Proof. Recall that $y_{{\mathbf r}} = F\left (\frac {\log r_1}{\log R}, \dots , \frac {\log r_K}{\log R}\right )$ , where $F(t_1, \dots , t_K) := \prod _{i=1}^K g(Kt_i)$ and $g(t) = \frac 1{1+t}$ for $t \le 1$ and $g(t) = 0$ otherwise. Given $u,v \ge 0$ with $|u-v| < \varepsilon $ , we have

$$ \begin{align*} \frac 1{1+Ku} = \frac{1+O(K\varepsilon)}{1+Kv}. \end{align*} $$

Let $u_i = \log r_i/\log R$ , $v_i = \log s_i/\log R$ and $\varepsilon _i = v_i-u_i$ . In part (i), $\varepsilon _i = 0$ for $i \ne j$ and $\varepsilon _j = \log A/\log R$ . Thus,

$$ \begin{align*} \frac 1{1+Kv_j} = \frac{1+O\left(\tfrac{K\log A}{\log R}\right)}{1+Ku_j}. \end{align*} $$

Multiplying by $\prod _{i\ne j} 1/(1+Ku_i)$ gives the result for (i).

Now consider part (ii), and let $\mathbf t$ be the vector with $t_i = [r_i,s_i]$ . Applying part (i) to each component in turn implies that

$$ \begin{align*} y_{\mathbf s}=y_{\mathbf t} + O\left(Ky_{{\mathbf s}} \sum_{i=1}^K \frac{\log\left([r_i,s_i]\right)/s_i}{\log R}\right) = y_{\mathbf t} + O\left(Ky_{{\mathbf s}}\frac{\log A}{\log R}\right). \end{align*} $$

The same holds for ${\mathbf r}$ and $\mathbf t$ , which implies (ii).

Proof. This is [Reference Goldston, Graham, Pintz and Yılıd rım6, Lemma 4] with $\kappa = \frac {1}{2}$ .

Proof. We would like to apply Lemma 6 for each variable. However, the variables $e_i$ are not independent, since $\mathbf e \in \mathcal {D}_K$ implies $(e_i,e_j) = 1$ for $i\neq j$ . Our first task is to decouple these variables with a negligible penalty.

Denote the sum evaluated in the lemma by

$$ \begin{align*}S = \sum_{\substack{\mathbf e \in \mathcal{D}_K \\ (e,Q) = 1}} \frac{\mu^2(e)}{f(e)} \prod_{i=1}^K G\left(\frac{\log e_i}{\log R}\right). \end{align*} $$

If $p|(e_r,e_j)$ for some $1\leq r< j \leq K$ , then we must have $p> D_0$ since $(e,Q) = 1$ . Thus, using also the fact that G is decreasing, we have that

$$ \begin{align*} \sum_{\substack{\mathbf e \in \mathbb N_K \\ (e,Q) = 1 \\ p\mid(e_i,e_j)}} &\prod_{j=1}^K \frac{\mu^2(e_j)}{f(e_j)} \prod_{i=1}^K G\left(\frac{\log e_i}{\log R}\right) \\ &\ll \frac{1}{f(p)^2} \sum_{\substack{\mathbf e \in \mathbb N_K \\ (e,Q) = 1}} \prod_{j=1}^K\frac{\mu^2(e_j)}{f(e_j)} G\left(\frac{\log e_r + \log p}{\log R}\right) G\left(\frac{\log e_j + \log p}{\log R}\right) \prod_{\substack{i=1 \\ i \neq r,j}}^K G\left(\frac{\log e_i}{\log R}\right) \\ & \ll (p + O(K))^{-2}S. \end{align*} $$

Writing

$$ \begin{align*}S' = \sum_{\substack{\mathbf e \in\mathbb{N}^ K \\ (e,Q) = 1}} \prod_{j=1}^K \frac{\mu^2(e_j)}{f(e_j)} \prod_{i=1}^K G\left(\frac{\log e_i}{\log R}\right) = \Big(\sum_{\substack{n \in \mathbb N \\ (n,Q) = 1}} \frac{\mu^2(n)}{f(n)} G\left(\frac{\log n}{\log R}\right)\Big)^K , \end{align*} $$

it follows that

$$ \begin{align*}S - S' \ll S \binom{K}{2}\sum_{p> D_0}\frac{1}{(p+O(K))^2} \ll_K \frac{S}{D_0}. \end{align*} $$

We now consider $S'$ in place of S and apply Lemma 6, where we take $g(p) = \frac 1{f(p)}$ and $\gamma (p) = \frac {p}{f(p)+1}$ for $p \nmid Q$ and $g(p) = \gamma (p) = 0$ for $p|Q$ . For this, we need a bound on the constant L from Lemma 6. If we did not have the restriction $g(p) = \gamma (p) = 0$ for $p|Q$ , then using the prime number theorem in arithmetic progressions, we could take L to be a constant, since

$$ \begin{align*}\sum_{\substack{w\leq p < z \\ p\equiv 3 \bmod 4}}\frac{\log p}{p + O(K)} - \frac{1}{2}\log{\left(\frac{z}{w}\right)} \ll_K 1. \end{align*} $$

It follows that in our case, we can choose L satisfying

$$ \begin{align*}L \ll_K 1 + \sum_{\substack{p \leq D_0 \\ p\equiv 3 \bmod 4}}\frac{\log p }{p} + \sum_{p \mid \alpha }\frac{\log p}{p}. \end{align*} $$

The first sum is $(1+o(1))\frac {1}{2}\log {D_0}$ . For the second sum, we have the bound

$$ \begin{align*}\sum_{p|\alpha} \frac{\log p}{p} \ll_{K} \log\log R \end{align*} $$

since $\alpha = R^{O(K)}$ and the sum is dominated by taking the smallest possible primes. This implies that

$$ \begin{align*}L \ll_{K} \frac{1}{2}\log D_0 + \log\log R \ll_{K} \log\log R. \end{align*} $$

We will write

$$ \begin{align*}\Omega_G =G_{\mathrm{max}}\cdot \left(\int_{t\geq 0} G(t)\frac{dt}{\sqrt{t}}\right)^{-1}, \end{align*} $$

where $G_{\mathrm {max}}$ is defined as in Lemma 6. Applying Lemma 6 successively K times, we get

$$ \begin{align*} S' &= \frac{c_\gamma^K}{\Gamma(1/2)^K}(\log R)^{K/2} \left(\int_{t_1, \dots, t_K \ge 0} \prod_{i=1}^K G(t_i) \frac{\mathrm{d}t_i}{\sqrt{t_i}}\right) \left(1 + \sum_{\ell = 1}^K \binom{K}{\ell} O_{K}\left( \frac{\Omega_G \log\log R}{\log R}\right)^\ell \right) \\ &= \frac{c_\gamma^K}{\Gamma(1/2)^K}(\log R)^{K/2} \left(\int_{t_1, \dots, t_K \ge 0} \prod_{i=1}^K G(t_i) \frac{\mathrm{d}t_i}{\sqrt{t_i}}\right) \left( 1 + O_{K,G}\left(\frac{ \log\log R}{\log R}\right)\right), \end{align*} $$

where $c_\gamma $ is given in Lemma 6 and satisfies

$$ \begin{align*}c_\gamma = (1 + o(1)) \prod_{p\mid \alpha}\left(1 - \frac{1}{p}\right)^{\frac{1}{2}} \prod_{p\leq D_0} \left(1 - \frac{1}{p}\right)^{\frac{1}{2}} = (1 + o(1)) \left(\frac{\phi(\alpha)}{\alpha}\right)^{\frac{1}{2}} \frac{e^{-\gamma/2}}{(\log D_0)^{\frac{1}{2}}} \end{align*} $$

Combining our estimates completes the proof.

Proof. By our choice of $y_{\mathbf r}$ , for any $\mathbf d \in \mathcal D_K$ , we have

(36) $$ \begin{align} |\lambda_{\mathbf d}| = \prod_{i=1}^K d_i \sum_{\substack{\mathbf r \in \mathcal D_K \\ \mathbf d|\mathbf r}} \frac{y_{\mathbf r}}{\phi(r)} = \frac{d}{\phi(d)}\sum_{\mathbf d|\mathbf r \in \mathcal D_K} \frac{F\left(\frac{\log r_1}{\log R},\dots, \frac{\log r_K}{\log R}\right)}{\phi(r/d)}. \end{align} $$

Since F is decreasing in each argument, we can bound $|\lambda _{\mathbf d}|$ above by replacing each $\frac {\log r_i}{\log R}$ in the argument of F with $\sigma _i = \frac {(\log r_i/d_i)}{\log R}$ .

We now apply Lemma 7 with $Q = dq_3W$ , which gives

$$ \begin{align*} &\sum_{\mathbf d|\mathbf r \in \mathcal D_K} \frac{F(\sigma_1, \dots, \sigma_K)}{\phi(r/d)} = \sum_{\mathbf d|\mathbf r\in \mathcal D_K} \frac 1{\phi(r/d)} \prod_{i=1}^K g(K\sigma_i) \\& \qquad \sim \left(\frac{\phi(d)}{d}\right)^{\frac{K}{2}} \left(\frac{e^{-\gamma /2}}{\Gamma(1/2)}\right)^K \left(\frac{\log R}{\log D_0}\right)^{K/2} \left(\int_{t_1, \dots, t_K \ge 0} \prod_{i=1}^K g(K t_i) \frac{\mathrm{d}t_i}{\sqrt{t_i}}\right). \end{align*} $$

Evaluating the integral gives

$$ \begin{align*} \int_{t_1, \dots, t_K \ge 0} \prod_{i=1}^K g(Kt_i) \frac{\mathrm dt_i}{\sqrt{t_i}} = \left(\frac{\pi}{2\sqrt{K}}\right)^K. \end{align*} $$

Substituting this expression back into (36), we get that

$$ \begin{align*} |\lambda_{\mathbf d}| \ll_K \left(\frac{\phi(d)}{d}\right)^{\frac{K}{2}-1} \left(\frac{\log R}{\log D_0}\right)^{K/2} \ll \left(\frac{\log R}{\log D_0}\right)^{K/2} \end{align*} $$

for $K\geq 4$ . This completes the proof of (34).

For the second claim, recall that

$$ \begin{align*} w_n(\mathcal L) &= \Big(\sum_{\substack{\mathbf d \in \mathcal D_K \\ d_i\mid\ell_i(n)}} \lambda_{\mathbf d}\Big)^2 \ll_K \left(\frac{\log R}{\log D_0}\right)^{K} \#\{\mathbf d \in \mathcal D_K : d_i\mid \ell_i(n)\}^2 \\ &\ll_K \left(\frac{\log R}{\log D_0}\right)^{K} \prod_{i=1}^K \prod_{\substack{p\mid \ell_i(n) \\ p \equiv 3 \bmod 4}}4, \end{align*} $$

as desired.

For the third claim, note that since $\lambda _{\mathbf d}$ is supported on $d=d_1 \cdots d_K < R$ , we have

$$ \begin{align*} w_n(\mathcal L) &\ll_K \left(\frac{\log R}{\log D_0}\right)^{K} \left(\sum_{d_1\cdots d_K < R} 1\right)^2 \ll_K R^{2+o(1)} \left(\sum_{d_1 \cdots d_K < R} \frac 1{d_1 \cdots d_K}\right)^2 \ll_K R^{2+o(1)}. \end{align*} $$

Proof. The proof of this lemma is for the most part identical to the proof in Appendix B of [Reference McGrath15], so here we will restrict ourselves to highlighting the differences in the argument.

In general, the application of the Selberg–Delange method is identical to that described in [Reference McGrath15], with the same arguments applying for bounding, for example, relevant analytic functions; the only change that need be made to McGrath’s arguments is replacing W by an arbitrary Q (which must divide the W that McGrath uses) everywhere. Thus, by following McGrath’s proof, we get that, for

$$ \begin{align*} K_1(s)^2 = \left(1-\frac 1{2^s}\right)^{-1}\prod_{p\equiv 3 \bmod 4}\left(1-\frac 1{p^{2s}}\right)^{-1} \end{align*} $$

and

$$ \begin{align*} G_1(Q,s) = \prod_{\substack{p|Q \\ p \equiv 1 \bmod 4 }} \left(1-\frac 1{p^s}\right)^{-1}, \end{align*} $$

we have

$$ \begin{align*} X_{x,Q} = \frac{K_1(1)G_1(Q)}{\Gamma(3/2)\sqrt{L(1,\chi_4)}}(\log v)^{\tfrac 12} + O_\varepsilon\left(\frac{(\log x)^{\varepsilon}}{(\log v)^{\tfrac 12}}\right), \end{align*} $$

which simplifies to the desired expression.

In the same manner, expressions for $Z^{(1)}_{x,Q}$ and $Z^{(2)}_{x,Q}$ can be derived.

