Proof. This is inspired by and similar to [Reference Matomäki and Shao15, Theorem 2.3] of Matomäki and Shao. See also [Reference Green and Tao4, Proposition 5.1], where similar ideas were applied for Roth-type problems. Take $\unicode[STIX]{x1D6FA}=\{\unicode[STIX]{x1D709}\in \mathbb{Z}_{N}:|\widehat{f_{1}}(\unicode[STIX]{x1D709})|\geqslant \unicode[STIX]{x1D700}_{0}\}\cup \{1\}$ , where $\unicode[STIX]{x1D700}_{0}$ will be chosen small enough in terms of $\unicode[STIX]{x1D6FF}$ , $\unicode[STIX]{x1D700}$ and $K_{0}$ . Condition (iii) tells us that $|\unicode[STIX]{x1D6FA}|\leqslant K_{0}\unicode[STIX]{x1D700}_{0}^{-3}+1$ . Let $\unicode[STIX]{x1D712}=\unicode[STIX]{x1D712}_{\unicode[STIX]{x1D6FA},\unicode[STIX]{x1D702}}:\mathbb{Z}\rightarrow \mathbb{R}_{{\geqslant}0}$ be as in the proposition (so that $\unicode[STIX]{x1D712}$ fulfills (2.1)). We will later choose $\unicode[STIX]{x1D702}$ to be small enough in terms of $\unicode[STIX]{x1D6FF}$ , $\unicode[STIX]{x1D700}$ and $K_{0}$ . Introduce the functions

$$\begin{eqnarray}\displaystyle g_{2}=\frac{1}{\Vert \unicode[STIX]{x1D712}\Vert _{1}}f_{2}\ast \unicode[STIX]{x1D712}\quad \text{and}\quad h_{2}=f_{2}-g_{2}. & & \displaystyle \nonumber\end{eqnarray}$$

We have

$$\begin{eqnarray}\displaystyle \widehat{g_{2}}=\frac{1}{\Vert \unicode[STIX]{x1D712}\Vert _{1}}\widehat{f_{2}}\widehat{\unicode[STIX]{x1D712}}\quad \text{and}\quad \widehat{h_{2}}=\widehat{f_{2}}\biggl(1-\frac{\widehat{\unicode[STIX]{x1D712}}}{\Vert \unicode[STIX]{x1D712}\Vert _{1}}\biggr), & & \displaystyle \nonumber\end{eqnarray}$$

so that in particular $|\widehat{h_{2}}(\unicode[STIX]{x1D709})|\leqslant 2|\widehat{f_{2}}(\unicode[STIX]{x1D709})|$ .

Next we estimate from above and below the average $(1/N)\sum _{n\in \mathbb{Z}_{N}}|f_{1}\ast h_{2}(n)|^{2}$ , starting with the lower bound. Owing to conditions (i) and (ii), for $n\in [0.9N,N]$ , we have

(2.3) $$\begin{eqnarray}\displaystyle f_{1}\ast g_{2}(n)=\frac{1}{\Vert \unicode[STIX]{x1D712}\Vert _{1}}f_{2}\ast \unicode[STIX]{x1D712}\ast f_{1}(n)\geqslant \frac{\unicode[STIX]{x1D6FF}}{N}\mathop{\sum }_{\substack{ n-2N/3<k<n-N/3 \\ k\in \mathbb{Z}_{N}}}f_{1}(k)\geqslant \unicode[STIX]{x1D6FF}^{2}\quad & & \displaystyle\end{eqnarray}$$

since $(N/3,N/2)\subset (n-2N/3,n-N/3)$ for $n\in [0.9N,N]$ . Denoting $T=\{n\in [0.9N,N]:f_{1}\ast f_{2}(n)<\unicode[STIX]{x1D6FF}^{2}/3\}$ and using the simple inequality $|a-b|^{2}\geqslant a^{2}/2-b^{2}$ and (2.3), we infer that

(2.4) $$\begin{eqnarray}\displaystyle \frac{1}{N}\mathop{\sum }_{n\in \mathbb{Z}_{N}}|f_{1}\ast h_{2}(n)|^{2} & {\geqslant} & \displaystyle \frac{1}{N}\mathop{\sum }_{n\in T}\biggl(\frac{1}{2}|f_{1}\ast g_{2}(n)|^{2}-|f_{1}\ast f_{2}(n)|^{2}\biggr)\nonumber\\ \displaystyle & {\geqslant} & \displaystyle \biggl(\frac{\unicode[STIX]{x1D6FF}^{4}}{2}-\biggl(\frac{\unicode[STIX]{x1D6FF}^{2}}{3}\biggr)^{2}\biggr)\frac{|T|}{N}\geqslant \frac{\unicode[STIX]{x1D6FF}^{4}}{10}\frac{|T|}{N}.\end{eqnarray}$$

When it comes to an upper bound, Parseval’s identity gives

$$\begin{eqnarray}\displaystyle \frac{1}{N}\mathop{\sum }_{n\in \mathbb{Z}_{N}}|f_{1}\ast h_{2}(n)|^{2} & = & \displaystyle \mathop{\sum }_{\unicode[STIX]{x1D709}\in \mathbb{Z}_{N}}|\widehat{f_{1}\ast h_{2}}(\unicode[STIX]{x1D709})|^{2}\nonumber\\ \displaystyle & = & \displaystyle \mathop{\sum }_{\unicode[STIX]{x1D709}\in \mathbb{Z}_{N}}|\widehat{f_{1}}(\unicode[STIX]{x1D709})\widehat{h_{2}}(\unicode[STIX]{x1D709})|^{2}\nonumber\\ \displaystyle & {\leqslant} & \displaystyle \unicode[STIX]{x1D700}_{0}^{1/2}\mathop{\sum }_{\unicode[STIX]{x1D709}\not \in \unicode[STIX]{x1D6FA}}|\widehat{f_{1}}(\unicode[STIX]{x1D709})|^{3/2}|\widehat{h_{2}}(\unicode[STIX]{x1D709})|^{2}+\mathop{\sum }_{\unicode[STIX]{x1D709}\in \unicode[STIX]{x1D6FA}}|\widehat{f_{1}}(\unicode[STIX]{x1D709})|^{2}|\widehat{h_{2}}(\unicode[STIX]{x1D709})|^{2}.\nonumber\end{eqnarray}$$

