2.1 Preliminary

In this section, we establish the key results involving the division polynomials, which are needed for proving the results stated in Section 1.2.

As usual, for any $\Psi \in {\mathbb F}_p(E)$ , we define

$$\begin{align*}\deg \Psi=\sum_{\substack{P\in E(\overline{{\mathbb F}_p})\cr \nu_P(\Psi)>0}} \nu_{P}(\Psi), \end{align*}$$

where $\nu _{P}(\Psi )$ is the multiplicity of P as a zero of $\Psi $ (and thus this is a finite sum over all zeros of $\Psi $ ).

Let us again recall R from (1.1). Note that [Reference Shparlinski and Stange13, Lemma 3.1] for $s=d$ implies that the sequence $\chi (\psi _n(P))$ is periodic with a period dividing R, as long as $\operatorname {ord} P\geq 3$ . Recall the following estimate from [Reference Bhakta and Shparlinski3, Lemma 2.3].

Lemma 2.1 Let $\chi $ be any non-principal multiplicative character of ${\mathbb F}_p^{*}$ , and R be as in (1.1). Assume that $\Psi $ is not a nontrivial power of a function in $\overline {{\mathbb F}_p(E)}$ . Then, we have the following estimate uniformly over $a\in {\mathbb Z:}$

$$ \begin{align*}\left|\sum_{n\leq R} \chi(\Psi(nP)){\mathbf{\,e}}_R(an)\right|\ll \deg \Psi \cdot \sqrt{p},\end{align*} $$

where the implied constant may depend on d.

We also need the following lemma to show that the functions of our requirements satisfy the condition of Lemma 2.1.

Lemma 2.2 Let $m_1,m_2,n_1,n_2$ be positive integers satisfying

(2.1) $$ \begin{align} \gcd(m_1n_1,m_2n_2)=\gcd(m_1n_2,m_2n_1). \end{align} $$

Suppose further that

(2.2) $$ \begin{align} m_1 \neq m_2 \quad \text{and} \quad n_1 \neq n_2. \end{align} $$

Then, the function

$$ \begin{align*}\Psi(Q)=\psi_{m_1n_1}\cdot \psi_{m_2n_2}\cdot \psi_{m_1n_2}^{-1}\cdot \psi_{m_2n_1}^{-1}(Q)\end{align*} $$

is not a nontrivial power of any function in $\overline {{\mathbb F}_p(E)}$ .

Proof For any integer h, note that $\psi _h$ has $h^2-1$ simple zeros (see [Reference Silverman15, Exercise III.3.7]).

Clearly, then the product $\psi _{m_1n_1} \cdot \psi _{m_2n_2}$ has

$$ \begin{align*}(m_1n_1)^2+(m_2n_2)^2-2\gcd(m_1n_1,m_2n_2)\end{align*} $$

many simple zeros, and $\gcd (m_1n_1,m_2n_2)-1$ many zeros of multiplicity $2$ . Similarly, $\psi _{m_1n_2} \cdot \psi _{m_2n_1}$ has

$$ \begin{align*}(m_1n_2)^2+(m_2n_1)^2-2\gcd(m_1n_2,m_2n_1)\end{align*} $$

many simple zeros, and $\gcd (m_1n_2,m_2n_1)-1$ many of zeros of multiplicity $2$ .

In particular, if $\Psi =G^{\nu }$ for some $\nu \geq 2$ and $G\in \overline {{\mathbb F}_p(E)}$ , then

$$\begin{align*}\widetilde{\Psi}=\Psi \cdot \psi^{2}_{m_1n_2}\cdot \psi^{2}_{m_2n_1} \end{align*}$$

has the property that each zero of $\widetilde {\Psi }$ has multiplicity at least two. Therefore, each simple zero of $\psi _{m_1n_1} \cdot \psi _{m_2n_2}$ must also be a simple zero of $\psi _{m_1n_2} \cdot \psi _{m_2n_1}$ , and vice versa. Therefore, the total number of simple zeros for both $\psi _{m_1n_1}\cdot \psi _{m_2n_2}$ and $\psi _{m_1n_2} \cdot \psi _{m_2n_1}$ coincides. We get a contradiction due to the imposed conditions at (2.1) and (2.2).

To establish the main result of this section, we need one more ingredient from [Reference Shparlinski and Stange13, Lemma 3.2], which demonstrates an almost multiplicative nature of $\chi (\psi _n(P))$ .

Lemma 2.3 Let $\chi $ be a multiplicative character. Then for any integers $m,n,$ we have

$$\begin{align*}\chi(\psi_{mn}(P))=\chi(\psi_m(nP))\chi(\psi_n(P))^{m^2}. \end{align*}$$

2.2 Correlations of character values

Lemma 2.4 Let R be as in (1.6), and $1\leq N\leq R$ be any integer. Then, for any distinct primes $\ell _1,\ell _2\leq R$ , we have

$$ \begin{align*}\left|\sum_{n \leq N} \chi(\psi_{\ell_1n}(P))\overline{\chi(\psi_{\ell_2n}(P))}\right|\ll \max\{\ell_1,\ell_2\}p^{\frac{1}{12}}R^{5/6}(\log R)(\log \log R)^{1/3},\end{align*} $$

where the implied constant may depend on d.

