# Diophantine approximations and quadratic polynomials

I am working on a problem these days and the following issue came up. I am not sure yet that I understand it's depth very well, so I would like to discuss a simple case. For those interested, the problem has applications in coding theory.

Consider a quadratic polynomial $f \left( x_1, x_2, x_3, x_4 \right)$ with real roots and coefficients drawn from a continuous distribution (and therefore irrationals with probability 1). Is there a strictly positive lower bound on $|f \left( x_1, x_2, x_3, x_4 \right)|$ if we constrain all $x_1, x_2, x_3, x_4$ to lie in $\mathbb{Z}$ ? In other words, is there a $\gamma > 0$ such that $\displaystyle |f \left( x_1, x_2, x_3, x_4 \right)| \geq \gamma~~ \text{for all} ~~ x_1, x_2, x_3, x_4 \in \mathbb{Z}$ ?

It seems to me that this is a Diophantine approximation - type problem. Note that one can show through a simple application of Khintchine-Groshev theorem that $|f \left( x_1, x_2, x_3, x_4 \right)|$ will be strictly positive for all $x_1, x_2, x_3, x_4 \in \mathbb{Z}$, in that case with probability 1. This is relatively straightforward.

• What does it mean for a polynomial in more than one variable to have real roots? – Qiaochu Yuan Mar 15 '12 at 22:42
• It means that there exist real numbers $x_1^*, x_2^*, x_3^*, x_4^*$ such that $f(x_1^*, x_2^*, x_3^*, x_4^*) = 0$. – Stewart Mar 15 '12 at 22:45

• another related theorem - the fractional parts of $\alpha n^2$ are equidistributed in the unit interval (by a refinemnet of the Kronecker lemma, proved by Weyl/Hardy, and by Frustenberg). I'm guessing that if you take the coeff. if $f$ to be drwan in some iid manner and to be say diophantine generic, you can even get a quantitative version of those results, that is to get estimate on the maximal sizes of $x_1,\ldots, x_4$ which are needed in order to get $f(x_{1},\ldots,x_{4})$ to be less than some epsilon. – Asaf Mar 15 '12 at 23:34
• @GH: My problem in its full generality involves $n$-degree polynomials with $n^2$ variables. So I think I will be able to show something there. Your suggestions seem to be critical, thank you very much. – Stewart Mar 15 '12 at 23:57