Here the first sum can be bounded with the Cauchy–Schwarz inequality and (iii), implying that

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D700}_{0}^{1/2}\mathop{\sum }_{\unicode[STIX]{x1D709}\not \in \unicode[STIX]{x1D6FA}}|\widehat{f_{1}}(\unicode[STIX]{x1D709})|^{3/2}|\widehat{h_{2}}(\unicode[STIX]{x1D709})|^{2} & {\leqslant} & \displaystyle \unicode[STIX]{x1D700}_{0}^{1/2}\biggl(\mathop{\sum }_{\unicode[STIX]{x1D709}\in \mathbb{Z}_{N}}|\widehat{f_{1}}(\unicode[STIX]{x1D709})|^{3}\biggr)^{1/2}\biggl(\mathop{\sum }_{\unicode[STIX]{x1D709}\in \mathbb{Z}_{N}}|\widehat{h_{2}}(\unicode[STIX]{x1D709})|^{4}\biggr)^{1/2}\nonumber\\ \displaystyle & {\leqslant} & \displaystyle 8\unicode[STIX]{x1D700}_{0}^{1/2}K_{0}.\nonumber\end{eqnarray}$$

The sum over $\unicode[STIX]{x1D709}\in \unicode[STIX]{x1D6FA}$ in turn can be bounded by using the fact that

$$\begin{eqnarray}\displaystyle \biggl|1-\frac{\widehat{\unicode[STIX]{x1D712}}(\unicode[STIX]{x1D709})}{\Vert \unicode[STIX]{x1D712}\Vert _{1}}\biggr|\leqslant 30\unicode[STIX]{x1D702}\quad \text{for every}~\unicode[STIX]{x1D709}\in \unicode[STIX]{x1D6FA}, & & \displaystyle \nonumber\end{eqnarray}$$

the proof of which is contained in the proof of Theorem 2.3 in [Reference Matomäki and Shao15, §4]. After this, we may again use the Cauchy–Schwarz inequality and (iii) to obtain

$$\begin{eqnarray}\displaystyle \mathop{\sum }_{\unicode[STIX]{x1D709}\in \unicode[STIX]{x1D6FA}}|\widehat{f_{1}}(\unicode[STIX]{x1D709})|^{2}|\widehat{h_{2}}(\unicode[STIX]{x1D709})|^{2} & {\leqslant} & \displaystyle (30\unicode[STIX]{x1D702})^{2}\mathop{\sum }_{\unicode[STIX]{x1D709}\in \unicode[STIX]{x1D6FA}}|\widehat{f_{1}}(\unicode[STIX]{x1D709})|^{2}|\widehat{f_{2}}(\unicode[STIX]{x1D709})|^{2}\nonumber\\ \displaystyle & {\leqslant} & \displaystyle 1000\unicode[STIX]{x1D702}^{2}K_{0}.\nonumber\end{eqnarray}$$

At this stage, we fix the choices $\unicode[STIX]{x1D700}_{0}=\unicode[STIX]{x1D702}=\unicode[STIX]{x1D6FF}^{8}\unicode[STIX]{x1D700}^{2}/10^{4}K_{0}^{2}$ , so that

(2.5) $$\begin{eqnarray}\displaystyle \frac{1}{N}\mathop{\sum }_{n\in \mathbb{Z}_{N}}|f_{1}\ast h_{2}(n)|^{2}\leqslant 8\unicode[STIX]{x1D700}_{0}^{1/2}K_{0}+1000\unicode[STIX]{x1D702}^{2}K_{0}\leqslant \frac{1}{10}\unicode[STIX]{x1D6FF}^{4}\unicode[STIX]{x1D700}. & & \displaystyle\end{eqnarray}$$

Combining (2.4) and (2.5) above, we discover that $|T|\leqslant 10\unicode[STIX]{x1D6FF}^{-4}\cdot {\textstyle \frac{1}{10}}\unicode[STIX]{x1D6FF}^{4}\unicode[STIX]{x1D700}N=\unicode[STIX]{x1D700}N$ , which concludes the proof.◻

Proof. Since $2^{J-1}\nmid n$ , we may write $n=2^{g}s$ , where $0\leqslant g\leqslant J-5$ and $s\not \equiv 0~(\text{mod}~16)$ . It is easy to check that every such $s$ may be written as $s=a^{\prime }+b^{\prime }$ with $a^{\prime }\equiv 2^{i}~(\text{mod}~2^{i+2})$ , $b^{\prime }\equiv 2^{j}~(\text{mod}~2^{j+2})$ for some $0\leqslant i,j\leqslant 3$ . Then $n=a+b$ with $a=2^{g}a^{\prime }$ , $b=2^{g}b^{\prime }$ is a representation of the desired form.◻

Proof. One easily sees that $x^{2}+y^{2}~(\text{mod}~27)$ attains all residue classes except those that are $\equiv 3~(\text{mod}~9)$ or $\equiv 6~(\text{mod}~9)$ as $x$ and $y$ vary. Now the lemma only states that every $m^{\prime }\not \equiv 3,6~(\text{mod}~9)$ is the sum of two numbers, each of which is $0,2,5$ or $8~(\text{mod}~9)$ and neither of which is $0~(\text{mod}~27)$ . This can quickly be verified by hand.◻

Proof of Theorem 1.1 assuming Propositions 3.2 and 3.3.

Given any small $\unicode[STIX]{x1D700}>0$ , we must show that once $N$ is large enough, the interval $[0.9N,N]$ contains at most $\unicode[STIX]{x1D700}N$ integers $m\equiv 0~(\text{mod}~2)$ , $m\not \equiv 5,8~(\text{mod}~9)$ that cannot be written as $m=p+q$ with $p$ and $q$ primes of the form $x^{2}+y^{2}+1$ .