Proof Let a be an integer. Denote

$$ \begin{align*}S_{\ell_1,\ell_2}(a)=\sum_{n \leq R} \chi(\psi_{\ell_1n}(P))\overline{\chi(\psi_{\ell_2n}(P))}{\mathbf{\,e}}_R\left(an\right).\end{align*} $$

For a parameter L to be determined later, we consider the set of integers

$$ \begin{align*}\mathcal{R}_{d}=\left\{1\leq r \leq L:~r\equiv 1 \pmod d,\ \gcd(r,\ell_1\ell_2R)=1\right\}.\end{align*} $$

Clearly by the same argument as in the proof of [Reference Bhakta and Shparlinski3, Lemma 2.4], we have

$$ \begin{align*}\sharp\,{\mathcal R}_{d} \ge \frac{\varphi(\ell_1\ell_2 R)(L-1)}{d\ell_1\ell_2 R}-\tau(\ell_1\ell_2 R). \end{align*} $$

Since $\tau (\ell _1\ell _2R)\ll \tau (R)$ and $\ell _1,\ell _2\leq R$ , by the standard estimates of the functions $\tau (n)$ and $\varphi (n)$ (see [Reference Hardy and Wright5, Theorems 317 and 328], for instance), we have

(2.3) $$ \begin{align} \sharp\,{\mathcal R}_{d} \gg \frac{L}{\log \log R}, \end{align} $$

provided that

(2.4) $$ \begin{align} L \ge \exp(0.7 \log p/\log \log p). \end{align} $$

Let us now set

$$ \begin{align*}W = \sum_{r\in {\mathcal R}_{d}} \sum_{n\leq R} \chi(\psi_{ \ell_1 r n}(P))\overline{\chi(\psi_{\ell_2 r n}(P))} {\mathbf{\,e}}_R(ar n).\end{align*} $$

Since each $r\in {\mathcal R}_{d}$ is co-prime to R, it is clear that

(2.5) $$ \begin{align} S_{\ell_1,\ell_2}(a)=\frac{1}{\sharp\,{\mathcal R}_d} W. \end{align} $$

Applying Cauchy–Schwarz combined with Lemma 2.3, we get

(2.6) $$ \begin{align} \begin{aligned} |W|^2 &\leq R \sum_{n\leq R} \left|\sum_{r\in {\mathcal R}_{d}}\chi(\psi_{\ell_1 r n}(P))\overline{\chi(\psi_{\ell_2 r n}(P))} {\mathbf{\,e}}_R(ar n)\right|^2\\ &= R \sum_{n\leq R} \left|\sum_{r\in {\mathcal R}_{d}}\chi(\psi_{\ell_1 r }(nP))\overline{\chi(\psi_{\ell_2 r }(nP))}\chi(\psi_n(P))^{\ell_1^2-\ell_2^2} {\mathbf{\,e}}_R(ar n)\right|^2\\ &= R \sum_{n\leq R} \left|\sum_{r\in {\mathcal R}_{d}}\chi(\psi_{\ell_1 r }(nP))\overline{\chi(\psi_{\ell_2 r }(nP))} {\mathbf{\,e}}_R(ar n)\right|^2\\ &\le R \sum_{r_1, r_2 \in {\mathcal R}_{d}}\left|\sum_{n \leq R} \chi(\varPsi_{r_1,r_2}(nP)) {\mathbf{\,e}}_R(a(r_1 - r_2)n)\right|, \end{aligned} \end{align} $$

where $\varPsi _{r_1,r_2}\in {\mathbb F}_p(E)$ is given by the following:

$$ \begin{align*}\varPsi_{r_1,r_2}(Q)= \psi_{\ell_1 r_1}(Q)\cdot \psi_{\ell_2 r_2}(Q)\cdot \psi_{\ell_2 r_1}^{-1}(Q)\cdot \psi_{\ell_1 r_2}^{-1}(Q).\end{align*} $$

Clearly, we have

$$\begin{align*}\deg \varPsi_{r_1,r_2} \ll (L\cdot \max\{\ell_1,\ell_2\})^2. \end{align*}$$

Now, for any $r_1\neq r_2\in {\mathcal R}_{d}$ , we have

(2.7) $$ \begin{align} \gcd(\ell_1 r_1,\ell_2 r_2)=\gcd(r_1,r_2)=\gcd(\ell_1r_2,\ell_2r_1). \end{align} $$

Therefore, Lemma 2.2 shows that $\varPsi _{r_1,r_2}$ is not a nontrivial power of any $G\in \overline {{\mathbb F}_p(E)}$ . Applying Lemma 2.1 to the inner sum in (2.6) for each non-diagonal terms $r_1\neq r_2$ , and trivially estimating the contributions from the diagonal terms $r_1=r_2$ , we obtain

$$ \begin{align*} |W|^2& \leq R^2 \sharp\, {\mathcal R}_d+R \sum_{r_1 \neq r_2 \in {\mathcal R}_d}\left|\sum_{n\leq R} \chi(\Psi_{r_1,r_2}(nP)) {\mathbf{\,e}}_R(a(r_1 - r_2)n)\right|\cr &\ll R^2 \sharp\, {\mathcal R}_d+R(\sharp\, {\mathcal R}_d)^2 \cdot \deg \varPsi_{r_1,r_2}\cdot \sqrt{p}\cr &\ll R^2 \sharp\, {\mathcal R}_{d}+R(\sharp\, {\mathcal R}_d)^2 (L\cdot \max\{\ell_1,\ell_2\})^2\cdot \sqrt{p}. \end{align*} $$

In particular, (2.3) and (2.5) imply that

$$ \begin{align*} |S_{\ell_1,\ell_2}(a)|\ll R L^{-1/2}(\log \log R)^{1/2}+R^{1/2}L\max\{\ell_1,\ell_2\}p^{1/4}. \end{align*} $$

The proof concludes, choosing $L=R^{1/3}p^{-1/6}(\log \log R)^{1/3}$ (clearly L satisfies (2.4) because of the condition (1.6)), and by the completing technique as in [Reference Iwaniec and Kowalski9, Section 12.2].