Proof. We have

$$ \begin{align*} S_1(\nu_0) &= \sum_{\substack{x\leq n \leq 2x \\ n\equiv 1\bmod 4 \\ n \equiv \nu_0 \bmod W}}w_n(\mathcal L) = \sum_{\substack{x\leq n \leq 2x \\ n\equiv 1\bmod 4 \\ n \equiv \nu_0 \bmod W}} \left(\sum_{\mathbf d \in \mathcal D_{K}} \lambda_{\mathbf d}\right)^2. \\ \end{align*} $$

By expanding the square and swapping the order of summation, we get that

$$ \begin{align*} S_1(\nu_0) &= \sum_{\substack{\mathbf d, \mathbf e \in \mathcal D_K}} \lambda_{\mathbf d}\lambda_{\mathbf e} \sum_{\substack{x \leq n \leq 2x \\ qn+a_i+qb_i \equiv 0 \bmod{[d_i,e_i]} \\ n\equiv 1 \bmod 4 \\ n \equiv \nu_0 \bmod W}} 1 \\ &= \sum_{\substack{ \mathbf d, \mathbf e \in \mathcal D_K}}\lambda_{\mathbf d}\lambda_{\mathbf e} \left(\frac{x}{4W}\prod_{i=1}^{K}\frac{1}{[d_i,e_i]} + O(1) \right) \\ &= \frac{x}{4W} \sum_{\mathbf d, \mathbf e \in \mathcal D_{K}} \frac{\lambda_{\mathbf d}\lambda_{\mathbf e}}{[\mathbf d, \mathbf e]} +O\left( \sum_{\substack{\mathbf d, \mathbf e \in \mathcal D_K}}\lambda_{\mathbf d}\lambda_{\mathbf e} \right), \end{align*} $$

where the second line follows from the first by the Chinese remainder theorem and the observation that $4$ , W and the $[d_i,e_i]$ are all pairwise relatively prime. In the notation of Lemma 9, the sum in the main term is precisely the sum $S_{\emptyset } = S_{\emptyset ,1,1,m}$ , which by Lemma 9 is equal to $B^KL_K(F)$ .

It remains only to bound the error term; We have

$$ \begin{align*}\left| \sum_{\mathbf d, \mathbf e\in\mathcal D_K}\lambda_{\mathbf d}\lambda_{\mathbf e} \right| \ll \lambda_{\mathrm{max}}^2 |\mathcal D_K|^2. \end{align*} $$

We use (34) to bound $\lambda _{\text {max}}$ , and the bound

(37) $$ \begin{align} |\mathcal D_K| \leq \sum_{n\leq R}\tau_K(n) \ll_\epsilon R^{1 + \epsilon}. \end{align} $$

Since $R = x^{\theta _2/2}$ with $\theta _2<\frac {1}{18}$ , we conclude

$$ \begin{align*}\left| \sum_{\mathbf d, \mathbf e\in\mathcal D_K}\lambda_{\mathbf d}\lambda_{\mathbf e} \right| \ll_{K,\epsilon} R^{2+\epsilon} = x^{\theta_2 + \epsilon} \ll x^{1/18 + \varepsilon}, \end{align*} $$

which is negligible. This completes the proof.

Proof. By definition of $\rho $ , we have that

$$ \begin{align*} \rho(\ell_m(n)) = \frac{r_2(\ell_m(n))}{\log v}\sum_{\substack{a\mid \ell_m(n) \\ p\mid a \Rightarrow p\equiv 1 \bmod 4 \\ a\leq v}} \frac{\mu(a)}{g_2(a)} \log \left( \frac{v}{a}\right). \end{align*} $$

Note that since $a|\ell _m(n)$ , which is relatively prime to q, we must have $(a,q) = 1$ for any a in the definition of $\rho (\ell _m(n))$ . Since a is only divisible by primes that are $1$ mod $4$ , the only nontrivial constraint on a is that $(a,q_1) = 1$ . By expanding the definitions of $w_n(\mathcal L)$ and $\rho $ in the expression for $S_2^{(m)}(\nu _0)$ and changing the order of summation, we get

(38) $$ \begin{align} S_2^{(m)}(\nu_0) = \frac 1{\log v} \sum_{\mathbf d, \mathbf e \in \mathcal D_K} \lambda_{\mathbf d}\lambda_{\mathbf e} \sum_{\substack{a \le v \\ (a,q_1)=1 \\ p|a \Rightarrow p \equiv 1 \bmod 4}} \frac{\mu(a)}{g_2(a)} \log \frac{v}{a} \sum_{\substack{x < n \le 2x \\ n \equiv 1 \bmod 4 \\ n\equiv \nu_0\bmod W \\ [d_i,e_i]\mid \ell_i(n) \forall i \\ a \mid \ell_m(n) }} r_2(\ell_m(n)). \end{align} $$

The parameter a is supported on integers whose prime factors are all $1 \bmod 4$ , whereas by the definition of $\mathcal D_K$ , each $[d_i,e_i]$ is only divisible by primes that are $3 \bmod 4$ . Thus, the $[d_i,e_i]$ ’s and a are pairwise coprime. Also, all prime factors of each $[d_i,e_i]$ are larger than $D_0$ , whereas each $qb_i < D_0$ , so each $[d_i,e_i]$ is coprime to each $qb_i$ .

Thus, by Lemma 10, (applied with $r = qW\prod _{i\ne m} [d_i,e_i]$ and with $d = a[d_m,e_m]$ ), we get

(39) $$ \begin{align} \sum_{\substack{x < n \le 2x \\ n \equiv 1 \bmod 4 \\ n \equiv \nu_0 \bmod W\\ [d_i,e_i]\mid \ell_i(n) \forall i \\ a \mid \ell_m(n) }} r_2(\ell_m(n)) &= \frac{ g_1\left( qW \prod_{i\neq m}[d_i,e_i]\right) g_2(a[d_m,e_m]) }{2 qW a \prod_i[d_i,e_i]} \pi qx \notag \\ &\quad + O_\epsilon\left(\left((WR^3v)^{1/2} + x^{1/3}\right)v^{1/2}x^{\epsilon}\right).\end{align} $$

Recall that $v = x^{\theta _1}$ , $R^2 = x^{\theta _2}$ , $\theta _1 + \theta _2 < 1/18$ , and $W \ll _\varepsilon x^{\varepsilon }$ for all $\varepsilon> 0$ . Thus,

$$ \begin{align*} & S_2^{(m)}(\nu_0)\\ &\quad = \frac {\pi x}{\log v} \sum_{\mathbf d, \mathbf e \in \mathcal D_K} \lambda_{\mathbf d}\lambda_{\mathbf e} \sum_{\substack{a \le v \\ (a,q_1) = 1 \\ p|a \Rightarrow p \equiv 1 \bmod 4}} \frac{\mu(a)}{g_2(a)} \log \frac{v}{a} \frac{ g_1\left(q W \prod_{i\neq m}[d_i,e_i]\right) g_2(a[d_m,e_m]) }{2 W a \prod_i[d_i,e_i]} + E, \end{align*} $$

where

$$ \begin{align*} |E| &\ll_\varepsilon \frac 1{\log v} \left|\sum_{\mathbf d, \mathbf e \in \mathcal D_K} \lambda_{\mathbf d}\lambda_{\mathbf e} \right| \sum_{\substack{a \le v \\ p|a \Rightarrow p \equiv 1 \bmod 4}} \frac{1}{g_2(a)} \log \frac{v}{a} x^{\frac{1}{3} + (\theta_1 + \theta_2)/2 + \epsilon} \\ & \ll \lambda_{\mathrm{max}}^2 |\mathcal D_K|^2 v x^{\frac{1}{3} + (\theta_1 + \theta_2)/2 + \epsilon} \ll_{\varepsilon'} x^{\frac{1}{3} + (\theta_1 + \theta_2)\frac{3}{2} + \epsilon'} \ll x^{\frac{5}{12} + \epsilon'}, \end{align*} $$

where we used use (34) to bound the $\lambda _{\mathrm {max}}$ and (37) for $|\mathcal D_K|$ .

It remains to consider the main term, which is given by

$$ \begin{align*} &\frac { \pi x}{\log v} \sum_{\mathbf d, \mathbf e \in \mathcal D_K} \lambda_{\mathbf d}\lambda_{\mathbf e} \sum_{\substack{a \le v \\ (a,q_1) = 1 \\ p|a \Rightarrow p \equiv 1 \bmod 4}} \frac{\mu(a)}{g_2(a)} \log\left( \frac{v}{a} \right) \frac{ g_1\left( qW \prod_{i\neq m}[d_i,e_i]\right) g_2(a[d_m,e_m]) } {2 W a \prod_i[d_i,e_i]} \\ = &\frac{\pi x g_1(qW)}{2W\log v} \sum_{\mathbf d, \mathbf e \in \mathcal D_K} \lambda_{\mathbf d} \lambda_{\mathbf e} \frac{g_1\left(\prod_{i\ne m}[d_i,e_i]\right) g_2([d_m,e_m])}{\prod_i[d_i,e_i]} \sum_{\substack{a \le v \\ (a,q_1) = 1 \\ p|a \Rightarrow p \equiv 1 \bmod 4}} \frac{\mu(a)}{a} \log \frac va. \end{align*} $$

The inside sum is exactly $X_{x,q_1}$ , as defined in Lemma 13. Also, all primes dividing $[d_m,e_m]$ are congruent to $3$ mod $4$ , so $g_2([d_m,e_m]) = \frac {1}{[d_m,e_m]}$ . We can thus rewrite the main term as

(40) $$ \begin{align} \frac{X_{x,q_1} \pi x g_1(qW) }{2 W \log v} \sum_{\mathbf d, \mathbf e \in \mathcal D_K} \lambda_{\mathbf d}\lambda_{\mathbf e} \prod_{i \ne m}\left(\frac{g_1([d_i,e_i])}{[d_i,e_i]}\right) \frac 1{[d_m,e_m]^2}. \end{align} $$

By Lemma 13, $X_{x,q_1} = (1+o(1))\frac {8A\sqrt {\log v}}{g_1(q_1)\pi }$ , and by Lemma 9 applied with $f(p) = \frac {g_1(p)}{p}$ and $g(p) = 1/p^2$ ,

$$ \begin{align*} \sum_{\mathbf d, \mathbf e \in \mathcal D_K} \lambda_{\mathbf d}\lambda_{\mathbf e} \prod_{i \ne m}\left(\frac{g_1([d_i,e_i])}{[d_i,e_i]}\right)\frac 1{[d_m,e_m]^2} = S_{\{m\}}=B^{K+1}L_{K;m}(F)(1+o(1)). \end{align*} $$

Then (40) can be rewritten as

$$ \begin{align*}\frac{\pi x g_1(qW)}{2 W \log v}\cdot \frac{8A\sqrt{\log v}}{g_1(q_1)\pi}B^{K+1}L_{K;m}(F)(1+o(1)). \end{align*} $$

By definition of $g_1$ ,

(41) $$ \begin{align} \frac{g_1(q_3W) \phi(q_3W)}{q_3W} = \frac{1}{2A^2} + O\left(\frac{1}{D_0}\right). \end{align} $$

Recalling also that $B = \frac {2A}{\pi }\frac {\phi (q_3W)}{q_3W}(\log R)^{\tfrac 12}$ , we get

$$ \begin{align*} S_2^{(m)}(\nu_0) = (1+o(1))\frac{4\sqrt{\frac{\log R}{\log v}} B^K x}{\pi W} L_{K;m}(F), \end{align*} $$

which, along with the fact from Lemma 4 that $L_{K;m}(F) = \frac {\pi ^2}{\pi +2} \frac {L_K(F)}{\sqrt K}$ , completes the proof.

Proof. We begin by expanding the definition of $w_n(\mathcal L)$ to rewrite $S_3^{(m_1,m_2)}(\nu _0)$ as

$$ \begin{align*} S_3^{(m_1,m_2)}(\nu_0) = \sum_{\mathbf d, \mathbf e \in \mathcal D_K} \lambda_{\mathbf d} \lambda_{\mathbf e} \sum_{\substack{x < n \le 2x \\ n \equiv 1 \bmod 4 \\ n \equiv \nu_0 \bmod W \\ [d_i,e_i]|\ell_i(n) \forall i}} \rho(\ell_{m_1}(n))\rho(\ell_{m_2}(n)). \end{align*} $$

Upon expanding the definition of $\rho $ and changing the order of summation, the inside sum over n is equal to

$$ \begin{align*} \frac 1{\log^2 v} & \sum_{\substack{a,b \le v \\ (a,q_1) = (b,q_1) = 1 \\ p|a,b \Rightarrow p \equiv 1 \bmod 4}} \frac{\mu(a)\mu(b)}{g_2(a)g_2(b)} \log \frac va \log \frac vb \\ &\quad \times\sum_{\substack{x < n \le 2x \\ n \equiv 1 \bmod 4 \\ n \equiv \nu_0 \bmod W \\ [d_i,e_i]|\ell_i(n) \forall i \\ a|(qn+a_{m_1} + qb_{m_1}) \\ b|(qn+a_{m_2} + qb_{m_2})}} r_2(qn+a_{m_1} + qb_{m_1})r_2(qn+a_{m_2} + qb_{m_2}). \end{align*} $$

Since $\ell _{m_1}(n)$ and $\ell _{m_2}(n)$ are always relatively prime to q, we must also always have $(a,q) = (b,q) = 1$ ; since a and b are only divisible by primes congruent to $1$ mod $4$ , this is equivalent to the constraint that $(a,q_1) = (b,q_1) = 1$ . For the inside sum to be nonzero, $[d_1, e_1], \dots , [d_K,e_K],$ and W must all be pairwise coprime, and each of these must be coprime to both a and b. Moreover, if any prime p divides $(a,b)$ , then $p|q(b_{m_2}-b_{m_1})$ . We thus have

$$ \begin{align*} S_3^{(m_1,m_2)}(\nu_0) &= \sum_{\mathbf d, \mathbf e \in \mathcal D_K} \frac {\lambda_{\mathbf d}\lambda_{\mathbf e}}{\log^2 v} \sum_{\substack{a,b \le v \\ (a,q_1) = (b,q_1) = 1 \\ p|a,b \Rightarrow p \equiv 1 \bmod 4}} \frac{\mu(a)\mu(b)}{g_2(a)g_2(b)} \log \frac va \log \frac vb \\ &\quad \times\sum_{\substack{\ell_{m_1}(x)< n \le \ell_{m_1}(2x) \\ n \equiv 1 \bmod 4 \\ n \equiv \alpha \bmod qW\prod_{i\neq m_1,m_2}[d_i,e_i] \\ a[d_{m_1},e_{m_1}]|n \\ b[d_{m_2},e_{m_2}]|n+h}} r_2(n)r_2(n+h), \end{align*} $$

where $\alpha $ is relatively prime to $qW\prod _{i\neq m_1,m_2}[d_i,e_i]$ and $h = q(b_{m_2} - b_{m_1})$ .

