Average of gcd of sum of two $k$th powers

Question

I am interested bounding the following quantity. Given fixed $k \in \mathbb{N}$, $a,b \in \mathbb{Z}$, $\sigma \in [0,1)$, and intervals $I_1,I_2 \subset \mathbb{Z}$ can we establish the bound

$$S = \sum_{x \in I_1,y \in I_2}\min\{(ax^k+by^k,q)^{1/k},q^{\sigma}\} \ll q^{\varepsilon}(|I_1||I_2|+q^{\sigma})?$$

The reason this bound seems like it might reasonable is because most of the time we have that $(ax^k,q) \neq (by^k,q)$, when this is case we have that $(ax^k+by^k,q)^{1/k} \ll \min\{(x,q),(y,q)\}$. We then get that $S \ll S_1+S_2$ where $$S_1 = \sum_{x \in I_1,y \in I_2}\min\{(x,q),(y,q),q^{\sigma}\} \ll q^{\varepsilon}(|I_1||I_2|+q^{\sigma}),$$ and $$S_2 = \sum_{\substack{x \in I_1,y \in I_2 \\ (ax^k,q) = (by^k,q)}}\min\{(ax^k+by^k,q)^{1/k},q^{\sigma}\}.$$ It feels like $S_2$ should be smaller than $S_1$ because of the GCD restriction, but I am having trouble seeing if this is actually the case or not.

One reason this might not be possible is if there is always a positive proportion of points $x,y$ in $I_1 \times I_2$ for which $(ax^k,q) = (by^k,q)$ and $(ax^k+by^k,q)^{1/k} \gg q^{\alpha}$ for some $\alpha > 0$, then we get the lower bound

$$S_2 \gg q^{\min\{\sigma,\alpha\}}|I_1||I_2|.$$

I am not sure if this is the case or not, I am not very familiar with these types of sums.

Answer 1

If $q$ is prime, then put $U_1 = \{(x,y) \in I_1 \times I_2 : q \vert ax^k + by^k\}$ and $U_2 = I_1 \times I_2 \setminus U_1$. Then $U_1$ is of the form $I_1 \times I_2 \cap L$, where $L$ is a union of at most $k$ lattices $\Lambda_\omega$ of the form

$$\displaystyle \Lambda_\omega = \{(x,y) \in \mathbb{Z}^2 : x \equiv \omega y \pmod{q}\}$$

where $a\omega^k + b \equiv 0 \pmod{q}$. We have

$$\displaystyle |I_1 \times I_2 \cap \Lambda_\omega| = \frac{|I_1||I_2|}{q} + O(\max\{|I_1|, |I_2|\}).$$

The contribution of the sum coming from $U_2$ is clearly $O(|I_1||I_2|)$. On the other hand,

$$\displaystyle \sum_{(x,y) \in U_1} \min\{\gcd(ax^k + by^k, q)^{1/k}, q^\sigma\} = \sum_{(x,y) \in U_1} \min\{q^{1/k}, q^\sigma\}$$ $$ \leq k \min\{q^{1/k}, q^\sigma\} \left(\frac{|I_1||I_2|}{q} + O(\max\{|I_1|, |I_2\})\right).$$

In general, if one of the lattices $\Lambda_{\omega}$ has a very short vector, which is the case say when $q = a + b$, then the big-$O$ term is sharp. Therefore in general I don't think this bound can be improved to completely remove the dependence on $\max\{|I_1|, |I_2|\}$, but probably in most interesting regimes this is harmless.

If you drop the requirement that $q$ is prime, then a bit more work has to be done to sort the sum by divisors of $q$, but basically the same type of bound can be obtained.

Answer 2

Let for $d \mid q$ take $$U_d(I_1,I_2) = \{x \in I_1,y \in I_2: ax^k+bx^k \equiv 0 \mod d\},$$ then one has that $$S \le \sum_{d |q}\min\{d^{1/k},q^{\sigma}\}\#U_d(I_1,I_2).$$ So the problem reduces to estimating $\#U_d(I_1,I_2)$, this may be done with exponential sums since \begin{align*} \#U_d(I_1,I_2) &= \frac{1}{d}\sum_{1 \le c \le d} \sum_{\substack{x \in I_1 \\ y \in I_2}}e\left(\frac{c}{d}(ax^k+by^k) \right) \\ &= \frac{1}{d}\sum_{1 \le c \le d} \sum_{1 \le x,y \le d}e\left(\frac{c}{d}(ax^k+by^k) \right)\left(\frac{|I_1||I_2|}{d^2}+ O \left(\frac{|I_1|+|I_2|}{d} \right) \right) \\ &\ll d^{-2/k}\left(|I_1||I_2|+ d(|I_1|+|I_2|) \right). \end{align*} Where the last bound comes from the well known bound $$\sum_{1 \le x \le d}e\left(\frac{c}{d}ax^k \right) \ll (a,d)^{1/k}d^{1-1/k} \ll d^{1-1/k}.$$ Thus we obtain the bound \begin{align*} S &\ll \sum_{d |q}\min\{d^{1/k},q^{\sigma}\}d^{-2/k}\left(|I_1||I_2|+ d(|I_1|+|I_2|) \right) \\ &\le |I_1||I_2|\sum_{d|q}d^{-1/k}+q^{\sigma+1-2/k}(|I_1|+|I_2|)\sum_{d \mid q}1 \\ &\ll q^{\epsilon}\left( |I_1||I_2|+ q^{\sigma+1-2/k}(|I_1|+|I_2|)\right). \end{align*} This bound is worse than the one Stanley hints at for $k>2$, however I have not been able to completely figure out how his argument generalizes to general $q$. For now this bound is actually sufficient for me since I am most interested in the case $k=2$.