Remark 2.5 Following the proof of Lemma 2.4, one observes that the assumption $\ell _1,\ell _2 \le R$ is in fact unnecessary. It was imposed only to justify the estimate

$$\begin{align*}\frac{\phi(\ell_1\ell_2R)}{d\ell_1\ell_2R} \gg \frac{1}{\log\log(\ell_1\ell_2R)} \gg \frac{1}{\log\log R}. \end{align*}$$

If, however, one of $\ell _1$ or $\ell _2$ exceeds R, the same conclusion continues to hold; simply because we have $1-\frac {1}{\ell }\ge 1/2$ for any prime $\ell $ . For simplicity of exposition, we have stated Lemma 2.4 under the condition $\ell _1,\ell _2 \le R$ , as this is the only case required in the proof of Theorem 1.1.

Moreover, if both $\ell _1,\ell _2$ are odd, and $\chi $ is quadratic, then just by Lemmas 2.1 and 2.3, we immediately have

$$ \begin{align*}\left|\sum_{n \leq N} \chi(\psi_{\ell_1n}(P))\overline{\chi(\psi_{\ell_2n}(P))}\right|\ll (\max\{\ell_1,\ell_2\})^2p^{1/2}(\log R).\end{align*} $$

Due to (1.4), of course, the bound above is better than Lemma 2.4, as long as $\ell _1,\ell _2$ are fixed.

Remark 2.6 If we take $m_1,m_2\leq R$ to be two distinct positive integers. Then, by the same argument in the proof of Lemma 2.1, we have the following weaker estimate:

$$ \begin{align*} \left|\sum_{n \leq N} \chi(\psi_{m_1n}(P))\overline{\chi(\psi_{m_2n}(P))}\right|\ll \max\{m_1,m_2\}p^{\frac{1}{12}}R^{5/6}(\log R)^{4/3}. \end{align*} $$

In particular, we have

(2.8) $$ \begin{align} \left|\sum_{N\le n \leq 2N} \chi(\psi_{m_1n}(P))\overline{\chi(\psi_{m_2n}(P))}\right|\ll \max\{m_1,m_2\}p^{\frac{1}{12}}R^{5/6}(\log R)^{4/3}. \end{align} $$

The reason for the weaker bound is that the identity (2.7) does not necessarily hold, unless $r_1,r_2$ are primes. In that case, ${\mathcal R}_d$ is essentially a set of primes, and we then have to argue as in the proof of [Reference Shparlinski and Stange13, Theorem 5.1], where the argument does not allow us to replace the factor $(\log R)^{1/3}$ by $(\log \log R)^{1/3}$ .

Following the argument in the remark above, we also have the following estimate for any m:

(2.9) $$ \begin{align} \left|\sum_{n \leq N} \chi(\psi_{m n}(P))\right|\ll m p^{\frac{1}{12}}R^{5/6}(\log R)^{4/3}. \end{align} $$

In fact, one can get here $(\log R)(\log \log R)^{1/3}$ instead of $(\log R)^{4/3}$ . This is because $\psi _{mr_1}\cdot \psi _{mr_2}^{-1}$ is not a power of any function in $\overline {{\mathbb F}_p}(E)$ for any integers $r_1\neq r_2$ .

However, the dependence of m in (2.9) would be too restrictive for us to prove Theorem 1.3. To overcome that obstacle, we use Lemma 2.3 to get a sum associated with the point $mP\in E({\mathbb F}_p)$ (see Lemma 2.8 below). To handle the extra twisting factors $\chi (\psi _m(P))^{n^2}$ arising from Lemma 2.3, we first need the following estimate, which essentially generalizes both [Reference Bhakta and Shparlinski3, Lemma 2.4] and [Reference Shparlinski and Stange13, Theorem 5.1].

Lemma 2.7 Let q and $1\leq k\leq q$ be any two fixed integers, and R be as in (1.6). Then for any integer $1\le N\leq R$ , we have

$$ \begin{align*}\left|\sum_{\substack{n\leq N\\ n\equiv k \pmod q}} \chi (\psi_n(P))\right| \ll p^{1/12}R^{5/6}(\log R)(\log \log R)^{1/3},\end{align*} $$

where the implied constant may depend on d.

Proof For a suitable parameter L, consider the set of integers

$$ \begin{align*}\mathcal{R}_{d,q}=\{r<L:r \equiv 1\pmod {dq},\ \gcd(r,R) = 1\},\end{align*} $$

where again, d is the order of $\chi $ .