We now apply Lemma 11 to estimate the inner sum, taking $r = qW\prod _{i\ne m_1,m_2}[d_i,e_i]$ , $c_1 = a[d_{m_1},e_{m_1}]$ and $c_2=b[d_{m_2},e_{m_2}]$ . Note that Lemma 11 does not require that $c_1$ and $c_2$ be relatively prime, but that the main term is $0$ unless $(c_1,c_2)|h$ . Thus,

$$ \begin{align*} & S_3^{(m_1,m_2)}(\nu_0) = \sum_{\mathbf d, \mathbf e \in \mathcal D_K} \frac {\lambda_{\mathbf d} \lambda_{\mathbf e}}{\log^2 v} \sum_{\substack{a,b \le v \\ (a,q_1) = (b,q_1) = 1 \\ p|a,b \Rightarrow p \equiv 1 \bmod 4}} \frac{\mu(a)\mu(b)}{g_2(a)g_2(b)} \log \frac va \log \frac vb \\ &\quad \times \Bigg[ \pi^2 q x \frac{g_1^2\left(qW\prod_{i\neq m_1,m_2}[d_i,e_i]\right)} {qW\prod_{i\neq m_1,m_2}[d_i,e_i]} \frac{g_2(a[d_{m_1},e_{m_1}])g_2(b[d_{m_2},e_{m_2}])}{a[d_{m_1},e_{m_1}] b [d_{m_2},e_{m_2}] } \\ &\quad \times \sum_{\left(t, 2qW\prod_{i\neq m_1,m_2}[d_i,e_i]\right) = 1}\frac{c_t(h)}{t^2}\frac{(a,t)(b,t)}{\Psi(a,t)\Psi(b,t)} \prod_{i=1,2} \frac{\left([d_{m_i},e_{m_i}],t\right) \chi\left(([d_{m_i},e_{m_i}]^2,t)\right) }{\Psi\left([d_{m_i},e_{m_i}],t\right)} \\ &\quad + O\left(x^{\frac{5}{6} + \epsilon}a^{\frac{1}{2}}b^{\frac{1}{2}} [d_{m_1},e_{m_1}]^{\frac{1}{2}}[d_{m_2},e_{m_2}]^{\frac{1}{2}} \right) \\ &\quad + O\Big( x^{\frac{3}{4} + \epsilon} \Big(qW\prod_{i\neq m_1,m_2}[d_i,e_i]\Big)^{\frac{1}{2}} a b [d_{m_1},e_{m_1}][d_{m_2},e_{m_2}] \Big) \Bigg]. \end{align*} $$

Taking absolute values and noting that $d_{m_i},e_{m_i}\ll R^{\frac {1}{K}}$ , the first error term is bounded by

$$ \begin{align*}\lambda_{\mathrm{max}}^2 |\mathcal D_K|^2 \left( x^{\frac{5}{6} + \epsilon} v^{\frac{3}{2}} R^{\frac{1}{K}} + x^{\frac{3}{4} + \epsilon} v^{2} R^{2}\right) \ll \lambda_{\mathrm{max}}^2 |\mathcal D_K|^2 x^{\frac{5}{6}}v^{2} R^{2}. \end{align*} $$

Using (34) to bound $\lambda _{\mathrm {max}}$ , (37) for $|\mathcal D_K|$ , and the fact that $\theta _1 + \theta _2 < \frac {1}{18}$ , we get that the error term is bounded by

$$ \begin{align*}\ll_{\epsilon'} x^{\frac{5}{6} + 2(\theta_1 + \theta_2) + \epsilon'} \ll x^{\frac{17}{18} + \epsilon'}, \end{align*} $$

which is negligible.

We return to the main term which, after some simplification, and recalling that $g_2(p) = \frac {1}{p}$ for $p\equiv 3 \bmod {4}$ , becomes

(42) $$ \begin{align} \begin{aligned} S_3^{(m_1,m_2)}(\nu_0) &\sim \frac{\pi^2 x g_1^2\left(qW\right)}{W \log^2 v} \sum_{\mathbf d, \mathbf e \in \mathcal D_K} \lambda_{\mathbf d} \lambda_{\mathbf e} \frac{ \prod_{i\neq m_1,m_2}g_1^2\left([d_i,e_i]\right) }{ [d_{m_1},e_{m_1}]^2 [d_{m_2},e_{m_2}]^2 \prod_{i\neq m_1,m_2}[d_i,e_i] } \\ &\times \sum_{\substack{a,b \le v \\ (a,q_1) = (b,q_1) = 1 \\ p|a,b \Rightarrow p \equiv 1 \bmod 4}} \frac{\mu(a)\mu(b)}{a b } \log \frac va \log \frac vb \sum_{\left(t, 2qW\prod_{i\neq m_1,m_2}[d_i,e_i]\right) = 1}\frac{c_t(h)}{t^2} \\ &\times \prod_{i=1,2} \frac{\left([d_{m_i},e_{m_i}],t\right) \chi\left(([d_{m_i},e_{m_i}]^2,t)\right) }{\Psi\left([d_{m_i},e_{m_i}],t\right) } \frac{(a,t)(b,t)}{\Psi(a,t)\Psi(b,t)}. \end{aligned} \end{align} $$

The sum over t is multiplicative, and can thus be written as a product $\Sigma _1 \times \Sigma _3$ , where $\Sigma _1$ ranges over integers divisible only by primes congruent to $1$ mod $4$ and $\Sigma _3$ ranges over integers divisible only by primes congruent to $3$ mod $4$ . That is,

$$ \begin{align*}\Sigma_3 = \sum_{\substack{t\\ p\mid t \Rightarrow p\equiv 3 \bmod 4 \\ (t,q_3W \prod_{i\neq m_1,m_2}[d_i,e_i])=1}} \frac{c_t(h)}{t^2}\prod_{i=1,2} \frac{\left([d_{m_i},e_{m_i}],t\right) \chi\left(([d_{m_i},e_{m_i}]^2,t)\right) }{\Psi\left([d_{m_i},e_{m_i}],t\right) }, \end{align*} $$

and

$$ \begin{align*}\Sigma_1 = \sum_{\substack{t \\ (t,q_1) = 1\\ p\mid t \Rightarrow p\equiv 1 \bmod 4}} \frac{c_t(h)}{t^2} \frac{(a,t)(b,t)}{\Psi(a,t)\Psi(b,t)}. \end{align*} $$

We have used the fact that in $\Sigma _3$ we have $(a,t) = (b,t) = \Psi (a,t) = \Psi (b,t) = 1$ , and in $\Sigma _1$ we have $\left ([d_{m_i},e_{m_i}],t\right ) = \chi \left (([d_{m_i},e_{m_i}]^2,t)\right ) = \Psi \left ([d_{m_i},e_{m_i}],t\right ) = 1$ .

We begin by considering $\Sigma _3$ . Using the definition $c_t(h) = \sum _{t_1|(t,h)} \mu (t/t_1) t_1$ , and swapping sums and relabeling via $t_2 = t/t_1$ , we get

$$ \begin{align*} \Sigma_3 = \sum_{\substack{t_1,t_2 \ {\mathfrak p}|t_1t_2 \Rightarrow p \equiv 3 \bmod 4 \\ (t_1t_2,q_3W\prod_{i\neq m_1,m_2}[d_i,e_i]) = 1 \ {\mathbf t}_1 \mid h}} \frac{t_1 \mu(t_2)}{(t_1t_2)^2} \prod_{i=1,2} \frac{\left([d_{m_i},e_{m_i}],t_1t_2\right) \chi\left(([d_{m_i},e_{m_i}]^2,t_1t_2)\right) }{\Psi\left([d_{m_i},e_{m_i}],t_1t_2\right) }. \end{align*} $$

Any prime $p|h$ with $p \equiv 3 \bmod 4$ divides $q_3W$ , and $t_1|h$ must be co-prime with $q_3W$ , so $t_1 = 1$ . Furthermore, $t_2$ is squarefree due to the term $\mu (t_2)$ , which implies that $\Psi ([d_{m_i},e_{m_i}],t_2) = 1$ . Thus,

$$ \begin{align*} \Sigma_3 &= \sum_{\substack{t_2\ {\mathfrak p}|t_2 \Rightarrow p \equiv 3 \bmod 4\\ (t_2,q_3W\prod_{i\neq m_1,m_2}[d_i,e_i]) = 1}} \frac{ \mu(t_2)}{t_2^2} \prod_{i=1,2} \left([d_{m_i},e_{m_i}],t_2\right) \chi\left(([d_{m_i},e_{m_i}]^2,t_2)\right) \\ &= \prod_{\substack{p\nmid q_3W\prod [d_i,e_i] \\ p\equiv 3 \bmod 4}} \left(1 - \frac{1}{p^2}\right) \prod_{\substack{p\nmid q_3W\prod_{i\neq m_1,m_2} [d_i,e_i] \\ p\mid \prod_{i=1,2}[d_{m_i},e_{m_i}] \\ p\equiv 3 \bmod 4}} \left(1 + \frac{1}{p}\right). \end{align*} $$

Since $q_3W$ is the product of all primes congruent to 3 mod 4 smaller than $D_0$ , we have

$$ \begin{align*} \Sigma_3 &= \prod_{\substack{p>D_0 \\ p\equiv 1 \bmod 4}}\left(1 - \frac{1}{p^2}\right) \prod_{i\neq m_1,m_2}\left(\prod_{p\mid [d_{i},e_{i}]} \left(1 - \frac{1}{p^2}\right)^{-1}\right) g_1([d_{m_1},e_{m_1}])g_1([d_{m_2},e_{m_2}]) \\ &= \left(1 + o(1)\right) g_1([d_{m_1},e_{m_1}])g_1([d_{m_2},e_{m_2}]) \prod_{i\neq m_1,m_2}f\left([d_i,e_i]\right), \end{align*} $$

where $f(n) = \prod _{p\mid n}\left (1 - \frac {1}{p^2}\right )^{-1}$ .

Plugging this back into (42), we get

$$ \begin{align*} &S_3^{(m_1,m_2)}(\nu_0) \\ &\quad \sim \frac{\pi^2 x g_1^2\left(qW\right)}{W \log^2 v} \sum_{\mathbf d, \mathbf e \in \mathcal D_K} \lambda_{\mathbf d} \lambda_{\mathbf e} \frac{ \prod_{i\neq m_1,m_2}g_1^2\left([d_i,e_i]\right)f\left([d_i,e_i]\right) \prod_{i=1,2}g_1\left([d_{m_i},e_{m_i}]\right) }{ [d_{m_1},e_{m_1}]^2 [d_{m_2},e_{m_2}]^2 \prod_{i\neq m_1,m_2}[d_i,e_i] } \\ &\qquad \times \sum_{\substack{a,b \le v \\ (a,q_1) = (b,q_1) = 1 \\ p|a,b \Rightarrow p \equiv 1 \bmod 4}} \frac{\mu(a)\mu(b)}{a b } \log \frac va \log \frac vb \sum_{\substack{t \\ (t,q_1) = 1\\ p\mid t \Rightarrow p\equiv 1 \bmod 4}} \frac{c_t(h)}{t^2} \frac{(a,t)(b,t)}{\Psi(a,t)\Psi(b,t)}. \end{align*} $$

In the notation of Lemma 9, the sum over $\mathbf d$ and $\mathbf e$ (which is independent of $a,b,t$ ) is of the form $S_J$ for $J = \{m_1,m_2\}$ , so that by Lemma 9, we get

$$ \begin{align*} S_3^{(m_1,m_2)}(\nu_0) &\sim \frac{\pi^2 x g_1^2\left(qW\right)}{W \log^2 v} B^{K+2}L_{K;m_1,m_2}(F) \\ &\quad\times \sum_{\substack{a,b \le v \\ (a,q_1) = (b,q_1) = 1 \\ p|a,b \Rightarrow p \equiv 1 \bmod 4}} \frac{\mu(a)\mu(b)}{a b } \log \frac va \log \frac vb \sum_{\substack{t \\ (t,q_1) = 1\\ p\mid t \Rightarrow p\equiv 1 \bmod 4}} \frac{c_t(h)}{t^2} \frac{(a,t)(b,t)}{\Psi(a,t)\Psi(b,t)}. \end{align*} $$

Switching the order of summation gives

(43) $$ \begin{align} S_3^{(m_1,m_2)}(\nu_0) &\sim \frac{\pi^2 x g_1^2\left(qW\right)}{W \log^2 v} B^{K+2}L_{K;m_1,m_2}(F) \\ &\quad \times \sum_{\substack{t \\ (t,q_1) = 1\\ p\mid t \Rightarrow p\equiv 1 \bmod 4}} \frac{c_t(h)}{t^2} \sum_{\substack{a,b \le v \\ (a,q_1) = (b,q_1) = 1 \\ p|a,b \Rightarrow p \equiv 1 \bmod 4}} \frac{\mu(a)\mu(b)}{a b } \frac{(a,t)(b,t)}{\Psi(a,t)\Psi(b,t)} \log \frac va \log \frac vb.\notag \end{align} $$

Denoting the inner sum as $\Sigma _{a,b}(t)$ , we can write

$$ \begin{align*}\Sigma_{a,b}(t) = \Bigg(\sum_{\substack{a \le v \\ (a,q_1)= 1 \\ p|a \Rightarrow p \equiv 1 \bmod 4}} \frac{\mu(a)(a,t)}{a\Psi(a,t)} \log \frac va \Bigg)^2. \end{align*} $$

The calculations from [Reference Hooley9, Lemma 6] (along much the same lines as Lemma 13) imply that

$$ \begin{align*}\Sigma_{a,b}(t) = \left(\frac{8A}{\pi g_1(q_1)} C(t) \log^{\frac{1}{2}} v + O\left( \frac{t^{\frac{1}{4}}}{\log^{\frac{1}{2}} v} \right)\right)^2, \end{align*} $$

where $C(t)$ is the constant in [Reference Hooley9, Lemma 6], given by

$$ \begin{align*} C(t) = \begin{cases}\prod_{\substack{p\mid t \\ p\equiv 1 \bmod 4}}\left(2 - \frac{1}{p}\right)^{-1} &\text{ if }p|t \text{ and }p \equiv 1 \bmod 4\text{ implies } p^2|t \\ 0 &\text{ otherwise.} \end{cases} \end{align*} $$

Note that in [Reference Hooley9, Lemma 6], the Landau–Ramanujan constant is normalized as $\sqrt {2}A$ , and that the statements of [Reference Hooley9, Lemma 5] and [Reference Hooley9, Lemma 6] are missing another factor of $\sqrt {2}$ .

