siegel's theorem with real coefficients
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).