Since $q,d$ are both fixed, by the same argument as in the proof of [Reference Bhakta and Shparlinski3, Lemma 2.4], we have

$$ \begin{align*}\sharp\,{\mathcal R}_{d,q} \gg \frac{L}{\log \log R},\quad \mathrm{provided~that}\quad L \ge \exp(0.7 \log p/\log \log p). \end{align*} $$

Then, for any $r\in \mathcal {R}_{d,q}$ ,

$$ \begin{align*}\sum_{\substack{n\leq R\\ n\equiv k \pmod q}} \chi(\psi_{n}(P))e_R(an)=\sum_{\substack{n\leq R\\ n\equiv k \pmod q}} \chi(\psi_{nr}(P))e_R(anr).\end{align*} $$

In particular, the desired sum is bounded by $\frac {1}{\sharp \, \mathcal {R}_{d,q}}\cdot |W_k|$ , where

$$ \begin{align*}W_k = \sum_{r \in \mathcal{R}_{d,q}}~\sum_{\substack{n\leq R\\ n\equiv k \pmod q}} \chi(\psi_{nr}(P))e_R(anr).\end{align*} $$

By Cauchy–Schwarz and Lemma 2.3, we have

$$ \begin{align*} |W_k|^2 &\le R \sum_{n\leq R}\left|\sum_{r \in \mathcal{R}_{d,q}} \chi(\psi_{r n}(P))e_R(ar n)\right|^2\\ &=R\sum_{n\leq R}\left|\sum_{r \in \mathcal{R}_{d,q}} \chi(\psi_{r}(nP))\chi(\psi_n(P))e_R(a r n)\right|^2\\ &= R\sum_{n\leq R}\left|\sum_{r \in \mathcal{R}_{d,q}} \chi(\psi_{r}(nP))e_R(a r n)\right|^2\\ &\le R \sum_{r_1,r_2 \in \mathcal{R}_{d,q}}\left|\sum_{n\leq R} \chi(\psi_{r_1}\cdot \psi_{r_2}^{-1}(nP))e_R(a(r_1-r_2) n)\right|. \end{align*} $$

From this point, we proceed exactly as in the proof of [Reference Bhakta and Shparlinski3, Lemma 2.4], applying the completing technique from [Reference Iwaniec and Kowalski9, Section 12.2] to conclude.

Now, we are ready to weaken the dependence of m in (2.9).

Lemma 2.8 Let R and $\varepsilon $ be as in Theorem 1.1, and $m\leq R$ be any positive integer satisfying $(m,R)\leq p^{\varepsilon /2}$ . Then for any integer $1\leq N\leq R$ , we have

$$ \begin{align*}\left|\sum_{n\leq N} \chi(\psi_{mn}(P))\right|\ll p^{1/12}R^{5/6}(\log R)(\log \log R)^{1/3},\end{align*} $$

where the implied constant may depend on d and $\varepsilon $ .

Proof By Lemma 2.3, we can rewrite the sum as

$$\begin{align*}\left|\sum_{n\leq N} \chi(\psi_n(mP))\chi(\psi_m(P))^{n^2}\right| \ll \sum_{k \leq d} \left| \sum_{\substack{n \leq N \cr n \equiv k \pmod{d}}} \chi(\psi_n(mP)) \right|. \end{align*}$$

Now, applying Lemma 2.7 to each of the inner sums with $q = d$ , and noting that

$$\begin{align*}\operatorname{ord}(mP) = \frac{\operatorname{ord}(P)}{(m, \operatorname{ord}(P))} \geq p^{1/2 + \varepsilon/2}, \end{align*}$$

the proof follows, because $mP$ satisfies (1.6) for any $p>C(\varepsilon )$ , where $C(\varepsilon )>0$ is some constant depending only on $\varepsilon $ .

3.1 Treatment for $U_0$

The fundamental lemma of combinatorial sieve [Reference Tenenbaum17, Theorem 4.4] shows that the set ${\mathcal A}_{0}(N,I)$ is small when x and y are widely separated parameters. However, we need an estimate for the sum of $\tau _{\nu }$ -bounded functions over this small set. We simply use the following estimate, readily available from [Reference Korolev and Shparlinski10, Equations (8-2) and (8-3)]:

(3.2) $$ \begin{align} |U_{0}|\ll N \exp\left(-\frac{\log N}{4\log x}\right) (\log N)^{\frac{\nu^2-1}{2}}+\frac{N(\log N)^{\nu-1}}{(\log y)^{\nu}}(\log x)^{\nu}. \end{align} $$

A similar analysis could also be found in [Reference Sun16, Lemma 2.2].

3.2 Treatment for $U_r$ for $r>0$

We can split the sum $U_r$ as

$$ \begin{align*}U_r = U_{r,1} + U_{r,2},\end{align*} $$

where $U_{r,1}$ denotes the contribution from the integers $n \in {\mathcal A}_r(N,I)$ whose prime divisors from the interval I all appear with multiplicity one. The remaining contribution, coming from integers where at least one prime divisor from I appears with multiplicity at least two, is denoted by $U_{r,2}$ .

Note that the sums $U_{r,2}$ are much easier to handle, as crudely using the submultiplicative property of $\tau _{\nu }(n)$ from (1.2) and (1.3) for $C=1$ , we have

(3.3) $$ \begin{align} \begin{aligned} \left|\sum_{r\ll \log N} U_{r,2}\right| &\leq \sum_{\substack{x< \ell\leq y\\ \ell~\mathrm{prime}}}~\sum_{1\leq s \leq \frac{N}{\ell^{2}}} \left|f(\ell^2 s)\right|\leq \sum_{\substack{x< \ell\leq y\\ \ell~\mathrm{prime}}} \tau_{\nu}(\ell^{2}) \sum_{1\leq s \leq \frac{N}{\ell^{2}}} \tau_{\nu}(s)\cr &\ll N (\log N)^{\nu-1}\sum_{\substack{x< \ell\\ \ell~\mathrm{prime}}} \frac{\tau_{\nu}(\ell^{2})}{\ell^2}\ll \frac{N(\log N)^{\nu-1}}{x \log x}. \end{aligned} \end{align} $$