Let $U$ and $W$ be given by (3.3) with $J=\lfloor 10/\unicode[STIX]{x1D700}\rfloor$ and $w\ll 1$ large enough. We start by showing that for any $m\in [0.9N,N]$ , $m\equiv 0~(\text{mod}~2)$ , $m\not \equiv 5,8~(\text{mod}~9)$ , $m\not \equiv 2~(\text{mod}~2^{J})$ , we may find integers $0\leqslant B_{1},B_{2}\leqslant W-1$ such that $m=B_{1}+B_{2}$ and the linear functions $Wn+B_{1}$ and $Wn+B_{2}$ are amenable. The integers $m\equiv 2~(\text{mod}~2^{J})$ can be disposed of since there are ${\leqslant}(\unicode[STIX]{x1D700}^{2}/10)N$ such integers up to $N$ .

To see that $B_{1}$ and $B_{2}$ exist, write $m=2m^{\prime }+2$ , so that $m^{\prime }\not \equiv 3,6~(\text{mod}~9)$ . Then $2^{J-1}\nmid m^{\prime }$ , so using Lemma 3.4 we may write $m^{\prime }\equiv a_{1}+a_{2}~(\text{mod}~2^{J})$ with $a_{1}\equiv 2^{i}~(\text{mod}~2^{i+2})$ , $a_{2}\equiv 2^{j}~(\text{mod}~2^{j+2})$ for some $0\leqslant i,j\leqslant J-3$ . Moreover, using Lemma 3.5, we may write $m^{\prime }\equiv a_{1}^{\prime }+a_{2}^{\prime }~(\text{mod}~3^{3})$ with $a_{1}^{\prime }$ and $a_{2}^{\prime }$ numbers such that $3^{3}\nmid a_{1}^{\prime }$ , $3^{3}\nmid a_{2}^{\prime }$ , $2a_{1}^{\prime }+1,2a_{2}+1^{\prime }\not \equiv 0~(\text{mod}~3)$ and the largest powers of $3$ dividing $a_{1}^{\prime }$ and $a_{2}^{\prime }$ have even exponents (take $a_{1}^{\prime }=x_{1}^{2}+x_{2}^{2}$ and $a_{2}^{\prime }=x_{3}^{2}+x_{4}^{2}$ in that lemma and notice that the largest power of $3$ dividing $x^{2}+y^{2}$ has an even exponent).

Now pick numbers $b_{p}$ for $5\leqslant p\leqslant w$ such that $b_{p}\not \equiv 0,1,m,m-1~(\text{mod}~p)$ . By the Chinese remainder theorem, we can find an integer $B$ such that $B\equiv 2a_{1}+1~(\text{mod}~2^{J})$ , $B\equiv 2a_{1}^{\prime }+1~(\text{mod}~3^{3})$ and $B\equiv b_{p}~(\text{mod}~p)$ for all $5\leqslant p<w$ . Therefore, we have found some integers $B_{1}:=B$ and $B_{2}:=m-B$ such that $m=B_{1}+B_{2}$ , $p\nmid B_{i}$ , $p\nmid B_{i}-1$ for $5\leqslant p<w$ , and $B_{1}-1$ and $B_{2}-1$ satisfy condition (iii) in the definition of amenability.

Therefore, we have a representation of any $m$ of the form above as

$$\begin{eqnarray}\displaystyle m\equiv B_{1}(m)+B_{2}(m)~(\text{mod}~W) & & \displaystyle \nonumber\end{eqnarray}$$

with $Wn+B_{1}(m)$ , $Wn+B_{2}(m)$ amenable linear functions and $0\leqslant B_{i}(m)\leqslant W-1$ (we use the notation $B_{i}(m)$ to emphasize that the $B_{i}$ depend on $m~(\text{mod}~W)$ ). For each $0\leqslant a\leqslant W-1$ , we denote

$$\begin{eqnarray}\displaystyle {\mathcal{B}}_{a}=\{m\in [0.9N,N]:m\equiv a~(\text{mod}~W)\}. & & \displaystyle \nonumber\end{eqnarray}$$

We will show that each ${\mathcal{B}}_{a}$ with $a\equiv 0~(\text{mod}~2)$ , $a\not \equiv 5,8~(\text{mod}~9)$ , $a\not \equiv 2~(\text{mod}~2^{J})$ contains at most $\unicode[STIX]{x1D700}(N/2W)$ values of $m\in [0.9N,N]$ that are not of the form $p+q$ with $p$ and $q$ primes of the form $x^{2}+y^{2}+1$ , and afterwards we sum this result over $a$ .

If $a$ satisfies the congruence conditions above, the polynomials $Wn+B_{1}(a)$ and $Wn+B_{2}(a)$ are amenable linear polynomials. Set $M^{\prime }=\lfloor N/W\rfloor$ and, for $\ell \in \{1,2\}$ , set

$$\begin{eqnarray}\displaystyle f_{\ell }(n) & = & \displaystyle (\log N)^{3/2}\biggl(\frac{\unicode[STIX]{x1D711}(W)}{W}\biggr)^{3/2}1_{Wn+B_{\ell }(a)\in \mathbb{P},~Wn+B_{\ell }(a)-1\in {\mathcal{S}}}\nonumber\\ \displaystyle & & \displaystyle \text{for}~n\in \biggl(\frac{M^{\prime }}{3},\frac{2M^{\prime }}{3}\biggr)\nonumber\end{eqnarray}$$

with ${\mathcal{S}}$ as in (3.2) and let $f_{\ell }(n)=0$ for $n\in [0,M^{\prime })\setminus (M^{\prime }/3,2M^{\prime }/3)$ .

