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edited Oct 5 at 16:51

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 $$\sum_{1 \le x \le d}e\left(\frac{c}{d}ax^k \right) \ll (a,d)d^{1-1/k} \ll d^{1-1/k}.$$$$\sum_{1 \le x \le d}e\left(\frac{c}{d}ax^k \right) \ll (a,d)^{1/k}d^{1-1/k} \ll d^{1-1/k}.$$ 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 $k=2$.

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 $$\sum_{1 \le x \le d}e\left(\frac{c}{d}ax^k \right) \ll (a,d)d^{1-1/k} \ll d^{1-1/k}.$$ 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 $k=2$.

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 $k=2$.

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created Oct 5 at 6:27

Daniel Flores

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 $$\sum_{1 \le x \le d}e\left(\frac{c}{d}ax^k \right) \ll (a,d)d^{1-1/k} \ll d^{1-1/k}.$$ 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 $k=2$.