Plugging this estimate back into (43), we get

$$ \begin{align*} S_3^{(m_1,m_2)}(\nu_0) \sim \frac{64 A^2 x g_1^2\left(q_3W\right) B^{K+2}}{W \log v} L_{K;m_1,m_2}(F) \Bigg(\sum_{\substack{t \\ (t,q_1) = 1\\ p\mid t \Rightarrow p\equiv 1 \bmod 4}} \frac{c_t(h)C^2(t)}{t^2} + O\left(E\right)\Bigg), \end{align*} $$

where

$$ \begin{align*} E &\ll \frac{1}{\log v} \sum_{\substack{t \\ (t,q_1) = 1\\ p\mid t \Rightarrow p\equiv 1 \bmod 4}} \frac{|c_t(h)|C(t)}{t^{\frac{7}{4}}} +\frac{1}{\log^2 v}\sum_{\substack{t \\ (t,q_1) = 1\\ p\mid t \Rightarrow p\equiv 1 \bmod 4}} &\frac{|c_t(h)|}{t^{\frac{3}{2}}}\\ & \ll \frac{1}{\log v} \sum_{\substack{t \\ (t,q_1) = 1\\ p\mid t \Rightarrow p\equiv 1 \bmod 4}} \frac{|c_t(h)|}{t^{\frac{3}{2}}}. \end{align*} $$

From [Reference Hooley9, Eq 15], we have that $E \ll \frac {1}{\log v} \sigma _{-\frac {1}{2}}(h)$ , which implies $E \ll \frac {\log \log x}{\log x}$ since $h \leq \eta (\log x)^{\frac {1}{2}}$ . As for the main term, from [Reference Hooley9, Eq 18], we have

$$ \begin{align*} \sum_{\substack{t \\ (t,q_1) = 1\\ p\mid t \Rightarrow p\equiv 1 \bmod 4}} \frac{c_t(h)C^2(t)}{t^2} = \prod_{\substack{p^\beta \| h\\ p\equiv 1 \bmod 4 \\ p\nmid q_1}} \left(1 + \frac{1}{(2p-1)^2}\left(1 - \frac{1}{p^{\beta -1}} - \frac{1}{p^\beta}\right)\right). \end{align*} $$

Thus,

$$ \begin{align*} & S_3^{(m_1,m_2)} \\ &\sim \frac{64 A^2 x g_1^2\left(q_3W\right) B^{K+2}}{W \log v} \prod_{\substack{p^\beta \| b_{m_2} - b_{m_1}\\ p\equiv 1 \bmod 4 \\ p\nmid q_1}} \left(1 + \frac{1}{(2p-1)^2}\left(1 - \frac{1}{p^{\beta -1}} - \frac{1}{p^\beta}\right)\right) L_{K;m_1,m_2}(F). \end{align*} $$

The product over $p^\beta \|(b_{m_2}-b_{m_1})$ is bounded above by V (defined in (15)), so that

$$ \begin{align*} S_3^{(m_1,m_2)}(\nu_0) \le (1+o(1))\frac{64 A^2 x g_1^2\left(q_3W\right) B^{K+2}}{W \log v} V L_{K;m_1,m_2}(F). \end{align*} $$

Finally, using the identity that $B = \frac {2A}{\pi }\frac {\phi (q_3W)(\log R)^{1/2}}{q_3W}$ as well as applying (41) and Lemma 4 completes the proof.

Proof. We first expand the definitions of $w_n(\mathcal L)$ and $\rho ^2(\ell _m(n))$ and swap the order of summation to write

(44) $$ \begin{align} S_4^{(m)}(\nu_0) = \frac 1{\log^2 v} \sum_{\mathbf d, \mathbf e \in \mathcal D_K} \lambda_{\mathbf d}\lambda_{\mathbf e} \sum_{\substack{a,b \le v \\ (a,q_1) = (b,q_1) = 1 \\ p|ab \Rightarrow p \equiv 1 \bmod 4}} \frac{\mu(a)\mu(b)}{g_2(a)g_2(b)} \log\frac va \log \frac vb \sum_{\substack{x < n \le 2x \\ n \equiv 1 \bmod 4 \\ n\equiv \nu_0 \bmod W \\ [d_i, e_i]\mid \ell_i(n) \forall i \\ [a,b]\mid \ell_m(n) \\ }} r_2^2(\ell_m(n)). \end{align} $$

The quantities $W, [d_1,e_1], \dots , [d_k,e_k]$ , q, and $[a,b]$ must be pairwise coprime because of the support of $\mathcal D_K$ . We use Lemma 12 in order to evaluate the inner sum; we will apply Lemma 12 with $r = qW\prod _{i \ne m} [d_i,e_i]$ and $d = [a,b][d_m,e_m]$ . The sum (44) can then be written as

(45) $$ \begin{align} &\frac 1{\log^2 v} \sum_{\mathbf d, \mathbf e \in \mathcal D_K} \lambda_{\mathbf d}\lambda_{\mathbf e} \sum_{\substack{a,b \le v \\ p|ab \Rightarrow p \equiv 1 \bmod 4}} \frac{\mu(a)\mu(b)}{g_2(a)g_2(b)} \log\frac va \log \frac vb \times \\ &\qquad \left( \frac{g_3(r)g_4(d)}{rd} \left( \log qx + A_2 + 2\sum_{p\mid r}g_5(p) -2\sum_{p\mid d}g_6(p)\right) qx + O_\epsilon\left( (qx)^{\frac{3}{4} + \theta_2 + \varepsilon}\right) \right).\notag \end{align} $$

Taking absolute values, the error term from (45) is bounded by

$$ \begin{align*}\ll_{\varepsilon} \lambda_{\mathrm{max}}^2|\mathcal D_K|^2v^2 (qx)^{\frac{3}{4} + \theta_2 + \varepsilon} \ll_{\varepsilon'} x^{\frac{3}{4} + 2(\theta_2 + \theta_1) + \varepsilon'}, \end{align*} $$

where we used (34) to bound the $\lambda _{\mathrm {max}}$ and (37) for $|\mathcal D_K|$ . This error term is negligible, since $\theta _1+\theta _2 < \frac {1}{18}$ .

We now evaluate the main term via a process that is identical to the one in [Reference McGrath15, Proposition 6.2, part (iv)]. Using the notation from Lemma 13, the main term of (45) is

$$ \begin{align*}(1 + o(1)) \frac{g_3(qW) B^{K+1} qx}{W \log^2 v} \left( Z_{x,q_1}^{(1)}\log x - 2Z_{x,q_1}^{(2)} \right) L_{K;m}(F). \end{align*} $$

By Lemma 13, this is equal to

$$ \begin{align*} & (1 + o(1)) \frac{g_3(qW) B^{K+1} qx}{W \log^2 v} \frac{8 A g_7(q_1) }{\phi(q_1)g_1(q_1)} \sqrt{\log v}(\log x + \log v) \\ & \qquad \times \prod_{\substack{p \equiv 1 \bmod 4 \\ p\nmid q_1}}\left( 1 + \frac{1}{(2p-1)^2}\right) L_{K;m}(F)\\ &\quad = (1 + o(1)) \frac{8 A g_3(qW)g_7(q_1)B^{K+1}\log^{\frac{3}{2}}v \left(\frac{\log x}{\log v} + 1 \right) qx} {qW \phi(q_1)g_1(q_1) \log^2 v } \\ &\qquad \times \prod_{\substack{p \equiv 1 \bmod 4 \\ p\nmid q_1}}\left( 1 + \frac{1}{(2p-1)^2}\right) L_{K;m}(F). \end{align*} $$

Recalling the definition of B and equation(41), we get that

$$ \begin{align*} &S_4^{(m)}(\nu_0) \\& \quad = (1+o(1)) \frac{g_3(q_1)g_7(q_1)}{\phi(q_1)g_1(q_1)} \frac{8 B^{k}\sqrt{\frac{\log R}{\log v}} \left(\frac{\log x}{\log v} + 1 \right)x} {W \pi } \prod_{\substack{p \equiv 1 \bmod 4 \\ p\nmid q_1}}\left( 1 + \frac{1}{(2p-1)^2}\right) L_{K;m}(F). \end{align*} $$

Observing that the factors dividing $q_1$ cancel and applying the identity from Lemma 4 that $L_{K;m}(F) = \frac {\pi ^2}{\pi +2}\frac {L_K(F)}{\sqrt K}$ completes the proof.

Proof. First note that if $u \ne v$ , then for some prime p, $\sigma _{p}({\mathbf u},{\mathbf v}) = 0$ , so these terms do not contribute. Thus, for a fixed ${\mathbf u} \in \mathcal D_K$ ,

$$ \begin{align*} \sum_{\substack{{\mathbf v} \in \mathcal D_K}} \frac{\prod_{p|uv}|\sigma_{p}({\mathbf u},{\mathbf v})|}{\phi(v)} = \prod_{p|u} \Big(\sum_{\substack{\mathbf w \in \mathcal D_K \\ w_i|p \forall i}} \frac{|\sigma_{p}({\mathbf u},\mathbf w)|}{\phi(w)}\Big) = 1. \end{align*} $$

This, along with the bound that $y_{{\mathbf u}}y_{{\mathbf v}} \ll y_{{\mathbf u}}^2 + y_{{\mathbf v}}^2$ , implies that

$$ \begin{align*} T \ll \frac{x}{W}\sum_{{\mathbf u},{\mathbf v} \in \mathcal D_K} \frac{y_{{\mathbf u}}^2 + y_{{\mathbf v}}^2}{\phi(u)\phi(v)}\prod_{\substack{p|uv }} |\sigma_{p}({\mathbf u},{\mathbf v})| &\ll \frac{x}{W}\sum_{{\mathbf u}\in\mathcal D_K} \frac{y_{{\mathbf u}}^2} {\phi(u)} \left( \sum_{\substack{{\mathbf v} \in \mathcal D_K}} \frac{\prod_{p|uv}|\sigma_{p}({\mathbf u},{\mathbf v})|}{\phi(v)} \right) \\ &\ll \frac{x}{W}\sum_{{\mathbf u}\in\mathcal D_K} \frac{y_{{\mathbf u}}^2} {\phi(u)}. \end{align*} $$

By Lemma 7,

$$ \begin{align*} \sum_{{\mathbf r} \in \mathcal D_K} \frac{y_{{\mathbf r}}^2}{\phi(r)} &= \sum_{{\mathbf r} \in \mathcal D_K} \frac{\mu(r)^2}{\phi(r)}\prod_{i=1}^K g\left(K\frac{\log r_1}{\log R}\right)^2 \ll \left(\frac{e^{-\gamma /2}}{\Gamma(1/2)}\right)^K \left(\frac{\log R}{\log D_0}\right)^{K/2} L_K(F), \end{align*} $$

as desired.

Proof of Proposition 18.