Concerning condition (ii) of the transference-type result, applying Proposition 3.2 to the function $\unicode[STIX]{x1D712}\equiv 1$ , we see that

$$\begin{eqnarray}\displaystyle \mathop{\sum }_{M^{\prime }/3<n<2M^{\prime }/3}f_{1}(n)\geqslant \frac{\unicode[STIX]{x1D6FF}_{0}}{10}M^{\prime }, & & \displaystyle \nonumber\end{eqnarray}$$

but we evidently get the same outcome with summation over $M^{\prime }/3<n<M^{\prime }/2$ (since one could clearly replace $n\sim N/3$ with $N/3<n<N/2$ in Proposition 3.2). This takes care of condition (ii).

Next, by Proposition 3.3,

$$\begin{eqnarray}\displaystyle \mathop{\sum }_{\unicode[STIX]{x1D709}\in \mathbb{Z}_{M^{\prime }}}|\widehat{f_{\ell }}(\unicode[STIX]{x1D709})|^{r}\leqslant K_{0} & & \displaystyle \nonumber\end{eqnarray}$$

for some absolute constant $K_{0}$ when $r\in \{3,4\}$ , so condition (iii) also holds.

Let then $\unicode[STIX]{x1D712}=\unicode[STIX]{x1D712}_{\unicode[STIX]{x1D6FA},\unicode[STIX]{x1D702}}:\mathbb{Z}_{M^{\prime }}\rightarrow \mathbb{R}_{{\geqslant}0}$ be as in Proposition 2.1 (with $\unicode[STIX]{x1D712}$ depending on $K_{0}$ and $\unicode[STIX]{x1D6FF}_{0}$ that appeared above), where $\unicode[STIX]{x1D6FA}\subset \mathbb{Z}_{M^{\prime }}$ satisfies $1\in \unicode[STIX]{x1D6FA}$ , $|\unicode[STIX]{x1D6FA}|\ll _{\unicode[STIX]{x1D700}}1$ and $1\ll _{\unicode[STIX]{x1D700}}\unicode[STIX]{x1D702}\leqslant 0.05$ . According to (2.1), $\unicode[STIX]{x1D712}$ is symmetric around the origin and

$$\begin{eqnarray}\displaystyle \mathop{\sum }_{\substack{ n\in [-M^{\prime }/2,M^{\prime }/2] \\ |n|\geqslant 0.1M^{\prime }}}\unicode[STIX]{x1D712}(n)\leqslant \biggl(\frac{\unicode[STIX]{x1D702}^{2}}{8}\biggr)^{|\unicode[STIX]{x1D6FA}|}M^{\prime }\leqslant \unicode[STIX]{x1D702}\biggl(\frac{\unicode[STIX]{x1D702}}{2}\biggr)^{|\unicode[STIX]{x1D6FA}|}M^{\prime }\leqslant 0.05\Vert \unicode[STIX]{x1D712}_{1}\Vert M^{\prime }. & & \displaystyle \nonumber\end{eqnarray}$$

Keeping this in mind and using Proposition 3.2, for $t\in (M^{\prime }/3,2M^{\prime }/3)$ we obtain

$$\begin{eqnarray}\displaystyle \mathop{\sum }_{n\sim M^{\prime }/3}f_{2}(n)\unicode[STIX]{x1D712}(t-n) & {\geqslant} & \displaystyle \unicode[STIX]{x1D6FF}_{0}\bigg(\mathop{\sum }_{n\sim M^{\prime }/3}\unicode[STIX]{x1D712}(t-n)-\frac{CM^{\prime }}{w^{1/3}}\bigg)\nonumber\\ \displaystyle & {\geqslant} & \displaystyle \frac{\unicode[STIX]{x1D6FF}_{0}}{10}\bigg(\mathop{\sum }_{n\in \mathbb{Z}_{M^{\prime }}}\unicode[STIX]{x1D712}(t-n)-\frac{CM^{\prime }}{w^{1/3}}\bigg)\nonumber\\ \displaystyle & {\geqslant} & \displaystyle \frac{\unicode[STIX]{x1D6FF}_{0}}{20}M^{\prime }\Vert \unicode[STIX]{x1D712}\Vert _{1}\nonumber\end{eqnarray}$$

for $w$ large enough, the final step coming from (2.1), since

$$\begin{eqnarray}\displaystyle \Vert \unicode[STIX]{x1D712}\Vert _{1}\geqslant \biggl(\frac{\unicode[STIX]{x1D702}}{2}\biggr)^{|\unicode[STIX]{x1D6FA}|}\geqslant \frac{1}{w^{0.1}} & & \displaystyle \nonumber\end{eqnarray}$$

for $w$ large enough. This means that condition (i) of the transference-type result holds with $\unicode[STIX]{x1D6FF}=\unicode[STIX]{x1D6FF}_{0}/20$ .

From the transference-type result (Proposition 2.1), we conclude that $f_{1}\ast f_{2}(n)>0$ for all $n\in [0.9M^{\prime },M^{\prime }]$ , $n\not \in T_{a}$ , where $T_{a}$ is some set of integers with $|T_{a}|\leqslant (\unicode[STIX]{x1D700}/2)M^{\prime }=\unicode[STIX]{x1D700}(N/2W)$ . This leads to $n\equiv n_{1}+n_{2}~(\text{mod}~M^{\prime })$ with

(3.5) $$\begin{eqnarray}\displaystyle Wn_{i}+B_{i}(a)\in \mathbb{P},\qquad Wn_{i}+B_{i}(a)-1\in {\mathcal{S}} & & \displaystyle\end{eqnarray}$$

for $n\in [0.9M^{\prime },M^{\prime }]$ , $n\not \in T_{a}$ . Since $n_{1},n_{2}\in (M^{\prime }/3,2M^{\prime }/3)$ , we can actually say that $n=n_{1}+n_{2}$ . What we showed at the beginning of the proof is that any $m\in {\mathcal{B}}_{a}$ , $m\in [0.9N+2W,N]$ with $m\equiv 0~(\text{mod}~2)$ , $m\not \equiv 5,8~(\text{mod}~9)$ and $m\not \equiv 2~(\text{mod}~2^{J})$ can be written as $m=Wn+B_{1}(a)+B_{2}(a)$ with $n\in [0.9M^{\prime },M^{\prime }]$ and $Wn+B_{1}(a)$ and $Wn+B_{2}(a)$ amenable (the interval $[0.9N,0.9N+2W]$ contains ${\leqslant}(\unicode[STIX]{x1D700}^{2}/10)N$ numbers and can hence be ignored). Then