3.2.1 Estimating $U_{r,1}$

Denote ${\mathcal A}^{\prime }_r(N,I)$ be the set of integers that appear as an index in the sum $U_{r,1}$ . Note that any $n\in {\mathcal A}^{\prime }_r(N,I)$ has exactly r representations of the form $n = \ell m$ , where the prime $\ell \in I$ , and integer $m\in {\mathcal A}^{\prime }_{r-1}(N/\ell ,I)$ , with $\gcd (m,\ell )=1$ . Due to the multiplicativity of f, we can write

$$ \begin{align*}U_{r,1}=r^{-1}\sum_{\ell \in I}~f(\ell)~\sum_{\substack{m\in {\mathcal A}^{\prime}_{r-1}(N/\ell,I)\cr{\mathrm{gcd}}(m, \ell) = 1}} f(m)\chi(\psi_{\ell m}(P)).\end{align*} $$

Now, dividing $I=(x,y]$ into $K=O(\log y)$ many dyadic intervals $I_k=(x_k,y_k]$ for $k=0,1,\ldots , K$ , as in [Reference Korolev and Shparlinski10, Equation (7-7)], we can write

(3.4) $$ \begin{align} U_{r,1}=\sum_{k=0}^{K} V_{k,r}, \end{align} $$

where

$$ \begin{align*} V_{k,r}&=r^{-1}\sum_{\ell\in I_k}~f(\ell)~\sum_{\substack{m\in {\mathcal A}^{\prime}_{r-1}(N/\ell, I)\cr{\mathrm{gcd}}(m, \ell) = 1}} f(m)\chi(\psi_{\ell m}(P))\cr &=r^{-1}\sum_{\substack{m\in {\mathcal A}^{\prime}_{r-1}(N/x_k, I)}}~f(m)~\sum_{\substack{\ell \in I_k\cap (1,N/m] \cr{\mathrm{gcd}}(m, \ell) = 1}} f(\ell) \chi(\psi_{\ell m}(P)). \end{align*} $$

Note that $|f(m)|\le \tau _{\nu }(m)$ , and for each $m\in {\mathcal A}^{\prime }_{r-1}(N/x_k,I),$ there exists at most r many $\ell \in I_k$ for which $\gcd (m,\ell )\neq 1$ . Therefore, applying (1.2) with $C=1$ , we can write

(3.5) $$ \begin{align} V_{k,r}\ll r^{-1}W_{k,r}+N(\log N)^{\nu-1}x_k^{-1}, \end{align} $$

where

$$ \begin{align*} W_{k,r}=\sum_{m\in {\mathcal A}^{\prime}_{r-1}\left(N/x_k,I\right)}\tau_{\nu}(m)\left|\sum_{\ell \in I_k\cap (1,N/m]}f(\ell)\chi(\psi_{\ell m}(P))\right|. \end{align*} $$

Crudely, summing over $m\leq N/x_k$ and applying Cauchy–Schwarz, we have

$$ \begin{align*} |W_{k,r}|^2&\leq Nx_k^{-1}(\log N)^{\nu^2-1} \sum_{m \leq N/x_k} \left| \sum_{\ell \in I_k \cap (1,N/m]} f(\ell)\, \chi(\psi_{\ell m}(P)) \right|^2 \cr &\leq Nx_k^{-1}(\log N)^{\nu^2-1} \sum_{\ell_1,\,\ell_2 \in I_k} \left| \sum_{m \leq N/{\max\{x_k,\, \ell_1,\, \ell_2\}}} \chi(\psi_{\ell_1 m}(P))\, \overline{\chi(\psi_{\ell_2 m}(P))} \right|, \end{align*} $$

where we are using above that $|f(\ell _1)f(\ell _2)|= O(1)$ , and the estimate from (1.2) for $C=2$ .

At this point, our argument deviates from the proof of [Reference Korolev and Shparlinski10, Theorem 2.1]. Instead of using [Reference Korolev and Shparlinski10, Corollary 4.2], we appeal to Lemma 2.4 to estimate the inner sums above for $\ell _1\neq \ell _2$ . Note that (1.4) implies (1.6) for any $p>C(\varepsilon )$ , where $C(\varepsilon )>0$ is some constant depending only on $\varepsilon $ . Moreover, we estimate the contributions from $\ell _1=\ell _2$ trivially, and get

$$ \begin{align*} \begin{aligned} \left|W_{k,r}\right|^2&\ll \left(N x_k^{-1}\left(y_kNx_k^{-1}+y^3_k p^{\frac{1}{12}}R^{5/6+o(1)}\right)\right)(\log N)^{\nu^2-1}\cr &\ll \left(N^2x_k^{-1}+Nx_k^2 p^{\frac{1}{12}}R^{5/6+o(1)}\right)(\log N)^{\nu^2-1}. \end{aligned} \end{align*} $$

Consequently, we deduce the following estimate from (3.4) and (3.5):

(3.6) $$ \begin{align} \begin{aligned} U_{r,1}&\ll r^{-1}\sum_{k=0}^{K}(W_{k,r}+rN(\log N)^{\nu-1}x_k^{-1})\cr &\ll r^{-1} \left( Nx^{-1/2} + y N^{1/2} p^{\frac{1}{24}}R^{5/12+o(1)}\right)(\log N)^{\frac{\nu^2-1}{2}}+N(\log N)^{\nu-1}x^{-1}. \end{aligned} \end{align} $$

3.3 Concluding the proof

From (3.1), we have

$$ \begin{align*} S_{f, \chi,P}(N)&\ll \sum_{\substack{r\ll \log N \cr r \neq 0}} |U_{r,1}|+\left|\sum_{\substack{r\ll \log N \cr r\neq 0}} U_{r,2}\right|+|U_{0}|. \end{align*} $$

Let us choose the parameters

$$ \begin{align*}x=(\log R)^{\nu^2+3},\quad \mathrm{and} \quad y=R^{\varepsilon/4}.\end{align*} $$

From (3.6), we have

$$ \begin{align*} \sum_{\substack{r\ll \log N \cr r \neq 0}}|U_{r,1}|&\ll \frac{N} {\log R}+y N^{1/2} p^{\frac{1}{24}}R^{5/12+o(1)}\ll \frac{N}{\log R}, \end{align*} $$

where the last inequality is due to the condition (1.5).