Expanding the square and swapping the order of summation gives

$$ \begin{align*} S_5^{(m)}(\nu_0) = \sum_{\substack{p < x^\xi \\ p \equiv 3 \bmod 4}} \sum_{\mathbf d, \mathbf e \in \mathcal{D}_K} \lambda_{\mathbf d}\lambda_{\mathbf e} \sum_{\substack{x < n \le 2x \\ n \equiv 1 \bmod 4 \\ n \equiv \nu_0 \bmod W \\ [d_i,e_i]|\ell_i(n) \\ p|\ell_m(n)}} 1. \end{align*} $$

By choice of $\nu _0 \bmod W$ , if $p|\ell _m(n)$ , then $p> D_0$ . Because of the support of $\mathcal {D}_K$ , if $\lambda _{\mathbf d} \ne 0$ and $\lambda _{\mathbf e} \ne 0$ , then any prime $p \equiv 3 \bmod 4$ can divide at most one of the $\ell _i(n)$ . Thus, if $p|\ell _m(n)$ , then $p \nmid \ell _i(n)$ for all $i \ne m$ , which implies that $(d_ie_i,p) = 1$ for all $i \ne m$ . By the Chinese remainder theorem, the inner sum is of the form $\frac xQ + O(1)$ , where $Q = 4W[d_m,e_m,p]\prod _{i \ne m} [d_i,e_i].$ Note that $Q < 4WR^2x^\xi $ , and for any fixed Q, there are $O(\tau _{3k+4}(Q))$ choices of $\mathbf d,\mathbf e, p$ giving rise to the modulus Q. Thus, the error term from the Chinese remainder theorem application and (34) makes a contribution that is

$$ \begin{align*} &\ll \sum_{Q < 4WR^2x^\xi} \tau_{3k+4}(Q) \lambda_{\mathrm{max}}^2 \ll_\varepsilon R^2x^{\xi}x^\varepsilon = x^{\theta_2 + \xi + \varepsilon}, \end{align*} $$

which is negligible for $\xi $ small.

The remaining term is given by

$$ \begin{align*} &\frac x{4W} \sum_{\substack{D_0 < p < x^\xi \\ p \equiv 3 \bmod 4}} \frac 1p \sum_{\substack{\mathbf d, \mathbf e \in \mathcal{D}_K \\ (d_ie_i,p) = 1 \forall i \ne m}} \frac{\lambda_{\mathbf d}\lambda_{\mathbf e}p}{[d_m,e_m,p]\prod_{i\ne m}[d_i,e_i]}. \end{align*} $$

Expanding the definitions of $\lambda _{\mathbf d}, \lambda _{\mathbf e}$ and rearranging, this is

(46) $$ \begin{align} \begin{aligned} &\frac{x}{4W}\sum_{\substack{D_0 < p < x^\xi \\ p \equiv 3 \bmod 4}} \frac 1p \sum_{\substack{\mathbf d, \mathbf e \in \mathcal{D}_K \\ (d_ie_i,p) = 1 \forall i \ne m}} \frac{\mu(d)\mu(e)dep}{[d_m,e_m,p]\prod_{i\ne m}[d_i,e_i]} \sum_{\substack{\mathbf r, \mathbf s \in \mathcal{D}_K \\ \mathbf d|\mathbf r,\mathbf e|\mathbf s}} \frac{y_{\mathbf r} y_{\mathbf s}}{\phi(r)\phi(s)} \\ = &\frac{x}{4W}\sum_{\substack{D_0 < p < x^\xi \\ p \equiv 3 \bmod 4}} \frac 1p \sum_{\substack{\mathbf r, \mathbf s \in \mathcal{D}_K}} \frac{y_{\mathbf r} y_{\mathbf s}}{\phi(r)\phi(s)} \sum_{\substack{\mathbf d, \mathbf e \in \mathcal{D}_K \\ (d_ie_i,p) = 1 \forall i \ne m \\ \mathbf d|\mathbf r, \mathbf e|\mathbf s}} \frac{\mu(d)\mu(e)dep}{[d_m,e_m,p]\prod_{i\ne m}[d_i,e_i]}. \end{aligned} \end{align} $$

The inside sum is multiplicative over $p'|rs$ ; write $\sigma _{p'}(\mathbf r, \mathbf s,p)$ for the $p'$ component. If $p'\ne p$ , then

$$ \begin{align*} \sigma_{p'}({\mathbf r},{\mathbf s},p) =\sum_{\substack{\mathbf d, \mathbf e \in \mathcal{D}_K \\ d_i,e_i \mid p' \; \forall i \\ \mathbf d|\mathbf r, \mathbf e|\mathbf s}} \frac{\mu(d)\mu(e)dep}{[d_m,e_m,p]\prod_{i\ne m}[d_i,e_i]} = \begin{cases} p'-1 &\text{ if } p'|({\mathbf r},{\mathbf s}) \\ -1 &\text{ if } p'|r, p'|s, p'\nmid ({\mathbf r},{\mathbf s}) \\ 0 &\text{ otherwise,} \end{cases} \end{align*} $$

where we recall that $({\mathbf r},{\mathbf s}) = \prod _{i}(r_i,s_i)$ . If $p'= p$ , then

$$ \begin{align*} \sigma_p({\mathbf r},{\mathbf s},p) = \sum_{\substack{\mathbf d, \mathbf e \in \mathcal{D}_K \\ d_m,e_m \mid p \\ d_i=e_i=1 \; \forall i \neq m \\ \mathbf d|\mathbf r, \mathbf e|\mathbf s}} \frac{\mu(d)\mu(e)dep}{[d_m,e_m,p]\prod_{i\ne m}[d_i,e_i]} =\begin{cases} (p-1)^2 &\text{ if } p|(r_m,s_m) \\ -(p-1) &\text{ if } p|r_ms_m, p\nmid(r_m,s_m) \\ 1 &\text{ otherwise.}\end{cases}. \end{align*} $$

Let $f_{\mathbf u}({\mathbf r}) = (r_1,\dots , r_m/(r_m,p), \dots , r_k)$ be the vector formed by removing a possible factor of p from $r_m$ . Then our expression (46) can be written as

$$ \begin{align*} &\frac{x}{4W}\sum_{\substack{D_0 < p < x^\xi \\ p \equiv 3 \bmod 4}} \frac 1p\sum_{\substack{\mathbf{u}, \mathbf s \in \mathcal{D}_K \\ (u_m,p) = 1}} \frac{y_{\mathbf s}}{\phi(s)} \sum_{\substack{{\mathbf r} \\ f_{\mathbf u}({\mathbf r}) = {\mathbf u}}} \frac{ y_{{\mathbf r}}}{\phi(r)} \prod_{p'|rs} \sigma_{p'}({\mathbf r},{\mathbf s},p). \end{align*} $$

We split the sum above into several parts. Let $\Sigma _1$ be the summands where $p|u_j$ for some $j \ne m$ , and let $\Sigma _2$ be the summands where $p\nmid u_i$ for all i. Define

(47) $$ \begin{align} T = \frac xW \sum_{\substack{{\mathbf u},{\mathbf v} \in \mathcal{D}_K}} \frac{y_{{\mathbf u}}y_{{\mathbf v}}}{\phi(u)\phi(v)} \prod_{\substack{p' \mid uv}}\left|\sigma_{p'}({\mathbf u},{\mathbf v})\right|, \end{align} $$

where

$$ \begin{align*} \sigma_{p'}({\mathbf u},{\mathbf v}) = \sum_{\substack{\mathbf d \mid {\mathbf u},\; \mathbf e \mid {\mathbf v} \\ d_i,e_i \mid p' \; \forall i }} \frac{\mu(d)\mu(e) d e}{[\mathbf d, \mathbf e]}.\end{align*} $$

We will bound both $\Sigma _1$ and $\Sigma _2$ in terms of T, showing first that $\Sigma _1 \ll \frac 1{D_0}T$ . We have

$$ \begin{align*} \Sigma_1 = \frac x{4W}\sum_{\substack{D_0 < p < x^\xi \\ p \equiv 3 \bmod 4}} \frac 1p \sum_{i\neq m} \sum_{\substack{\mathbf u, \mathbf s \in \mathcal{D}_K \\ (u_m,p) = 1 \\ p|u_i}} \frac{y_{\mathbf s}}{\phi(s)} \sum_{\substack{{\mathbf r} \\ f_{\mathbf u}({\mathbf r}) = {\mathbf u}}} \frac{y_{{\mathbf r}}}{\phi(r)} \sigma_p({\mathbf r},{\mathbf s}, p)\prod_{\substack{p'\mid rs \\ p' \ne p}}\sigma_{p'}({\mathbf r},{\mathbf s},p). \end{align*} $$

Given $\mathbf u$ , there is only one vector ${\mathbf r}$ such that $f_{\mathbf u}({\mathbf r}) = {\mathbf u}$ ; namely, ${\mathbf r} = {\mathbf u}$ . Thus,

$$ \begin{align*} \Sigma_1 = \frac x{4W}\sum_{\substack{D_0 < p < x^\xi \\ p \equiv 3 \bmod 4}} \frac 1p \sum_{i\neq m} \sum_{\substack{\mathbf u, \mathbf s \in \mathcal{D}_K \\ (u_m,p) = 1 \\ p|u_i}} \frac{y_{\mathbf s} y_{{\mathbf u}}}{\phi(s)\phi(u)} \sigma_p({\mathbf u},{\mathbf s}, p)\prod_{\substack{p'\mid us \\ p' \ne p}}\sigma_{p'}({\mathbf u},{\mathbf s}). \end{align*} $$

Denote by ${\mathbf u}'$ the vector obtained from ${\mathbf u}$ by removing all factors of p. Then $\phi (u) = (p-1)\phi (u')$ and $\sigma _p({\mathbf u},{\mathbf s},p) = \sigma _p({\mathbf u}',{\mathbf s},p) = \mu ((s_m,p))\phi ((s_m,p))$ , which is independent of ${\mathbf u}$ because we already require $(u_m,p) = 1$ . Thus,

$$ \begin{align*} \Sigma_1 = \frac x{4W} \sum_{{\mathbf s} \in \mathcal{D}_K} \frac{y_{{\mathbf s}}}{\phi(s)} \sum_{\substack{D_0 < p < x^\xi \\ p \equiv 3 \bmod 4}} \frac 1p \sum_{\substack{{\mathbf u}' \in \mathcal{D}_K \\ p\nmid u_i^{\prime}}} \frac{1}{\phi(u')} \sigma_p({\mathbf u}',{\mathbf s},p) \prod_{\substack{p'\mid u's\\ p'\ne p}}\sigma_{p'}({\mathbf u}',{\mathbf s}) \sum_{\substack{{\mathbf u} \in \mathcal{D}_K\\ {\mathbf u}\rightarrow{\mathbf u}'}} \frac{y_{{\mathbf u}}}{(p-1)}. \end{align*} $$

By Lemma 5, we have $y_{{\mathbf u}} = y_{{\mathbf u}'}\left ( 1+ O\left (K\xi \right )\right )$ . By assumption, $K\xi \ll 1$ , so (recalling that the weights $y_{{\mathbf r}}$ are nonnegative), $y_{{\mathbf u}} \ll y_{{\mathbf u}'}$ and

$$ \begin{align*} \Sigma_1 \ll \frac xW (K-1) \sum_{{\mathbf s} \in \mathcal{D}_K} \frac{y_{{\mathbf s}}}{\phi(s)} \sum_{\substack{D_0 < p < x^\xi \\ p \equiv 3 \bmod 4}} \frac 1 {p(p-1)} \sum_{\substack{{\mathbf u}' \in \mathcal{D}_K \\ p\nmid u_i^{\prime}}} \frac{ y_{{\mathbf u}'}}{\phi(u')} |\sigma_p({\mathbf u}',{\mathbf s},p)| \prod_{\substack{p'\mid u's\\ p'\ne p}}|\sigma_{p'}({\mathbf u}',{\mathbf s})|. \end{align*} $$

To bound $\Sigma _1$ , we now further split it into subsums. First, let $T_1$ consist of all those terms with ${\mathbf s}$ such that $p \nmid s_i$ for all i. In this case, $\sigma _p({\mathbf u}',{\mathbf s}, p) = 1$ , so

$$ \begin{align*} T_1 \ll \frac xW(K-1) \sum_{\substack{D_0 < p < x^\xi \\ p \equiv 3 \bmod 4}} \frac 1 {p(p-1)} \sum_{\substack{{\mathbf s} \in \mathcal{D}_K \\ p\nmid s_i \; \forall i}} \frac{y_{{\mathbf s}}}{\phi(s)} \sum_{\substack{{\mathbf u}'\in \mathcal{D}_K \\ p\nmid u_i^{\prime} \; \forall i}} \frac{y_{{\mathbf u}'}}{\phi(u')} \prod_{\substack{p'\mid u's}}\left|\sigma_{p'}({\mathbf u}',{\mathbf s})\right|. \end{align*} $$

Dropping the requirement that $p\nmid s_i$ , $p\nmid u^{\prime }_i$ only increases $T_1$ . The sum over p is then independent of the rest of the expression, and converges to a constant that is $\ll \frac 1{D_0}$ , which in turn implies that $T_1 \ll \frac K{D_0}T$ , where T is defined in (47).

Now consider $T_2$ , the terms ${\mathbf s}$ in $\Sigma _1$ such that $p\mid s_i$ for some $i\neq m$ . In this case, $\sigma _p({\mathbf u}',{\mathbf s}, p) = 1$ , so

$$ \begin{align*} T_2 \ll \frac xW (K-1)^2 \sum_{D_0 < p < x^\xi} \frac 1 {p(p-1)} \sum_{\substack{{\mathbf s} \in \mathcal{D}_K \\ p\mid s_1}} \frac{y_{{\mathbf s}}}{\phi(s)} \sum_{\substack{{\mathbf u}' \in \mathcal{D}_K \\ p\nmid u_i^{\prime} \; \forall i}} \frac{y_{{\mathbf u}'}}{\phi(u')} \prod_{\substack{p'\mid u's \\ p' \ne p}}\left|\sigma_{p'}({\mathbf u}',{\mathbf s})\right|. \end{align*} $$

Let ${\mathbf s}'$ be the vector obtained by removing the factor of p from ${\mathbf s}$ . Once again, $y_{{\mathbf s}} \ll y_{{\mathbf s}'}$ , so

$$ \begin{align*} T_2 \ll \frac xW (K-1)^2 \sum_{\substack{D_0 < p < x^\xi \\ p \equiv 3 \bmod 4}} \frac 1 {p(p-1)^2} \sum_{\substack{{\mathbf s}'\in \mathcal{D}_K \\ p\nmid s_i \; \forall i}} \frac{y_{{\mathbf s}'}}{\phi(s')} \sum_{\substack{{\mathbf u}'\in \mathcal{D}_K \\ p\nmid u_i^{\prime} \; \forall i}} \frac{y_{{\mathbf u}'}}{\phi(u')} \prod_{\substack{p'\mid u's}}\left|\sigma_{p'}({\mathbf u}',{\mathbf s})\right|. \end{align*} $$

Once more, we can remove the constraints that $p\nmid s_i$ and $p\nmid u^{\prime }_i$ and evaluate the sum over p to get that $T_2 \ll \frac {K^2}{D_0^2} T$ .