$$\begin{eqnarray}\displaystyle m=(Wn_{1}+B_{1}(a))+(Wn_{2}+B_{2}(a)) & & \displaystyle \nonumber\end{eqnarray}$$

for some $n_{1}$ and $n_{2}$ satisfying (3.5) whenever $m\in {\mathcal{B}}_{a}\setminus T_{a}^{\prime }$ , $m\in [0.9N+2W,N]$ , $m\equiv 0~(\text{mod}~2)$ , $m\not \equiv 5,8~(\text{mod}~9)$ and $m\not \equiv 2~(\text{mod}~2^{J})$ , where $T_{a}^{\prime }=\{a+W\unicode[STIX]{x1D70F}:\unicode[STIX]{x1D70F}\in T_{a}\}$ satisfies $|T_{a}^{\prime }|\leqslant \unicode[STIX]{x1D700}(N/2W)$ . Since

$$\begin{eqnarray}\displaystyle \mathop{\sum }_{\substack{ 0\leqslant a\leqslant W-1 \\ a\equiv 0~(\text{mod}~2) \\ a\not \equiv 5,8~(\text{mod}~9) \\ a\not \equiv 2~(\text{mod}~2^{J})}}|T_{a}|\leqslant W\cdot \unicode[STIX]{x1D700}\frac{N}{2W}=\frac{\unicode[STIX]{x1D700}}{2}N, & & \displaystyle \nonumber\end{eqnarray}$$

we conclude that all but ${\leqslant}(\unicode[STIX]{x1D700}/2+\unicode[STIX]{x1D700}^{2})N\leqslant \unicode[STIX]{x1D700}N$ even integers $m\in [0.9N,N]$ satisfying $m\not \equiv 5,8~(\text{mod}~9)$ can be written as $m=p+q$ with $p,q$ primes of the form $x^{2}+y^{2}+1$ .◻

Proof of Proposition 4.2 assuming Proposition 4.1.

Our function $\unicode[STIX]{x1D6FD}(\cdot )$ fulfills the same axioms as in the paper of Green and Tao (except the pointwise lower bound, which is not used for the proof of [Reference Green and Tao4, Proposition 4.2]). Therefore, the proof of [Reference Green and Tao4, Proposition 4.2] goes through in this setting.◻

Proof of Proposition 3.3 assuming Proposition 4.1.

We already know that if Proposition 4.1 is true, so is Proposition 4.2. We choose $a_{n}=g(n)/\unicode[STIX]{x1D6FD}(n)$ whenever $\unicode[STIX]{x1D6FD}(n)\neq 0$ and $a_{n}=0$ otherwise. Since $0\leqslant g(n)\leqslant f(n)\leqslant \unicode[STIX]{x1D705}_{1}^{-1}\unicode[STIX]{x1D6FD}(n)$ in the notation of Proposition 3.3, from Proposition 4.2 we immediately derive

$$\begin{eqnarray}\displaystyle \biggl(\mathop{\sum }_{\unicode[STIX]{x1D709}\in \mathbb{Z}_{N}}|\widehat{g}(\unicode[STIX]{x1D709})|^{r}\biggr)^{1/r} & {\leqslant} & \displaystyle C_{r}\biggl(\frac{1}{N}\mathop{\sum }_{\substack{ n\leqslant N \\ \unicode[STIX]{x1D6FD}(n)\neq 0}}\frac{g(n)^{2}}{\unicode[STIX]{x1D6FD}(n)}\biggr)^{1/2}\nonumber\\ \displaystyle & {\leqslant} & \displaystyle C_{r}\biggl(\frac{\unicode[STIX]{x1D705}_{1}^{-2}}{N}\mathop{\sum }_{n\leqslant N}\unicode[STIX]{x1D6FD}(n)\biggr)^{1/2}\leqslant C_{r}\unicode[STIX]{x1D705}_{1}^{-1}\unicode[STIX]{x1D705}_{2}^{1/2}\nonumber\end{eqnarray}$$

by part (ii) of Proposition 4.1. ◻

Proof of Proposition 4.1.

We first introduce some notation. For a prime $p$ , let ${\mathcal{A}}_{p}\subset \mathbb{Z}_{p}$ denote the residue classes $(\text{mod}~p)$ that are sifted away when looking for primes of the form $x^{2}+y^{2}+1\equiv B~(\text{mod}~W)$ . In other words,

$$\begin{eqnarray}\displaystyle {\mathcal{A}}_{p}=\left\{\begin{array}{@{}ll@{}}\emptyset \quad & \text{for}~p\leqslant w,\\ \{0\}\quad & \text{for}~p\equiv 1~(\text{mod}~4),~p>w,\\ \{0,1\}\quad & \text{for}~p\equiv -1~(\text{mod}~4),~p>w.\end{array}\right. & & \displaystyle \nonumber\end{eqnarray}$$

Further, for square-free $d$ , let

$$\begin{eqnarray}\displaystyle {\mathcal{A}}_{d}=\mathop{\bigcap }_{p\mid d}{\mathcal{A}}_{p}, & & \displaystyle \nonumber\end{eqnarray}$$

where ${\mathcal{A}}_{d}$ is interpreted as a subset of $\mathbb{Z}_{d}$ . Set also ${\mathcal{A}}_{1}=\mathbb{Z}_{1}$ and ${\mathcal{A}}_{d}=\emptyset$ when $d$ is not square-free. For $d\geqslant 2$ , we have $|{\mathcal{A}}_{d}|=\unicode[STIX]{x1D714}(d)$ , where $\unicode[STIX]{x1D714}(\cdot )$ is a multiplicative function supported on the square-free integers and having the values

