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Note: I posted this on math.stackexchange.com earlier (original post here: https://math.stackexchange.com/questions/1471331/showing-that-a-particular-area-is-small), but it received no responses and very few views, so I thought it might be appropriate for here.

Suppose that $F(x,y) \in \mathbb{Z}[x,y]$ is an irreducible binary form of degree $d \geq 3$. Let $B$ be a positive parameter, considered to be large (i.e. one can consider it to be sufficiently large as to dwarf the coefficients of $F$). Let $\delta > 0$ be a small parameter. I wish to inquire whether the area

$$\displaystyle A_F(B; \delta) = m(\{(x,y) \in \mathbb{R}^2 : |F(x,y)| \leq B, \max\{|x|, |y|\} > B^{1/d} (\log B)^\delta \})$$

is small with respect to $B^{2/d}$. In particular, I would like to know whether $A_F(B;\delta) = o_{F,\delta}(B^{2/d})$ holds.

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  • $\begingroup$ What happens for $\delta=0$? $\endgroup$ Commented Oct 9, 2015 at 15:10
  • $\begingroup$ @IlyaBogdanov then it would depend on the coefficients of $F$. In particular, in general the area will be $\Omega(B^{2/d})$. $\endgroup$ Commented Oct 9, 2015 at 15:16

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Since $F$ is irreducible, after a suitable rotation (not affecting the condition on $\max(|x|,|y|)$ too much) we have $$ F(x,y)=\alpha\prod_{i=1}^d(x-c_iy), $$ where $c_i$ are distinct complex numbers.

Now, for every $y$ with $|y|=a>B^{1/d}\log^\delta B$, if $(x,y)\in A_F(B,\delta)$ then $|x-c_iy|\leq B^{1/d}$ for some $i$. Then for all $j\neq i$ we have $$ |x-c_iy|+|c_iy-c_jy|\geq |x-c_jy|\geq |c_iy-c_jy|-|x-c_iy|, $$ so $$ |x-c_jy|=(1+o(1))|c_j-c_i|a. $$ Therefore, our $x$ satisfies the condition iff $|x-c_iy|\leq \mu Ba^{1-d}$ for some constant $\mu$. Thus, on the line with this fixed $y$-coodrinate $A_F(B,\delta)$ has some (say, $k$) segments, corresponding to all real $c_i$, each segment of length $(1+o(1))2\mu Ba^{1-d}$. Integrating over $a$, we get $$ \sim 2k\cdot 2\mu B\int_{B^{1/d}\log^\delta B}^{+\infty}\frac{da}{a^{d-1}} =\frac{4k\mu B}{(d-2)B^{1-2/d}\log^{\delta(d-2)}B} =\frac{4k\mu}{d-2}\frac{B^{2/d}}{\log^{\delta(d-2)}B}. $$ This is an estimate for the part of $A_F(B,\delta)$ with large $|y|$. The same estimate works for large $|x|$.

Thus $A_F(B,\delta)=\Theta_F(B^{2/d}/\log^{\delta(d-2)}B)$.

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