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The following question is an attempt to find a lower bound for the value of a polynomial at integer points. It is something that I originally thought about while trying to understand how it would be possible to approach this MO question about polynomials representing the nonnegative integers, but does seem to be very interesting in its own right. I guess this would fall under the study of Diophantine approximation/geometry, which I am not at all expert on. So it could even be a known conjecture or theorem, or obviously false. Maybe someone on MO will be able to say?

As is well known, the Thue-Siegel-Roth theorem says that, for an irrational algebraic number $\alpha$, there are only finitely many pairs of integers $p,q$ satisfying the inequality $\vert p/q-\alpha\vert\le q^{-2-\epsilon}$. Here, $\epsilon$ is any fixed positive real number. This is easily seen to be equivalent to the following lower bound on the growth of homogeneous and irreducible polynomials $f\in\mathbb{Q}[X,Y]$ of degree $d > 1$. For all but at most finitely many integer pairs $x,y$, the inequality $$ \vert f(x,y)\vert\ge\vert y\vert^{d-2-\epsilon}\qquad\qquad{\rm(1)} $$ holds. The argument is very simple. We can decompose $f(x,y)$ as $y^d\prod_{i=1}^d(x/y-\alpha_i)$ for distinct irrational algebraic numbers $\alpha_i$. As $x/y-\alpha_i$ can only be made arbitrarily small for at most one $i$ at a time, (1) is equivalent to applying the Thue-Siegel-Roth theorem to $x/y-\alpha_i$.

Now for my question. Is there an extension of (1) to non-homogeneous polynomials $f$? Now, I realize that it cannot possibly carry directly across to the non-homogeneous case in the same form. For one thing, a simple change of variables allows us to replace $f$ by a polynomial of arbitrarily large degree, such as $\tilde f(x,y)=f(x+y^r,y)$, which would invalidate any inequality depending on the degree of $f$ and, similarly, the right hand side of (1) would change form under changes of variables. To guess how this can be fixed, we can look at Siegel's theorem, which says that $f(x,y)=a$ has only finitely many integer solutions in $x,y$ whenever $f-a$ defines a curve over $\mathbb{Q}$ of genus $g$ at least one. It seems reasonable then, that a generalization of (1) should involve the genus $g$ of the curve defined by $f-a$, for typical rationals $a$, and not the degree.

So, to be precise, my question is whether there is an increasing and unbounded function $\phi\colon\mathbb{N}\to\mathbb{R}$ with the following property: If $f\in\mathbb{Q}[X,Y]$ is such that $f-a$ defines a curve of genus $g$ (for all but finitely many $a$), there exists a nonconstant polynomial $h\in\mathbb{Q}[X,Y]$ such that the inequality $$ \vert f(x,y)\vert\ge \vert h(x,y)\vert^{\phi(g)-\epsilon}\qquad\qquad{\rm(2)} $$ holds on $\mathbb{Z}\times\mathbb{Z}$ outside of a finite set (*). We might even hope that $h(x,y)=y$ under a change of variables, but that seems like a bit much to ask. Comparing with the case where $f$ is homogeneous of degree d, the genus of $f-a$ for nonzero $a$ is given by $g=(d-1)(d-2)/2$ and we have $\phi(g)=(-1+\sqrt{1+8g})/2$, although I doubt that precise form would hold in the non-homogeneous case.

(*) Edit: It is easy to construct polynomials which vanish or degenerate into something simpler on a given finite set of curves. E.g., $f=\prod_{i=1}^n(X-a_i)$ is zero on the curves $x=a_i$. So I should say that (2) holds on $\mathbb{Z}\times\mathbb{Z}$ outside of a finite set of curves (of genus zero).

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OK so I just can't get to the bottom of this question at all. Let's consider $f(x,y)=x^4+y^4-C$ for $C$ any constant. Now $f-a$ will define a curve of genus 3 for all but finitely many $a$. But you want $\phi$ to depend only on the genus and hence to be independent of $C$?? So now whatever $\phi$ you choose, let me just let $x$ and $y$ be grotesquely huge, let me set $C=x^4+y^4-7$, and I have a polynomial $f$ and huge $x,y$ with $f(x,y)=7$ and I just don't get what you are possibly asking. What am I missing? –  Kevin Buzzard Nov 28 '10 at 11:20
    
Kevin - I'm not ruling out that this could be a dumb idea, but your comment doesn't seem to be an issue. I probably just worded it badly. Regardless of C, in your example if $\phi$ is less than 4 then $f(x,y)\ge\vert y\vert^\phi$ for all large enough x or y. As we restrict to integers, this holds for all but finitely many x,y. I didn't include the words "all but finitely many" when stating my second inequality, as they can be absorbed into h anyway. But, probably clearer if I did. –  George Lowther Nov 28 '10 at 12:28
    
@Kevin: The choice of h(x,y) (and the finite number of exceptions) will depend on f. –  George Lowther Nov 28 '10 at 12:37
    
@George: I see now---I hadn't understood the question properly. The point is that one problematic choice of x and y isn't a problem. –  Kevin Buzzard Nov 28 '10 at 23:45

1 Answer 1

up vote 6 down vote accepted

What you are asking is for a strong effective form of Siegel's theorem. There are a number of conjectures in this direction, all considered very hard. For example, if $f=x^3-y^2$, what you are asking is essentially Hall's conjecture: http://en.wikipedia.org/wiki/Hall%27s_conjecture . In general, it follows from Vojta's conjectures (LNM 1239). Nothing close is known. For certain classes of polynomials (general cubics, $y^m - f(x)$, a few others) you can get a lower bound of the type $\log \log \log \max\{x,y\}$ give or take a log or two, from the theory of linear forms in logarithms.

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Felipe - Thanks for that. Hall's conjecture asks for f to be greater than $\vert x\vert^{1/2-\epsilon}$ for all but finitely many $x$. I was a bit non-committal in what the relation between $phi$ and g is, so it is almost, but not quite, implied by a positive answer to my question. In any case, this does show that my question is very difficult and implies or almost implies some open problems. –  George Lowther Nov 28 '10 at 17:23
    
And going in the other direction, I need to read up on Vojta's conjectures. Are you saying that it implies a full positive answer to my question? Trying to work out if my question implies and is implied by open problems, in which case I think I should accept this and add the open-problem tag. –  George Lowther Nov 28 '10 at 17:27
    
Yes, Vojta's conjecture should imply a positive answer to yours. Vojta shows how to get Hall's conjecture from his and the argument should generalize. –  Felipe Voloch Nov 28 '10 at 18:05

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