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D714}(p)=\left\{\begin{array}{@{}ll@{}}0\quad & \text{for}~p\leqslant w,\\ 1\quad & \text{for}~p\equiv 1~(\text{mod}~4),~p>w,\\ 2\quad & \text{for}~p\equiv -1~(\text{mod}~4),~p>w.\end{array}\right. & & \displaystyle \nonumber\end{eqnarray}$$

For later use, we also define

(4.2) $$\begin{eqnarray}\displaystyle {\mathcal{K}}_{1}=\mathbb{Z}_{1},\qquad {\mathcal{K}}_{p}=\mathbb{Z}_{p}\setminus {\mathcal{A}}_{p},\qquad {\mathcal{K}}_{d}=\mathop{\bigcap }_{p\mid d}{\mathcal{K}}_{p}\quad \text{for}~\unicode[STIX]{x1D707}(d)^{2}=1 & & \displaystyle\end{eqnarray}$$

and let ${\mathcal{K}}_{d}=\mathbb{Z}_{d}$ for $\unicode[STIX]{x1D707}(d)=0$ .

Let the Selberg sieve coefficients $\unicode[STIX]{x1D70C}_{d}$ (not the same as sieve weights) be given by

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D70C}_{d} & = & \displaystyle \unicode[STIX]{x1D707}(d)\frac{G_{d}(z)}{G_{1}(z)},\quad \text{where}~z=N^{0.1},~G_{d}(z)=\mathop{\sum }_{\substack{ \unicode[STIX]{x1D6FF}\leqslant z \\ [d,\unicode[STIX]{x1D6FF}]\leqslant z}}h(\unicode[STIX]{x1D6FF}),\nonumber\\ \displaystyle h(\unicode[STIX]{x1D6FF}) & = & \displaystyle \mathop{\prod }_{p\mid \unicode[STIX]{x1D6FF}}h(p)\quad \text{and}\quad h(p)=\frac{\unicode[STIX]{x1D714}(p)}{p-\unicode[STIX]{x1D714}(p)}.\nonumber\end{eqnarray}$$

The above notations are otherwise the same as in [Reference Ramaré and Ruzsa18, §4], except that $\unicode[STIX]{x1D706}_{d}$ there has been replaced with $\unicode[STIX]{x1D70C}_{d}$ and ${\mathcal{L}}_{d}$ with ${\mathcal{A}}_{d}$ . We define

(4.3) $$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D6FD}(n)=G_{1}(z)\bigg(\mathop{\sum }_{\substack{ d\mid P(z) \\ Wn+B\in {\mathcal{A}}_{d}}}\unicode[STIX]{x1D70C}_{d}\bigg)^{2}, & & \displaystyle\end{eqnarray}$$

where

$$\begin{eqnarray}\displaystyle P(z)=\mathop{\prod }_{w<p<z}p. & & \displaystyle \nonumber\end{eqnarray}$$

In [Reference Ramaré and Ruzsa18], the factor $G_{1}(z)$ does not appear in their definition of $\unicode[STIX]{x1D6FD}(n)$ , but this is just a normalization constant. In (4.3), the condition $m\in {\mathcal{A}}_{d}$ means that $m~(\text{mod}~d)\in {\mathcal{A}}_{d}$ . Now we can check parts (i)–(v) of Proposition 4.1.

For part (i), first observe that if $Wn+B=x^{2}+y^{2}+1\in \mathbb{P}\cap ({\mathcal{S}}+1)$ with $n\sim N/3$ , then $x^{2}+y^{2}+1\not \equiv 0~(\text{mod}~p)$ for $w<p<z=N^{0.1}$ and $x^{2}+y^{2}\not \equiv 0~(\text{mod}~p)$ for $p\equiv -1~(\text{mod}~4)$ , $w<p<z$ , since $(x,y)\mid 6^{J}$ . This means that if $Wn+B=x^{2}+y^{2}+1\in \mathbb{P}\cap ({\mathcal{S}}+1)$ with $n\sim N/3$ , then $\unicode[STIX]{x1D6FD}(n)=G_{1}(z)$ . Now the assertion follows from

$$\begin{eqnarray}\displaystyle G_{1}(z)\geqslant 10^{-10}\mathop{\prod }_{w<p<z}\biggl(1-\frac{\unicode[STIX]{x1D714}(p)}{p}\biggr)^{-1}\geqslant 10^{-20}(\log N)^{3/2}(\log w)^{-3/2}. & & \displaystyle \nonumber\end{eqnarray}$$

Part (ii) in turn follows by applying the Selberg sieve [Reference Iwaniec and Kowalski10, Ch. 7] to estimate

$$\begin{eqnarray}\displaystyle G_{1}(z)\mathop{\sum }_{n\leqslant N}\bigg(\mathop{\sum }_{\substack{ d\mid P(z) \\ Wn+B\in {\mathcal{A}}_{d}}}\unicode[STIX]{x1D70C}_{d}\bigg)^{2} & {\leqslant} & \displaystyle 10^{10}(\log N)^{3/2}(\log w)^{-3/2}\nonumber\\ \displaystyle & & \displaystyle \times \,\biggl(N\mathop{\prod }_{w<p<z}\biggl(1-\frac{\unicode[STIX]{x1D714}(p)}{p}\biggr)+z^{3}\biggr)\nonumber\\ \displaystyle & {\leqslant} & \displaystyle 10^{20}(\log N)^{3/2}(\log w)^{-3/2}\cdot N\biggl(\frac{\log w}{\log z}\biggr)^{3/2}\nonumber\\ \displaystyle & {\leqslant} & \displaystyle 10^{30}N.\nonumber\end{eqnarray}$$

Part (iii) is verified as follows. From the definition of $\unicode[STIX]{x1D70C}_{d}$ , it is clear that $|\unicode[STIX]{x1D70C}_{d}|\leqslant 1$ , so that

(4.4) $$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D6FD}(n)\leqslant G_{1}(z)\bigg(\mathop{\sum }_{\substack{ d\mid P(z) \\ Wn+B\in {\mathcal{A}}_{d}}}1\bigg)^{2}. & & \displaystyle\end{eqnarray}$$