From (3.3), we have

$$ \begin{align*}\left|\sum_{\substack{r\ll \log N\\ r\neq 0}} U_{r,2}\right|\ll \frac{N(\log N)^{\nu-1}}{x \log x}\ll \frac{N}{\log R}.\end{align*} $$

The proof is now complete, as (3.2) implies $|U_0|\ll \varepsilon ^{-\nu }N\frac {(\log \log R)^{\nu }}{\log R}.$

4.1 Proof of Theorem 1.2

To prove this, we do not quite follow the techniques of the previous section; rather, we simply apply Lemma 2.7. Suppose that $\psi $ is a Dirichlet character modulo q. Then, we can write

$$ \begin{align*} S_{\psi,\chi,P}(N)=\sum_{k\leq q}\psi(k)\sum_{\substack{n\leq N\\ n\equiv k \pmod q}} \chi(\psi_{n}(P)). \end{align*} $$

The proof follows, immediately applying Lemma 2.7 to each of the q many inner sums.

4.2 Proof of Theorem 1.3

Let us first write

$$ \begin{align*}S_{\mu^2,\chi,P}(N) = \sum_{d\leq N^{1/2}} \mu(d)\sum_{n\leq N/d^2}\chi(\psi_{d^2n}(P)).\end{align*} $$

Then, we can split the sum as

(4.1) $$ \begin{align} S_{\mu^2,\chi,P}(N)=T_{1}+T_{2}, \end{align} $$

where we denote

$$ \begin{align*}T_{1}= \sum_{\substack{d\leq N^{1/2}\cr\gcd(d,R)\leq p^{\varepsilon/4}}} \mu(d)\sum_{n\leq N/d^2}\chi(\psi_{d^2n}(P)),\quad T_2=S_{\mu^2,\chi,P}(N)-T_1.\end{align*} $$

To estimate $T_1$ , first we split $T_1$ into $\log N$ many type I sums of the form

$$ \begin{align*}S(D,N)=\sum_{\substack{D\leq d<2D\\ \gcd(d,R)\leq p^{\varepsilon/4}}} \mu(d)\sum_{n\leq N/d^2}\chi(\psi_{d^2n}(P)).\end{align*} $$

As for any such d above, we have $\gcd (d^2,R)\leq p^{\varepsilon /2}$ , Lemma 2.8 implies that

(4.2) $$ \begin{align} |S(D,N)|\ll Dp^{1/12}R^{5/6}(\log R)(\log \log R)^{1/3}. \end{align} $$

Estimating rather trivially, we also have

(4.3) $$ \begin{align} |S(D,N)|\ll N/D. \end{align} $$

In particular, choosing $D_0$ such that

$$ \begin{align*}D_0^2p^{1/12}R^{5/6}(\log R)(\log \log R)^{1/3}=N,\end{align*} $$

and applying (4.2) for $D\le D_0$ , and (4.3) for $D\ge D_0$ , we finally derive

$$ \begin{align*}|T_1|\ll N^{1/2}p^{1/24}R^{5/12}(\log R)^{3/2}(\log \log R)^{1/6}.\end{align*} $$

On the other hand, we have

$$ \begin{align*} \begin{aligned} |T_2|&=\left|\sum_{\substack{d\leq N^{1/2}\cr\gcd(d,R)> p^{\varepsilon/4}}} \mu(d)\sum_{n\leq N/d^2}\chi(\psi_{d^2n}(P))\right|\le \sum_{\substack{d\leq N^{1/2}\cr\gcd(d,R)> p^{\varepsilon/4}}} \sum_{n\leq N/d^2}~1.\cr &\le N\sum_{\substack{d\le N^{1/2}\cr \gcd(d,R)>p^{\varepsilon/4}}}\frac{1}{d^2} + \sharp\,\{d\le N^{1/2}:\gcd(d,R)>p^{\varepsilon/4}\}. \end{aligned} \end{align*} $$

To bound the first term above, note that if $\gcd (d,R)>p^{\varepsilon /4}$ , then there exists a divisor $g\mid R$ with $g>p^{\varepsilon /4}$ such that $g\mid d$ . In particular, we have

$$\begin{align*}\sum_{\substack{d\le N^{1/2}\cr \gcd(d,R)>p^{\varepsilon/4}}}\frac{1}{d^2} \le \sum_{\substack{g\mid R\\ g>p^{\varepsilon/4}}} \sum_{\substack{d\le N^{1/2}\cr g\mid d}}\frac{1}{d^2} \ll \sum_{\substack{g\mid R\\ g>p^{\varepsilon/4}}}\frac{1}{g^2} \ll \frac{\tau(R)}{p^{\varepsilon/2}}, \end{align*}$$

where $\tau (R)=\tau _{1}(R)$ denotes the number of divisors of R.

