1
$\begingroup$

Let $f(x,y) = a^2 x^2 - (b^2 - 2ac)xy + c^2 y^2$ be a positive definite binary quadratic form with co-prime integer coefficients such that $b \ne 0$. For a given pair of integers $(u,v)$ and a prime $p | f(u,v)$, we see that $u$ is a quadratic residue modulo $p$ if and only if $v$ is because

$$\displaystyle b^2 uv \equiv (ax + cy)^2 \pmod{p}.$$

Define the function $r(u,v)$ by

$$\displaystyle r(u,v) = \sum_{m | f(u,v)} \left(\frac{u}{m} \right) = \sum_{m | f(u,v)} \left(\frac{v}{m}\right).$$

Here $\left(\frac{\cdot}{n}\right)$ denotes the Jacobi symbol. What is the asymptotic value of

$$\displaystyle \sum_{f(u,v) \leq X} r(u,v)?$$

$\endgroup$
  • 1
    $\begingroup$ How important is the height condition $f(u,v) \leq X$? If this were instead $u,v \leq X$, the problem would be slightly easier and could possibly be dealt with using some version of the Chebotarev density theorem. $\endgroup$ – Daniel Loughran May 18 '17 at 16:11
  • $\begingroup$ I suppose averaging over a box is a start, but the problem I am considering is over an ellipse $\endgroup$ – Stanley Yao Xiao May 19 '17 at 19:59
  • $\begingroup$ Your problem is roughly a divisor sum problem for values of binary forms. There are quite a few papers in the literature on this kind of problem. You should be able to get an asymptotic formula after expanding up $r$ and changing the order of summation. The divisibility condition $m \mid f(u,v)$ can be interpreted as a kind of lattice condition for some expanding region. This is method due to Daniel (see his paper "On the divisor-sum problem for binary forms"). I would recommend that you speak with Roger about this as he is very familiar with these techniques. $\endgroup$ – Daniel Loughran May 20 '17 at 13:37
3
$\begingroup$

I will work with $f=u^2+v^2$ and the region $|u|,|v|\leq X$ but the argument easily generalises. The numbers $m$ go up to $X^2$ but by Dirichlet's divisor trick we can assume that they go up to $\sqrt{X^2}$. Then the congruence $u^2+v^2\equiv 0 \mod{m}$ can be written equivalently as $u\equiv \alpha v \mod{m}$ for some $\alpha \mod{m}$ that satisfies $\alpha^2+1\equiv 0 \mod{m}$. Then the contribution of these $m$ is
$$ \sum_{1\leq m \ll X} \sum_{\substack{\alpha \mod{m}\\\alpha^2+1\equiv 0 \mod{m}}} \left(\frac{\alpha}{m}\right) \sum_{\substack{|u|,|v|\leq X\\ u\equiv \alpha v \mod{m}}}1 .$$ The sum over $u,v$ is really the number of integer planar points in a lattice, therefore it equals $c_0 X^2m^{-1}+O(X|\mathbf{y}|^{-1}+1)$, where $\mathbf{y}$ comes from the first successive minima and it is of size $\ll \sqrt{m}$ by Minkowski's theorem. The main term gives $$X^2 \sum_{m\ll X}\frac{1}{m}\sum_{\substack{\alpha \mod{m}\\\alpha^2+1\equiv 0 \mod{m}}} \left(\frac{\alpha}{m}\right)$$ which are the first terms in a convergent series coming from some $L$-function without a pole at $1$.The $O(1)$ term gives $\ll X^{1+\epsilon}$ because the sum over $\alpha$ is of course $\ll m^\epsilon$, although a bound of the form $\ll \tau(m)^A$ for some fixed $A>1$ can also be proved (and needed when $f$ has larger degree). The most interesting idea of Daniel comes from how to handle the sum over $\mathbf{y}$. He showed that while $\mathbf{y}\ll \sqrt{m}$, it is possible to show that on average over $m$ the vector $\mathbf{y}$ that depends on both $m,\alpha$ attains its maximum value $\sqrt{m}$, thus providing a crucial saving in the error term (crucial when the degree of $f$ is higher). Here is how: note that $\mathbf{y}=(y_1,y_2)$ belongs to the lattice hence $m$ divides $y_1^2+y_2^2$. Thus the contribution is $$\ll X \sum_{|\mathbf{y}|\ll \sqrt{X}} \frac{1}{|\mathbf{y}|}\sum_{d|y_1^2+y_2^2}1\ll X^{1+\epsilon}\sum_{|\mathbf{y}|\ll \sqrt{X}} \frac{1}{|\mathbf{y}|}\ll X^{3/2+\epsilon}.$$ Dealing with higher degrees forces to separately consider the contribution of $m$ lying extremely close to $\sqrt{x^{\deg(f)}}$, say $m \in (\sqrt{X^{\deg(f)}} (\log X)^{-100},\sqrt{X^{\deg(f)}}]$. These will go into the final error term and for this one needs a more complicated approach related to the distribution of divisors in small intervals. It was a surprise when the paper of Daniel came out because people believed one could not go higher than $\deg(f)\leq 3$ which was done by Greaves in a very long paper. It would be interesting to see if the approach of Daniel works for weighted projective cases, for example $$\sum_{1\leq u,v \leq X} \sum_{d|u^2+v^4}1,$$ this has not been done previously.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.