Note that if $Wn+B\in {\mathcal{A}}_{p}$ for some $w<p\leqslant z$ , then $p\mid Wn+B$ or $p\mid Wn+B-1$ , so that $p$ can be chosen in at most $\unicode[STIX]{x1D708}(Wn+B)+\unicode[STIX]{x1D708}(Wn+B-1)$ ways, where $\unicode[STIX]{x1D708}(\cdot )$ is the number of distinct prime factors. Since $d$ is square-free and a product of such primes $p$ , $d$ can be chosen in at most $2^{\unicode[STIX]{x1D708}(Wn+B)+\unicode[STIX]{x1D708}(Wn+B-1)}\ll N^{\unicode[STIX]{x1D700}/3}$ ways in (4.4). Therefore, (4.4) is $\ll (\log N)^{3/2}N^{(2/3)\unicode[STIX]{x1D700}}\ll N^{\unicode[STIX]{x1D700}}$ .

Part (iv), which is the most crucial part concerning pseudorandomness, was verified in [Reference Ramaré and Ruzsa18]. Namely, our set of primes of the form $Wn+B=x^{2}+y^{2}+1$ is “sufficiently sifted” in the sense of the definition given on [Reference Ramaré and Ruzsa18, pp. 1 and 2] (to see that, take in that paper $A$ to be the set of primes of the form under consideration up to $N$ and $\unicode[STIX]{x1D705}={\textstyle \frac{3}{2}}$ ). This property is all that is needed to obtain (iv) with the bound $v(a/q)\ll q^{-1/2}$ , by [Reference Ramaré and Ruzsa18, formula (4.1.19)]. It is clear that this can be replaced with the stronger bound $v(a/q)\ll q^{\unicode[STIX]{x1D700}-1}$ , since we have defined the sets ${\mathcal{K}}_{d}$ in (4.2) so that [Reference Ramaré and Ruzsa18, formula (4.1.18)] holds for $\unicode[STIX]{x1D709}=\unicode[STIX]{x1D700}/2$ , instead of just some $0<\unicode[STIX]{x1D709}<{\textstyle \frac{1}{2}}$ .

We are then left with part (v). Equations (4.1.13) and (4.1.21) of [Reference Ramaré and Ruzsa18] reveal that (4.1) holds when $v(a/q)$ is defined for $(a,q)=1$ by

$$\begin{eqnarray}\displaystyle v\biggl(\frac{a}{q}\biggr)=G_{1}(z)\mathop{\sum }_{q\mid [d_{1},d_{2}]}\frac{\unicode[STIX]{x1D70C}_{d_{1}}^{\ast }\unicode[STIX]{x1D70C}_{d_{2}}^{\ast }}{[d_{1},d_{2}]}|{\mathcal{K}}_{[d_{1},d_{2}]}|\cdot \frac{\mathop{\sum }_{b\in {\mathcal{K}}_{q}}e(ab/q)}{|{\mathcal{K}}_{q}|} & & \displaystyle \nonumber\end{eqnarray}$$

with

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D70C}_{\ell }^{\ast }=\mathop{\sum }_{d\equiv 0~(\text{mod}~\ell )}\unicode[STIX]{x1D707}\biggl(\frac{d}{\ell }\biggr)\unicode[STIX]{x1D707}(d)\unicode[STIX]{x1D70C}_{d}, & & \displaystyle \nonumber\end{eqnarray}$$

where the set ${\mathcal{K}}_{d}$ is given by (4.2). As in [Reference Ramaré and Ruzsa18, formula (4.1.17)], we have

$$\begin{eqnarray}\displaystyle \biggl|\mathop{\sum }_{b\in {\mathcal{K}}_{q}}e\biggl(\frac{ab}{q}\biggr)\biggr|=\biggl|\mathop{\sum }_{b\in \mathbb{Z}_{q}\setminus {\mathcal{K}}_{q}}e\biggl(\frac{ab}{q}\biggr)\biggr|\leqslant |\mathbb{Z}_{q}\setminus {\mathcal{K}}_{q}|\leqslant \mathop{\prod }_{p^{\unicode[STIX]{x1D6FC}}\mid \mid q}(p^{\unicode[STIX]{x1D6FC}}-|{\mathcal{K}}_{p^{\unicode[STIX]{x1D6FC}}}|), & & \displaystyle \nonumber\end{eqnarray}$$

which immediately gives $v(a/q)=0$ unless $q$ is square-free and $(q,W)=1$ . In addition, by formula (4.1.13) of the same paper (with the right-hand side multiplied by $G_{1}(z)$ ), we have

(4.5) $$\begin{eqnarray}\displaystyle v\biggl(\frac{a}{q}\biggr)=G_{1}(z)w_{q}^{\#}\cdot \frac{\mathop{\sum }_{b\in {\mathcal{K}}_{q}}e(ab/q)}{|{\mathcal{K}}_{q}|}, & & \displaystyle\end{eqnarray}$$

where by (4.1.14) we have

$$\begin{eqnarray}\displaystyle w_{q}^{\#}=\frac{1}{G_{1}(z)}\mathop{\sum }_{\unicode[STIX]{x1D6FF}\leqslant z}h(\unicode[STIX]{x1D6FF})\unicode[STIX]{x1D70C}_{z}(q,\unicode[STIX]{x1D6FF}), & & \displaystyle \nonumber\end{eqnarray}$$

and $\unicode[STIX]{x1D70C}_{z}(q,\unicode[STIX]{x1D6FF})$ satisfies (4.1.15). Putting $q=1$ into (4.1.15), we clearly get $w_{1}^{\#}=1/G_{1}(z)$ , so that $v(1)=1$ by (4.5).◻

Proof of Theorem 1.3.