On the other hand, we also have

$$ \begin{align*} \sharp\,\{d\le N^{1/2}:\gcd(d,R)>p^{\varepsilon/4}\} &\le \sum_{\substack{g\mid R\\ g>p^{\varepsilon/4}}} \sharp\,\{d\le N^{1/2}: g\mid d\}\ll \frac{N^{1/2}\tau(R)}{p^{\varepsilon/4}}. \end{align*} $$

Therefore, we derive

(4.4) $$ \begin{align} |T_2|\ll \frac{N\tau(R)}{p^{\varepsilon/2}}+\frac{N^{1/2}\tau(R)}{p^{\varepsilon/4}}. \end{align} $$

Note that if $N<p^{\varepsilon /2}$ , then the desired estimate at Theorem 1.3 holds trivially, as $R\ge p^{1/2}$ by (1.4). Hence, we may assume that $N\ge p^{\varepsilon /2}$ . Consequently, from (4.4), we obtain

$$ \begin{align*}|T_2|\ll Np^{-\varepsilon/2}\exp((\log 2+o(1))\log R/\log \log R),\end{align*} $$

where we use the well-known bound for $\tau (R)$ (see [Reference Hardy and Wright5, Theorem 317]). The proof now follows from (4.1).

5.1 Preliminaries

5.1.1 Smooth numbers in a smaller range

We first need the following estimates.

Lemma 5.1 Let $\alpha =\alpha (N,y)$ be as in (1.8). For any $1\le L\le N$ , we have

(5.1) $$ \begin{align} \Psi(N/L,y) \ll \frac{1}{L^{\alpha}} \Psi(N,y),\quad \mathrm{and} \end{align} $$

(5.2) $$ \begin{align} \Psi(L, y) \ll L^{\alpha}N^{o(1)}. \end{align} $$

Proof For (5.1), see [Reference Harper6, Section 2]. To prove (5.2), simply note that

(5.3) $$ \begin{align} \Psi(L, y) \ll \left(\frac{L}{N}\right)^{\alpha}\Psi(N,y)= L^{\alpha}N^{o(1)}, \end{align} $$

where, of course, we are using (1.7) for the last equality.

5.1.2 A decomposition of large smooth numbers

Let $L_0 \leq N$ be a parameter to be chosen later. Note that any $n\in {\mathcal S}(N,y)$ with $n> L_0$ can be uniquely written as

$$ \begin{align*}n = \ell m,\quad \mathrm{with}\quad L_0<\ell \le P(\ell)L_0 \quad \mathrm{and}\quad p(m) \geq P(\ell),\end{align*} $$

where $P(\ell )$ denotes the largest prime factor of $\ell $ and $p(m)$ denotes the smallest prime factor of m. Indeed, one can obtain such a decomposition by arranging all the prime factors of n as $p_1\le p_2\le \cdots \le p_K$ , and keep collecting the numbers $p_1,p_1p_2,\ldots $ , until we reach a number $\ell $ that just crosses $L_0$ . See also [Reference Shao, Shparlinski and Wijaya12, p. 9].

To show the uniqueness of the decomposition, write $n = p_1 \dots p_K$ with $p_1 \le p_2 \le \cdots \le p_K$ . Then $\ell $ must be of the form $p_1 \dots p_S$ and m of the form $p_{S+1} \dots p_K$ for some $S \le K$ . Now, the condition $L_0 \le \ell \le P(\ell )L_0$ implies

$$\begin{align*}L_0 \le p_1 \dots p_S \qquad \text{and} \qquad p_1 \dots p_{S-1} < L_0. \end{align*}$$

Hence, S is uniquely determined by $L_0$ , which proves the uniqueness of the decomposition.

Then, we can write

$$ \begin{align*}S_{\Psi_y,\chi,P}(P)=\sum_{\substack{L_0 < \ell \leq P(\ell) L_0 \cr \ell \in {\mathcal S}(N, y)}} \, \sum_{\substack{m \in {\mathcal S}(N/\ell, y) \cr p(m) \geq P(\ell)}} \chi(\psi_{\ell m}(P)) + O(L_0).\end{align*} $$

5.2 Concluding the proof

Note that the range of $\ell $ is given by $(L_0,yL_0]$ . After the dyadic partition of this range, we see that there is some $L \in (L_0, yL_0]$ such that

(5.4) $$ \begin{align} S_{\Psi_y,\chi,P}(P) \ll U \log N + L_0, \end{align} $$

where

$$ \begin{align*}U= \sum_{\substack{L < \ell \leq \min\left\{P(\ell)L_0, 2L\right\} \cr \ell \in {\mathcal S}(N, y)}} \, \sum_{\substack{m \in S(N/\ell, y) \cr p(m) \geq P(\ell)}} \chi(\psi_{\ell m}(P)).\end{align*} $$

We now argue exactly as in the proof of [Reference Shao, Shparlinski and Wijaya12, Theorem 1.3], and obtain

$$ \begin{align*} \begin{aligned} U^2 \ll & (\log N)^2 \Psi(L,y) \cr &\quad\quad\times \sum_{q \leq y} \, \sum_{m_1,m_2 \in {\mathcal S}(N/L,y)} \left| \sum_{L/q < \ell \le 2L/q} \chi(\psi_{q\ell m_1}(P)) \overline{\chi(\psi_{q\ell m_2}(P)} \right|. \end{aligned} \end{align*} $$

For the rest of the proof, we set $\alpha = \alpha (N,y)$ . Following the same argument as in [Reference Shao, Shparlinski and Wijaya12], the contribution $Y_1$ from the diagonal terms with $m_1=m_2$ is

$$ \begin{align*} Y_1&\ll (\log N)^2 \Psi(L,y) \sum_{q \leq y} \Psi(N/L, y) L/q\\ &\ll L N^{-\alpha} \Psi(N,y)^2 N^{o(1)}, \end{align*} $$

where the inequality above follows combining (5.1) and the first inequality in (5.3), and also recalling that $\log y\le \log N=N^{o(1)}$ .