Finally, consider $T_3$ , the subsum of $\Sigma _1$ with those ${\mathbf s}$ such that $p\mid s_m$ . In this case, $\sigma _p({\mathbf u}',{\mathbf s}, p) = -(p-1)$ . A similar computation gives

$$ \begin{align*} T_3 \ll \frac xW \sum_{\substack{D_0 < p < x^\xi \\ p \equiv 3 \bmod 4}} \frac 1 {p(p-1)^2} \sum_{\substack{{\mathbf s}'\in \mathcal{D}_K \\ p\nmid s_i \; \forall i}} \frac{y_{{\mathbf s}'}}{\phi(s')} \sum_{\substack{{\mathbf u}'\in \mathcal{D}_K \\ p\nmid u_i^{\prime} \; \forall i}} \frac{y_{{\mathbf u}'}}{\phi(u')} (p-1) \prod_{\substack{p'\mid u's}}\left|\sigma_{p'}({\mathbf u}',{\mathbf s})\right| \ll \frac{1}{D_0}T. \end{align*} $$

Thus, $|\Sigma _1| \ll T_1 + T_2 + T_3 \ll \frac {K^2}{D_0} T$ .

Now consider $\Sigma _2$ , given by

$$ \begin{align*} \Sigma_2 = \frac x{4W} \sum_{\substack{D_0 < p < x^\xi \\ p \equiv 3 \bmod 4}} \frac 1p \sum_{\substack{{\mathbf u},{\mathbf s} \in \mathcal D_K \\ (u_i,p) = 1 \forall i}} \frac{y_{{\mathbf s}}}{\phi(s)} \sum_{\substack{{\mathbf r} \in \mathcal D_K \\ f_{{\mathbf u}}({\mathbf r}) = {\mathbf u}}} \frac{y_{{\mathbf r}}}{\phi(r)} \prod_{p'|rs} \sigma_{p'}({\mathbf r},{\mathbf s},p). \end{align*} $$

Observe that for fixed ${\mathbf u}, {\mathbf s} \in \mathcal {D}_K$ with $p\nmid u_i$ for all i,

(48) $$ \begin{align} \sum_{\substack{{\mathbf r} \in \mathcal{D}_K \\ f_{{\mathbf u}}({\mathbf r})={\mathbf u}}} \frac{\sigma_p({\mathbf r},{\mathbf s},p)}{\phi(r)} = \frac{\mu((s_m,p))\phi((s_m,p))}{\phi(u)}\left(1 -\frac{p-1}{p-1}\right) = 0. \end{align} $$

We substitute $y_{{\mathbf r}} = y_{{\mathbf u}} + (y_{{\mathbf r}}-y_{{\mathbf u}})$ into $\Sigma _2$ . By (48), the $y_{{\mathbf u}}$ do not contribute, leaving only the contribution from $(y_{{\mathbf r}}-y_{{\mathbf u}})$ . The only terms remaining have ${\mathbf r} \ne {\mathbf u}$ , so that $p|r_m$ . Thus,

$$ \begin{align*} \Sigma_2= &\frac x{4W}\sum_{\substack{D_0 < p < x^\xi \\ p \equiv 3 \bmod 4}} \frac 1p\sum_{\substack{\mathbf r, \mathbf s \in \mathcal{D}_K \\ p|r_m}} \frac{y_{\mathbf s}(y_{{\mathbf r}}-y_{{\mathbf u}})}{\phi(r)\phi(s)}\prod_{p'|rs} \sigma_{p'}({\mathbf r},{\mathbf s},p). \end{align*} $$

By running the same argument for ${\mathbf s}$ and a tuple ${\mathbf v}$ obtained from ${\mathbf s}$ by removing a factor of p from $s_m$ (including bounding the terms where $p|v_i$ for some $i \ne m$ by $\frac {K^2}{D_0}T$ using identical arguments to the bound on $\Sigma _1$ ), we can also replace $y_{{\mathbf s}}$ by $y_{{\mathbf s}}-y_{{\mathbf v}}$ . By Lemma 5, we have

$$ \begin{align*} (y_{{\mathbf r}}-y_{{\mathbf u}})(y_{{\mathbf s}}-y_{{\mathbf v}}) \ll y_{{\mathbf u}}y_{{\mathbf v}}K^2 \frac{(\log p)^2}{(\log R)^2}, \end{align*} $$

so $\Sigma _2$ is given by

$$ \begin{align*} \Sigma_2 =&\frac x{4W}\sum_{\substack{D_0 < p < x^\xi \\ p \equiv 3 \bmod 4}} \frac 1p \sum_{\substack{{\mathbf r},{\mathbf s} \in \mathcal{D}_K \\ p|(r_m,s_m)}} \frac{(y_{{\mathbf r}}-y_{{\mathbf u}})(y_{{\mathbf s}}-y_{{\mathbf v}})}{\phi(r)\phi(s)} \prod_{p'|rs} \sigma_{p'}({\mathbf r},{\mathbf s},p) + O\left(\frac{K^2T}{D_0}\right) \\\ll &\frac{xK^2}{W} \sum_{\substack{D_0 < p < x^\xi \\ p \equiv 3 \bmod 4}} \frac 1p\left(\frac{\log p}{\log R}\right)^2 \sum_{\substack{{\mathbf u}, {\mathbf v} \in \mathcal{D}_K \\(uv,p) = 1}} y_{{\mathbf u}}y_{{\mathbf v}} \prod_{\substack{p'|uv \\ p'\ne p}} |\sigma_{p'}({\mathbf u},{\mathbf v})|\sum_{\substack{{\mathbf r},{\mathbf s} \in \mathcal{D}_K \\ r_m = pu_m \\ s_m = pv_m \\ r_i = u_i \forall i\neq m \\ s_i = v_i \forall i\neq m}}\frac{|\sigma_p({\mathbf r},{\mathbf s},p)|}{\phi(r)\phi(s)} + \frac{K^2T}{D_0}. \end{align*} $$

The sum over ${\mathbf r}$ and ${\mathbf s}$ is equal to $\phi (u)^{-1}\phi (v)^{-1}$ , so

$$ \begin{align*} \Sigma_2 &\ll \frac{xK^2}{W} \sum_{\substack{D_0 < p < x^\xi\\ p \equiv 3 \bmod 4}} \frac 1p\left(\frac{\log p}{\log R}\right)^2 \sum_{\substack{{\mathbf u}, {\mathbf v} \in \mathcal{D}_K \\ (uv,p) = 1}} \frac{y_{{\mathbf u}}y_{{\mathbf v}}}{\phi(u)\phi(v)} \prod_{\substack{p'|uv \\ p'\ne p}} |\sigma_{p'}({\mathbf u},{\mathbf v})| + \frac{K^2T}{D_0} \\ & \ll \frac{xK^2\xi^2}{W\theta_2^2} \sum_{\substack{{\mathbf u}, {\mathbf v} \in \mathcal{D}_K}} \frac{y_{{\mathbf u}}y_{{\mathbf v}}}{\phi(u)\phi(v)} \prod_{\substack{p'|uv}} |\sigma_{p'}({\mathbf u},{\mathbf v})| + \frac{KT}{D_0} \ll \left(K^2\frac{\xi^2}{\theta_2^2} + \frac{K^2}{D_0}\right) T. \end{align*} $$

Altogether, we get that $S_5^{(m)}(\nu _0) \ll \left (\frac {K^2\xi ^2}{\theta _2^2} + \frac {K^2}{D_0}\right )T$ . The contribution from the $\frac {K^2}{D_0}$ term vanishes as x grows large. The quantity T is evaluated in Lemma 19, giving

$$ \begin{align*} S_5^{(m)}(\nu_0) &\ll \frac{xK^2\xi^2}{W\theta_2^2} \left(\frac{e^{-\gamma /2}}{\Gamma(1/2)}\right)^K \left(\frac{\log R}{\log D_0}\right)^{K/2} L_K(F). \end{align*} $$

From the definition of B and Mertens’ theorem, we get

$$ \begin{align*}\left(\frac{e^{-\gamma /2}}{\Gamma(1/2)}\right)^K \left(\frac{\log R}{\log D_0}\right)^{K/2} \sim B^k, \end{align*} $$

which completes the argument.

Proof. We will apply Selberg’s sieve to bound the function $\mathbf 1_{S(\xi )}(\ell (n))$ , while also evaluating the sum over sieve weights $w_n(\mathcal L)$ . We begin by defining the additional sieve weights.

Recall that $S(\xi )$ denotes the set of integers such that for all primes $p < x^{\xi }$ with $p \equiv 3 \bmod 4$ , either $p \nmid n$ or $p^2|n$ . Thus, for each prime $p \equiv 3 \bmod 4$ , with $D_0 < p < x^{\xi }$ , we sieve by the set $\mathcal A_p$ of integers $x \le n \le 2x$ such that $p \mid \ell (n)$ but $p^2\nmid \ell (n)$ . The sieving set $\mathcal A_p$ has density function

$$ \begin{align*} g(p) &= \begin{cases} \frac{1}{p} - \frac{1}{p^2}& p> D_0 \text{ and } p\equiv3 \bmod 4 \\ 0 & \text{otherwise.} \end{cases} \end{align*} $$

We extend both $g(p)$ and $\mathcal A_p$ multiplicatively to squarefree d, so that

$$ \begin{align*}|r_d| := |\mathcal{A}_d - g(d) x| \ll \tau(d). \end{align*} $$

We will use the upper bound Selberg sieve

$$ \begin{align*} \mathbf 1_{\mathcal S(\xi)}(\ell(n)) \le \sum_{f|\ell(n)} \mu^+(f)|\mathcal A_f|, \end{align*} $$

where

$$ \begin{align*} \mu^+(f) = \frac{1}{\tilde{\lambda_{1}}^2}\sum_{[d_0,e_0] = f}\tilde{\lambda_{d_0}} \tilde{\lambda_{e_0}}, \end{align*} $$

and $\tilde {\lambda _d}$ is a sequence of weights defined as follows. Define ‘diagonalizing vectors’ $\tilde {y_{r_0}}$ via

$$ \begin{align*} \tilde{y_{r_0}} := \begin{cases} 1 &\text{ if } (r_0,q_3W) = 1, r_0 < x^{\xi}, \text{ and } p|r_0 \Rightarrow p \equiv 3 \bmod 4 \\ 0 &\text{ otherwise,}\end{cases} \end{align*} $$

and define $\tilde {\lambda _{d_0}}$ to be

(49) $$ \begin{align} \tilde{\lambda_{d_0}} := \mu(d_0) \frac{d_0^2}{\phi(d_0)} \sum_{d_0|r_0} \frac{\tilde{y_{r_0}}}{\phi(r_0)}. \end{align} $$

By Möbius inversion, we also have the relation that

$$ \begin{align*} \tilde{y_{r_0}} = \mu(r_0)\phi(r_0)\sum_{r_0|d_0} \frac{\tilde{\lambda_{d_0}}\phi(d_0)}{d_0^2}. \end{align*} $$

Note that $\tilde {\lambda _{d_0}}$ is supported on squarefree $d_0$ with $(d_0,q_3W) = 1$ , $d_0<x^{\xi }$ , and $d_0$ only divisible by primes congruent to $3$ mod $4$ . Also, with this choice,

(50) $$ \begin{align} \tilde{\lambda_{1}} = \sum_{\substack{r_0 < x^{\xi} \\ (r_0,q_3W) = 1 \\ p|r_0 \Rightarrow p \equiv 3 \bmod 4}} \frac{1}{\phi(r_0)} \gg \sqrt{\frac{\xi \log x}{\log D_0}}.\end{align} $$

For what follows, we will fix the notation that

$$ \begin{align*} \phi_{\omega^{*}}(n) = n\prod_{p\mid n}\left(1 - \frac{K+1 }{p}\right), \end{align*} $$

and define further ‘cross’-diagonalizing vectors

(51) $$ \begin{align} y_{{\mathbf r},r_0} := \mu(r_0r)\phi_{\omega^{*}}(r_0r) \sum_{\substack{{\mathbf r}|\mathbf d \\ r_0|d_0\\ (d_0,d) = 1}} \frac{\lambda_{\mathbf d}\tilde{\lambda}_{d_0}\phi(d_0)}{dd_0^2}, \end{align} $$

which satisfy the inverse relation that

(52) $$ \begin{align} \lambda_{\mathbf d}\tilde{\lambda_{d_0}} = \frac{\mu(d_0d)dd_0^2}{\phi(d_0)} \sum_{\substack{\mathbf d|\mathbf r \\ d_0|r_0 \\ (r_0,r) = 1}} \frac{y_{\mathbf r,r_0}}{\phi_{w^{*}}(r_r)}. \end{align} $$

We are now ready to apply Selberg’s sieve, which gives that

$$ \begin{align*} S_6^{(b)}(\nu_1) &\leq \frac{1}{\tilde{\lambda_{1}}^2} \sum_{d_0,e_0, \mathbf d, \mathbf e} \tilde{\lambda_{d_0}} \tilde{\lambda_{e_0}} \lambda_{\mathbf d} \lambda_{\mathbf e} \sum_{\substack{x < n \le 2x \\ [d_i,e_i]|\ell_i(n) \\ p\mid d_0e_0 \Rightarrow p\|\ell(n) \\ n \equiv 1 \bmod 4 \\ n \equiv \nu_1 \bmod q_3^2W^2}} |\mathcal A_{[d_0,e_0]}| \\ &= \frac{1}{\tilde{\lambda_{1}}^2} \sum_{\substack{d_0,e_0, \mathbf d, \mathbf e \\ (d_0e_0,de) = 1}} \tilde{\lambda_{d_0}} \tilde{\lambda_{e_0}} \lambda_{\mathbf d} \lambda_{\mathbf e} \left(\frac{xg([d_0,e_0])}{4q_3^2W^2 \prod_{i=1}^K [d_i,e_i]} + O(\tau([d_0,e_0]))\right). \end{align*} $$

The contribution from the $O(\tau ([d_0,e_0]))$ term satisfies

$$ \begin{align*} \frac 1{\tilde{\lambda_1}^2} \sum_{\substack{d_0,e_0,\mathbf d, \mathbf e \\ (d_0e_0,de) = 1}} |\tilde{\lambda_{d_0}}||\tilde{\lambda_{e_0}}||\lambda_{\mathbf d}||\lambda_{\mathbf e}| \tau([d_0,e_0]) &\ll R^{2+o(1)} \frac 1{\tilde{\lambda_1}^2} \sum_{\substack{d_0,e_0}} |\tilde{\lambda_{d_0}}||\tilde{\lambda_{e_0}}| \tau([d_0,e_0]). \end{align*} $$

By construction of $\tilde {\lambda _{1}}$ and $\tilde {\lambda _{d_0}}$ , we always have that $|\tilde {\lambda _{d_0}}|/|\tilde {\lambda _1}| \le \frac {d_0^2}{\phi (d_0)^2}$ . Thus, this term contributes $\ll _\varepsilon R^{2+\varepsilon }x^{2\xi +\varepsilon } \ll x^{\theta _2 + 2\xi +2\varepsilon }$ , which is negligible since $\xi < 1/4$ .

The remaining ‘main’ term is given by

(53) $$ \begin{align} \frac{x}{4q_3^2W^2\tilde{\lambda_{1}}^2} \sum_{\substack{d_0,e_0 \\(d_0,e_0,q_3W) = 1}} \tilde{\lambda_{d_0}}\tilde{\lambda_{e_0}} g([d_0,e_0]) \sum_{\substack{\mathbf d, \mathbf e \in \mathcal D_K\\ (de,d_0e_0) = 1}} \frac{\lambda_{\mathbf d}\lambda_{\mathbf e}}{\prod_{i=1}^K [d_i,e_i]}. \end{align} $$

Substituting the formula (52) for $\lambda _{\mathbf d}\tilde {\lambda _{d_0}}$ into (53), we get that (53) is equal to

$$ \begin{align*} &= \frac{x}{4q_3^2W^2\tilde{\lambda_{1}}^2} \sum_{\substack{\mathbf d, \mathbf e \in \mathcal{D}_K \\ d_0,e_0 \\ (d_ie_i, d_je_j) = 1 \; \forall 0\leq i < j \leq k}} \frac{\mu(d_0 d)\mu(e_0 e)d_0^2 d e_0^2 e g([d_0, e_0])}{[\mathbf{d}, \mathbf{e}]\phi(d_0)\phi(e_0)} \\ &\quad \times\sum_{\substack{\mathbf d \mid \mathbf r , \mathbf e \mid \mathbf s \\ d_0\mid r_0, e_0\mid s_0 }} \frac{y_{{\mathbf r},r_0} y_{{\mathbf s},s_0}}{\phi_{\omega^{*}}(r_0r) \phi_{\omega^{*}}(s_0s)} \\ &= \frac{x}{4q_3^2W^2\tilde{\lambda_{1}}^2} \sum_{\substack{\mathbf r, \mathbf s \in \mathcal{D}_K \\ r_0,s_0 \\ (r_0s_0, rs) = 1}} \frac{y_{{\mathbf r},r_0} y_{{\mathbf s},s_0}}{\phi_{\omega^{*}}(r_0r) \phi_{\omega^{*}}(s_0s)} \\ &\quad \times \sum_{\substack{\mathbf d \mid \mathbf r,\; \mathbf e \mid \mathbf s \\ d_0 \mid r_0,\; e_0\mid s_0 \\ (d_ie_i, d_je_j) = 1 \; \forall 0\leq i < j \leq k}} \frac{\mu(d_0 d)\mu(e_0 e)d_0^2 d e_0^2 e g([d_0, e_0])}{[\mathbf{d}, \mathbf{e}]\phi(d_0)\phi(e_0)}. \end{align*} $$

The inner sum is multiplicative over $p\mid rsr_0s_0$ , where the pth factor is given by

$$ \begin{align*}\sigma_p(\mathbf r, \mathbf s, r_0, s_0) = \begin{cases} p-1 & p\mid r_i,\; p\mid s_i,\; i\geq 1 \\ \frac{p^2}{p-1} - 1 & p\mid r_0,\; p\mid s_0\; \\ - 1 & p\mid r_i,\; p\mid s_j,\; i\neq j, \;i,j\geq 0\\ 0 & p \text{ divides exactly one of }rr_0 \text{ and } ss_0. \\ \end{cases} \end{align*} $$

The product $\prod _{p\mid rsr_0s_0}\sigma _p(\mathbf r, \mathbf s, r_0, s_0)$ is $0$ unless $rr_0 = ss_0$ . Then using the bound $y_{{\mathbf r},r_0} y_{{\mathbf s},s_0} \leq y_{{\mathbf r},r_0}^2 + y_{{\mathbf s},s_0}^2$ we see (by symmetry) that (53) is

(54) $$ \begin{align} &\le \frac{x}{4q_3^2W^2\tilde{\lambda_{1}}^2} \sum_{{\mathbf r} , r_0} \frac{y_{{\mathbf r},r_0}^2}{\phi_{\omega^{*}}^2(rr_0)} \sum_{\substack{{\mathbf s}, s_0 \\ ss_0 = rr_0}} \prod_{p \mid rr_0}|\sigma_p({\mathbf r},{\mathbf s},r_0,s_0)| \nonumber \\ &=\frac{x}{4q_3^2W^2\tilde{\lambda_{1}}^2} \sum_{{\mathbf r} , r_0} \frac{y_{{\mathbf r},r_0}^2}{\phi_{\omega^{*}}^2(rr_0)} \prod_{p\mid r} \left(K + p - 1\right) \prod_{p\mid r_0} \left(K + \frac{p^2}{p-1}- 1\right) \nonumber \\ &\le \frac{x}{4q_3^2W^2\tilde{\lambda_{1}}^2} \sum_{{\mathbf r} , r_0} y_{{\mathbf r},r_0}^2 \prod_{p\mid rr_0} \left( \frac{K + \frac{p^2}{p-1}- 1}{(p - K -1)^2} \right) \nonumber \\ &\ll \frac{x}{4q_3^2W^2\tilde{\lambda_{1}}^2} \sum_{{\mathbf r} , r_0} \frac{y_{{\mathbf r},r_0}^2}{\prod_{p\mid rr_0}\left(p + O(K) \right)}. \end{align} $$

In order to estimate this sum, we wish to express $y_{{\mathbf r},r_0}$ in terms of $y_{{\mathbf r}}$ and $\tilde {y_{r_0}}$ . This is very similar to the computation done in [Reference Maynard14, Proposition 9.4]. Writing $y_{{\mathbf r},r_0}$ as in (51) and using the definition of $\lambda _{\mathbf d}$ in (31) and $\tilde {\lambda _{d_0}}$ in (49), we get

$$ \begin{align*} y_{{\mathbf r},r_0} &= \mu(r_0r)\phi_{\omega^{*}}(r_0r) \sum_{\substack{{\mathbf r}|\mathbf d \\ r_0|d_0\\ (d_0,d) = 1}} \frac{\lambda_{\mathbf d}\tilde{\lambda_{d_0}}\phi(d_0)}{dd_0^2} \\ &= \mu(r_0r)\phi_{\omega^{*}}(r_0r) \sum_{r_0\mid d_0}\mu(d_0) \sum_{d_0 \mid f_0} \frac{\tilde{y_{f_0}}}{\phi(f_0)} \sum_{\substack{{\mathbf r} \mid \mathbf d \\ (d, d_0) = 1}} \mu(d)\sum_{\mathbf d \mid {\mathbf f}}\frac{y_{{\mathbf f}}}{\phi(f)} \\ &= \mu(r_0r)\phi_{\omega^{*}}(r_0r) \sum_{\substack{f_0, {\mathbf f} \\ r_0\mid f_0 , {\mathbf r} \mid {\mathbf f}}} \frac{y_{{\mathbf f}} \tilde{y_{f_0}}}{\phi(f_0)\varphi(f)} \sum_{\substack{d_0, \mathbf d \\ r_0\mid d_0, {\mathbf r} \mid \mathbf d \\ d_0 \mid f_0 , \mathbf d \mid {\mathbf f} \\ (d,d_0) = 1}}\mu(d)\mu(d_0). \end{align*} $$

The inner sum is 0 unless every prime dividing one of f and $f_0$ but not the other is a divisor of $rr_0$ ; in that case, the inner sum is $\pm 1$ . Thus, using the fact that $y_{\mathbf {r}} \geq y_{\mathbf {f}}$ (since F is decreasing), as well as the fact that $\tilde {y_{r_0}} \geq \tilde {y_{f_0}}$ , we get that

$$ \begin{align*} y_{{\mathbf r},r_0} \leq \phi_{\omega^{*}}(r_0r) y_{\mathbf{r}} \tilde{y_{r_0}} \sum_{\substack{f_0 \\ r_0 \mid f_0 \\ (f_0, q_3W) = 1}} \sum_{\substack{ {\mathbf f} \in \mathcal{D}_K\\ {\mathbf r} \mid {\mathbf f} \\ f f_0 / (f,f_0)^2 \mid rr_0}}\frac{\mu^2(f_0)}{\phi(f_0)\phi(f)}. \end{align*} $$

Let $f_0 = r_0 f_0^{\prime } g_0$ and $f_i = r_i f_i^{\prime } g_i$ for $1\leq i \leq K$ , where $f_i^{\prime } = f_i / (f_i, r r_0)$ is the largest divisor of $f_i$ that is relatively prime to $rr_0$ . In particular, $g_0\mid r$ and $g_i\mid r_0$ for $1\leq i \leq K$ . Since $f f_0 / (f,f_0)^2 \mid rr_0$ , we must have $f_0^{\prime } = \prod _{i=1}^{K} f_i^{\prime }$ . Thus, $y_{{\mathbf r},r_0}$ is bounded by

$$ \begin{align*} y_{{\mathbf r},r_0} &\leq \phi_{\omega^{*}}(r_0r) y_{\mathbf{r}} \tilde{y_{r_0}} \frac{1}{\phi(r_0 r)} \sum_{{\mathbf f}'\in \mathcal D_K} \frac{1}{\phi^2(f')} \sum_{\substack{\mathbf g \in \mathcal D_K \\ g_i\mid r_0 \forall 1\leq i \leq k}}\frac{1}{\phi(g)} \sum_{\substack{g_0 \mid r}}{\frac{1}{\phi(g_0)}} \\ &\leq \phi_{\omega^{*}}(r_0r) y_{\mathbf{r}} \tilde{y_{r_0}} \prod_{\substack{p> D_0 \\ p\equiv 3 \bmod{4}}}\left( 1 + \frac{K}{(p-1)^2}\right) \prod_{\substack{p\mid r_0 }}\left( 1 + \frac{K}{(p-1)}\right) \prod_{\substack{p\mid r }}\left( 1 + \frac{1}{(p-1)}\right). \end{align*} $$

The first product is $\ll O_K(1)$ . By the definition of $\phi _{\omega ^{*}}$ , we then have

$$ \begin{align*} y_{{\mathbf r},r_0} \ll y_{\mathbf{r}} \tilde{y_{r_0}} \prod_{p\mid r_0} \left(1 + \frac{K}{p-1}\right)\left( 1 - \frac{K+1}{p}\right) \prod_{p\mid r} \left(1 + \frac{1}{p-1}\right)\left( 1 - \frac{K+1}{p}\right), \end{align*} $$

which in turn implies that $y_{\mathbf r,r_0} \ll y_{\mathbf r} \tilde {y_{r_0}}$ because both products are $\le 1$ .