To estimate the contribution $Y_2$ from the non-diagonal terms $m_1\neq m_2$ , we apply (2.8) instead of [Reference Shao, Shparlinski and Wijaya12, Lemma 2.2], and obtain

$$ \begin{align*} Y_2 & \ll (\log N)^2 \Psi(L,y)\cdot y \Psi(N/L, y)^2\cdot N/L\cdot p^{\frac{1}{12}}R^{5/6+o(1)} \cr & \ll y p^{1/12}R^{5/6} L^{-(\alpha+1)} \Psi(N,y)^2 N^{1+o(1)}, \end{align*} $$

where the last inequality follows by combining both parts of Lemma 5.1.

In particular, recalling that $L_0<L\le yL_0$ , we have

$$ \begin{align*}U^2& \ll Y_1 + Y_2 \le (LN^{-\alpha} + yp^{1/12}R^{5/6}L_0^{-(\alpha+1)}N) \Psi(N,y)^2 N^{o(1)} \cr & \le y(L_0N^{-\alpha} + p^{1/12}R^{5/6}L_0^{-(\alpha+1)}N) \Psi(N,y)^2 N^{o(1)}. \end{align*} $$

Choosing $L_0 = p^{\frac {1}{12(\alpha +2)}} R^{\frac {5}{6(\alpha +2)}}N^{\frac {\alpha +1}{\alpha +2}}$ , we derive

$$ \begin{align*}U \le y^{1/2} p^{\frac{1}{24(\alpha+2)}} R^{\frac{5}{12(\alpha+2)}}N^{\frac{1-(\alpha^2+\alpha)}{2(\alpha+2)}} \Psi(N,y) N^{o(1)}.\end{align*} $$

Hence, (5.4) gives

$$ \begin{align*} S_{f_{y},\chi,P}(N)&\ll y^{1/2} p^{\frac{1}{24(\alpha+2)}} R^{\frac{5}{12(\alpha+2)}}N^{\frac{1-(\alpha^2+\alpha)}{2(\alpha+2)}} \Psi(N,y) N^{o(1)}\cr &\qquad\qquad\qquad\qquad+p^{\frac{1}{12(\alpha+2)}} R^{\frac{5}{6(\alpha+2)}}N^{\frac{\alpha+1}{\alpha+2}}. \end{align*} $$

Setting $\gamma =\frac {\alpha ^2+\alpha -1}{2(\alpha +2)}$ , we have $\alpha -2\gamma =\frac {\alpha +1}{\alpha +2}$ , and hence we finally deduce

$$ \begin{align*} |S_{f_{y},\chi,P}(N)| &\ll y^{1/2} p^{\frac{1}{24(\alpha+2)}} R^{\frac{5}{12(\alpha+2)}} N^{-\gamma} \Psi(N,y) N^{o(1)} \cr &\qquad\qquad\qquad\qquad + p^{\frac{1}{12(\alpha+2)}} R^{\frac{5}{6(\alpha+2)}} N^{\alpha - 2\gamma}. \end{align*} $$

If $p^{\frac {1}{24(\alpha +2)}} R^{\frac {5}{12(\alpha +2)}}> N^{\gamma }$ , then the stated bound at (1.9) is trivial. So, let us assume that $p^{\frac {1}{24(\alpha +2)}} R^{\frac {5}{12(\alpha +2)}}\le N^{\gamma }$ . In this case, the second term above is dominated by the first term because $\Psi (N,y)N^{o(1)}=N^{\alpha }$ , and the proof concludes.

Remark 5.2 Following the arguments in Section 3.3, the main bottleneck in improving Theorem 1.1 lies in the estimation of $U_0$ in (3.2). Note that the integers in the set ${\mathcal A}_0(N,I)$ can be written as products of the form

$$\begin{align*}(\textit{x}-{\rm smooth}) \times (\textit{y}-{\rm rough}). \end{align*}$$

This structure suggests that one might attempt to adapt the approach used in the proof of Theorem 1.4. However, as one may anticipate, complications arise for the integers whose x-smooth parts are small.

Remark 5.3 Consider the classical function $r_0(n)$ , the characteristic function of integers that are sums of two squares. It is well-known (see [Reference Tenenbaum17, p. 98, Equation (4.90)], for instance) that

$$\begin{align*}\sum_{n \leq N} r_0(n) \ll \frac{N}{(\log N)^{1/2}}. \end{align*}$$

Following the argument used in the proof of Theorem 1.1, one can easily deduce that

(5.5) $$ \begin{align} |S_{r_0,\chi,P}| \ll \frac{N}{(\log N)^{1/2}} \cdot \frac{\log \log R}{\log R}, \end{align} $$

where R and $\varepsilon $ be as in (1.4), and the implied constant in (5.5) may depend on d and $\varepsilon $ .

Moreover, the same saving compared to the trivial estimate likely holds for many other multiplicative functions, supported sparsely on the integers and taking values in the intervals $[0,1]$ .