Plugging this into (54), we get that

(55) $$ \begin{align} S_6^{(b)}(\nu_1) \ll \frac{x}{q_3^2W^2\tilde{\lambda_{1}}^2} \Bigg(\sum_{\substack{r_0 \leq x^{\xi} \\ (r_0,W) = 1}} \frac{\tilde{y_{r_0}}^2}{\prod_{p\mid r_0}(p+O(K))}\Bigg) \left(\sum_{\textbf{r}\in\mathcal{D}_K} \frac{y_{\textbf{r}}^2}{\prod_{p\mid r}(p+O(K))}\right). \end{align} $$

Recalling that $\tilde {y_{r_0}} = 1$ for $r_0 \le x^{\xi }$ and $(r_0,q_3W) = 1$ , we have

$$ \begin{align*}\sum_{\substack{r_0 \leq x^{\xi} \\ (r_0,q_3W) = 1}} \frac{\tilde{y_{r_0}}^2}{\prod_{p\mid r_0}(p+O(K))} \ll \left(\frac{\xi\log x}{\log D_0}\right)^{1/2}. \end{align*} $$

We can bound the sum over $\mathbf r$ using Lemma 7. From the definition of $L_K(F)$ , we then get

$$ \begin{align*}\sum_{\textbf{r}\in\mathcal{D}_K} \frac{y_{\textbf{r}}^2}{\prod_{p\mid r}(p+O(K))} \ll \left(\frac{e^{-\gamma /2}}{\Gamma(1/2)}\right)^K \left(\frac{\log R}{\log D_0}\right)^{K/2} L_K(F). \end{align*} $$

Using these estimates as well as the bound (50) on $\lambda _{\mathbf r}$ , equation (55) becomes

$$ \begin{align*} S_6^{(b)}(\nu_1) &\ll_K \frac{x}{q_3^2W^2} \frac{\log D_0}{\xi\log x} \left(\frac{\xi \log x}{\log D_0}\right)^{1/2} \left(\frac{\log R}{\log D_0}\right)^{K/2}L_K(F) \\ &\ll \frac{x}{q_3^2W^2}\xi^{-1/2}\theta_2^{-1/2} \left(\frac{\log R}{\log D_0}\right)^{\frac{K-1}{2}} L_K(F), \end{align*} $$

as desired.

Proof. We first consider the number of $\mathbf b \in \mathcal B$ that will produce an admissible tuple $\mathcal L (\mathbf b)$ . If $\mathcal L (\mathbf b)$ is not admissible, then there is some prime $p\leq K$ , $p\equiv 3 \bmod 4$ , $p\nmid q$ such that $\prod _{\ell \in \mathcal L (\mathbf b)}\ell (n)$ is always divisible by p. In order to prevent this situation, we can consider only those $\mathbf b$ for which each $b_i$ , $i\geq 2$ satisfies $q + a_i + qb_i \not \equiv 0,1 \bmod {p}$ for all $p\leq K$ , $p\equiv 3 \bmod 4$ , $p\nmid 2q$ . Having excluded two congruence classes for each prime p, together with the linear form corresponding to $b_1$ , the tuple $\mathcal {L}(\mathbf b)$ cannot cover all of the congruence classes mod p.

Thus, for each $2\leq i \leq K$ , we can choose $b_i$ from a set of size

$$ \begin{align*}\frac{\eta}{8q}\sqrt{\log x} \prod_{\substack{p\mid W \\ 2<p\leq K } } \left(1 - \frac{2}{p}\right) \gg_K \frac{\eta}{q}\sqrt{\log x} \end{align*} $$

while ensuring that the resulting $\mathbf b$ is admissible. It follows that there are at least $\gg _K (\frac {\eta }{q})^{K-1}(\log x)^{(K-1)/2}$ choices of $\mathbf b \in \mathcal B$ with $\mathcal L(\mathbf b)$ admissible.

For each $\mathbf b$ with $\mathcal L(\mathbf b)$ , we now consider the sum over $\nu _0$ . For fixed $\mathbf b$ , this is bounded by

$$ \begin{align*}\sum_{\substack{\nu_0 \bmod W \\ (\ell(\nu_0),W) = 1 \forall \ell \in \mathcal L}} 1 \gg W \prod_{\substack{p\mid W \\2<p\leq K }}\left(\frac{1}{p}\right) \prod_{\substack{p\mid W \\ p>K }}\left(1 - \frac{K}{p}\right) \gg_K W \left(\frac{\phi(W)}{W}\right)^K, \end{align*} $$

which, along with the number of choices of $\mathbf b$ yielding admissible tuples, completes the proof.

Proof. By definition, $\tilde {N_{p^2}}(\mathcal L,b)$ almost consists of all elements of a certain set of congruence classes modulo p when lifted to $\mathbb Z/p^2\mathbb Z$ , with the possible exception of one congruence class $\nu $ modulo $p^2$ such that $\ell ^b(\nu ) \equiv 0 \bmod p^2$ . In particular, this implies that

$$ \begin{align*} \tilde{N_{p^2}}(\mathcal L,b) = pN_p(\mathcal L,b) - E, \end{align*} $$

where E is either $0$ or $1$ . Thus,

$$ \begin{align*} \prod_{\substack{p \mid W}} \frac{1-\tilde{N_{p^2}}(\mathcal L,b)/p^2}{(1-1/p)^{K+1}} \le \prod_{\substack{p \mid W}} \frac{1-N_p(\mathcal L,b)/p + 1/p^2}{(1-1/p)^{K+1}}, \end{align*} $$

since the numerator of each factor in the product is either unchanged or has increased.

We can then rewrite the right-hand side as

$$ \begin{align*} &\prod_{\substack{p|W}} \frac{1-N_p(\mathcal L,b)/p + 1/p^2}{(1-1/p)^{K+1}} = \prod_{\substack{p|W \\ p \le K+1}} \frac{1-N_p(\mathcal L,b)/p + 1/p^2}{(1-1/p)^{K+1}} \prod_{\substack{p|W \\ p> K+1}} \frac{1-N_p(\mathcal L,b)/p + 1/p^2}{(1-1/p)^{K+1}} \\ &\qquad \qquad \qquad \le \prod_{\substack{p \le K+1}} \frac{1-1/p}{(1-1/p)^{K+1}}\prod_{\substack{p|W \\ p > K+1}} \frac{(1-N_p(\mathcal L,b)/p)}{(1-1/p)^{K+1}} \left(1 + \frac 1{p(p-N_p(\mathcal L,b))} \right)\\ &\qquad \qquad \qquad \ll_K \prod_{\substack{p|W \\ p > K+1}} \frac{(1-N_p(\mathcal L,b)/p)}{(1-1/p)^{K+1}} \prod_{\substack{p|W \\ p > K+1}} \left(1+\frac 1{p(p-K-1)}\right). \end{align*} $$

The second Euler product converges to a constant dependent only on K when extended over all primes $p> K+1$ , which completes the proof.

Proof of Proposition 22.

We begin by applying Lemma 23 to bound the left-hand side by

$$ \begin{align*} \ll_K \sum_{\substack{\mathbf b \in \mathcal B \\ \mathcal L(\mathbf b) \text{adm.}}} \sum_{\substack{b \le \eta \sqrt{\log x} \\ qn + b \not\in \mathcal L}} \prod_{\substack{p|W \\ p> K +1}} \frac{1-N_p(\mathcal L,b)/p}{(1-1/p)^{K+1}}, \end{align*} $$

where $N_p(\mathcal L,b)$ is the number of congruence classes $\nu \bmod p$ such that $p|\ell (\nu )$ for some $\ell \in \mathcal L$ or such that $p|\ell ^b(\nu )$ .

Let $\Delta (\mathcal L, b)$ denote the product

$$ \begin{align*} \Delta(\mathcal L, b) := \prod_{1 \le i_1 < i_2 \le K}(q(b_{i_2}-b_{i_1}) + a_{i_2}-a_{i_1}) \prod_{1 \le i \le K} (b-qb_i - a_i). \end{align*} $$

Thus, $1 \le N_p(\mathcal L,b) \le K+1$ , with equality on the right if and only if $p \nmid \Delta (\mathcal L,b)$ . Define $a(p,N_p)$ via

$$ \begin{align*} \frac{1-N_p/p}{(1-1/p)^{K+1}} = 1 + a(p,N_p), \end{align*} $$

and for squarefree r, define $a_{\mathcal L,b}(r)$ multiplicatively via $ a_{\mathcal L,b}(r) = \prod _{p|r} a(p,N_p(\mathcal L,b)), $ so that

$$ \begin{align*} \prod_{p|W}\frac{1-N_p(\mathcal L,b)/p}{(1-1/p)^{K+1}} = \sum_{r|W} a_{\mathcal L,b}(r). \end{align*} $$

By the same reasoning as in Gallagher’s proof of equation (3) in [Reference Gallagher5], for a constant x to be fixed later and for all $\varepsilon> 0$ , we have

(57) $$ \begin{align} &\sum_{\substack{\mathbf b \in \mathcal B \\ \mathcal L(\mathbf b) \text{ adm.}}} \sum_{\substack{b \le \eta \sqrt{\log x} \\ qn + b \not\in \mathcal L}} \prod_{\substack{p|W \\ p> K +1}} \frac{1-N_p(\mathcal L,b)/p}{(1-1/p)^{K+1}} \\ &\qquad = \sum_{r \le x} \sum_{\substack{\mathbf b \in \mathcal B \\ \mathcal L(\mathbf b) \text{ adm.}}} \sum_{\substack{b \le \eta \sqrt{\log x} \\ qn + b \not\in \mathcal L}} a_{\mathcal L,b}(r) + O_{K,\varepsilon}(\eta^K(\log x)^{K/2}(x\eta\log x)^{\varepsilon}/x).\notag \end{align} $$

The inner sums over $\mathbf b \in \mathcal B$ and b are equal to

$$ \begin{align*} \sum_{(N_p)_p} \prod_{\substack{p|r \\ p>K+1}} a(p,N_p) \left\{ \sideset{}{'}\sum 1 + O((\eta\sqrt{\log x})^{K-1})\right\}, \end{align*} $$

where each $(N_p)_p$ is a vector with positive integer entries for each prime $p|r$ with $p>K+1$ , and where $\sideset {}{'}\sum $ denotes the number of ways to choose values $b_2, \dots , b_K$ and b, not necessarily distinct, such that each $b_i \equiv 3 \bmod 4$ , such that $1 < qb_i \le \frac {\eta }{2}\sqrt {\log x}$ for $2 \le i \le jM$ , such that $\frac {\eta }{2}< qb_i \le \eta \sqrt {\log x}$ for $jM+1 \le i \le K$ , such that $b \le \eta \sqrt {\log x}$ , and most crucially, such that $b_1, \dots , b_K, b$ occupy precisely $N_p$ congruence classes modulo p for each $p|r$ . Recall that $b_1 = 3$ is fixed for all $\mathbf b \in \mathcal B$ .

By the Chinese remainder theorem, for $r \le \eta \sqrt {\log x}$ ,

$$ \begin{align*} \sideset{}{'}\sum = \left\{\left(\frac{\eta\sqrt{\log x}}{8qr}\right)^{K-1}\left(\frac{\eta\sqrt{\log x}}r\right) + O\left(\frac{\eta\sqrt{\log x}}{r}\right)^{K-1}\right\} \times \prod_{p|r} \binom{p-1}{N_p-1}\sigma(K,N_p), \end{align*} $$

where $\sigma (K,N_p)$ denotes the number of surjective maps from $\{1,\dots , K\}$ onto $\{1,\dots ,N_p\}$ . Thus, the inner sum is

$$ \begin{align*} \left(\frac{\eta\sqrt{\log x}}{8qr}\right)^{K-1}\left(\frac{\eta\sqrt{\log x}}{r}\right)A(r) + O\left(\left(\frac{\eta\sqrt{\log x}}{r}\right)^{K-1}B(r)\right) + O\left((\eta\sqrt{\log x})^{K-1}C(r)\right), \end{align*} $$

where

$$ \begin{align*} A(r) &= \sum_{(N_p)_{p|r}} \prod_{p|r} a(p,N_p)\binom{p}{N_p} \sigma(K,N_p), \\ B(r) &= \sum_{(N_p)_{p|r}} \prod_{p|r} |a(p,N_p)|\binom{p}{N_p} \sigma(K,N_p), \text{ and} \\ C(r) &= \sum_{(N_p)_{p|r}} \prod_{p|r} |a(p,N_p)|. \end{align*} $$

One can show via a combinatorial argument (identical to the one performed in [Reference Gallagher5]) that $A(r) = 0$ whenever $r> 1$ . Also by the same arguments as in [Reference Gallagher5], we have $B(r) \le C^{\omega (r)}\frac {r^K}{\phi (r)}$ and $C(r) \le C^{\omega (r)}\frac {r}{\phi (r)}$ for a suitable constant C.

Altogether, we get that

$$ \begin{align*} \sum_{\substack{\mathbf b \in \mathcal B \\ \mathcal L(\mathbf b) \text{ adm.}}} &\sum_{\substack{b \le \eta \sqrt{\log x} \\ qn + b \not\in \mathcal L}} \prod_{\substack{p|W \\ p> K +1}} \frac{1-N_p(\mathcal L,b)/p}{(1-1/p)^{K+1}} \\&= \left(\frac{\eta\sqrt{\log x}}{8q}\right)^{K-1}(\eta\sqrt{\log x}) + O\left((\eta\sqrt{\log x})^{K-1} \sum_{r \le x} \frac{C^{\omega(r)}r}{\phi(r)}\right) \\& \quad + O(\eta^K(\log x)^{K/2}(x\eta\log x)^{\varepsilon}/x) \\&= \frac{(\eta\sqrt{\log x})^{K}}{(8q)^{K-1}} +O_{K,\varepsilon,q}((\eta\sqrt{\log x})^{K-1/2 + \varepsilon}), \end{align*} $$

choosing $x = (\eta \sqrt {\log x)})^{1/2}.$ This completes the proof.