This is very similar to the proof of [Reference Green and Tao4, Theorem 1.2]. Let ${\mathcal{A}}\subset \mathscr{P}^{\ast }$ have positive upper density in $\mathscr{P}^{\ast }$ . Then there is $\unicode[STIX]{x1D6FF}>0$ (which may be assumed small) such that $|{\mathcal{A}}\cap (N/3,2N/3)|\geqslant \unicode[STIX]{x1D6FF}|\mathscr{P}^{\ast }\cap (N/3,2N/3)|$ for $N\in {\mathcal{N}}$ , where ${\mathcal{N}}$ is some infinite set of positive integers. Let $W$ , $w$ and $J$ be as in (3.3) with $J=\lfloor 10/\unicode[STIX]{x1D6FF}\rfloor$ .

Let $S_{B}=S\cap \{Wn+B:n\geqslant 1\}$ for any set $S$ and integer $B$ . Note that if $n=x^{2}+y^{2}+1\in (N/3,2N/3)$ is a prime with $(x,y)=1$ and $N\geqslant 10W$ , then $(n,W)=(n-1,s(W))=1$ and $(n-1,3)=1$ , $4\nmid n-1$ . Therefore,

$$\begin{eqnarray}\displaystyle \mathop{\sum }_{\substack{ 1\leqslant B\leqslant W \\ Wn+B~\text{amenable}}}\biggl|{\mathcal{A}}_{B}\cap \biggl(\frac{N}{3},\frac{2N}{3}\biggr)\biggr| & =\biggl|{\mathcal{A}}\cap \biggl(\frac{N}{3},\frac{2N}{3}\biggr)\biggr|\geqslant \unicode[STIX]{x1D6FF}\biggl|\mathscr{P}^{\ast }\cap \biggl(\frac{N}{3},\frac{2N}{3}\biggr)\biggr| & \displaystyle \nonumber\end{eqnarray}$$

for $N\geqslant 10W$ and $N\in {\mathcal{N}}$ , so using the pigeonhole principle and the lower bound for $|\mathscr{P}^{\ast }\cap (N/3,2N/3)|$ coming from Proposition 3.2 with $\unicode[STIX]{x1D712}\equiv 1$ , we can find a value of $B\in [1,W]$ such that the polynomial $Wn+B$ is amenable and

(4.6) $$\begin{eqnarray}\displaystyle \biggl|{\mathcal{A}}_{B}\cap \biggl(\frac{N}{3},\frac{2N}{3}\biggr)\biggr|\geqslant \unicode[STIX]{x1D6FF}_{1}\cdot \unicode[STIX]{x1D6FF}(\log w)^{3/2}\frac{N}{W(\log N)^{3/2}} & & \displaystyle\end{eqnarray}$$

for $N\in {\mathcal{N}}^{\prime }$ with ${\mathcal{N}}^{\prime }$ an infinite set of positive integers and for some small absolute constant $\unicode[STIX]{x1D6FF}_{1}>0$ , since the Chinese remainder theorem shows that there are ${\leqslant}10^{10}W(\log w)^{-3/2}$ amenable functions $Wn+B$ with $1\leqslant B\leqslant W$ .

Next set

$$\begin{eqnarray}\displaystyle g(n) & = & \displaystyle \unicode[STIX]{x1D6FF}_{2}(\log N)^{3/2}(\log w)^{-3/2}1_{{\mathcal{A}}_{B}\cap (N/3,2N/3)}(n)\nonumber\\ \displaystyle & & \displaystyle \text{for}~N\in {\mathcal{N}}^{\prime }~\text{and}~1\leqslant n\leqslant N\nonumber\end{eqnarray}$$

with $\unicode[STIX]{x1D6FF}_{2}>0$ small and extend $g$ periodically to $\mathbb{Z}_{N}$ . The assertion of the theorem will follow from the Green–Tao transference principle [Reference Green and Tao4, Proposition 5.1] as soon as we check formulas (5.3)–(5.6) of that paper for the functions $g(n)$ and $\unicode[STIX]{x1D708}(n)=\unicode[STIX]{x1D6FD}(n)1_{[1,N]}(n)$ (extended periodically to $\mathbb{Z}_{N}$ ) with $\unicode[STIX]{x1D6FD}(\cdot )$ given by Proposition 4.1. We know (5.3) from Proposition 4.1 and (5.6) from Proposition 3.3. Formula (5.5) follows from the properties (i)–(v) of $\unicode[STIX]{x1D6FD}(n)$ just as in [Reference Green and Tao4, Ch. 6]. We are left with (5.4), which follows (for a different value of $\unicode[STIX]{x1D6FF}$ ) for $N\in {\mathcal{N}}^{\prime }$ from (4.6). Now, as mentioned, [Reference Green and Tao4, Proposition 5.1] yields the result, since any triple of the form $(a,a+d+j_{1}N,a+2d+j_{2}N)$ is an arithmetic progression in $\mathbb{Z}$ if $a,a+2d+j_{1}N,a+2d+j_{2}N\in (N/3,2N/3)$ .◻

Proof. This is [Reference Matomäki and Shao15, Lemma 7.4]. ◻

Proof. Using Möbius inversion, the sum in question (without absolute values) becomes

(5.6) $$\begin{eqnarray}\displaystyle \mathop{\sum }_{d\mid q}\unicode[STIX]{x1D707}(d)\mathop{\sum }_{k\mid s(q)}\unicode[STIX]{x1D707}(k)\mathop{\sum }_{\substack{ n~(\text{mod}~q) \\ WVn\equiv -b~(\text{mod}~d) \\ WVn\equiv -(b-1)~(\text{mod}~k)}}e\biggl(\frac{a}{q}n\biggr). & & \displaystyle\end{eqnarray}$$

Now consider the sum

(5.7) $$\begin{eqnarray}\displaystyle \mathop{\sum }_{\substack{ n~(\text{mod}~q) \\ WVn\equiv -b~(\text{mod}~d) \\ WVn\equiv -(b-1)~(\text{mod}~k)}}e\biggl(\frac{a}{q}n\biggr). & & \displaystyle\end{eqnarray}$$

Note that the sum is non-empty only if $(d,k)=1$ . Let $x_{1},\ldots ,x_{R(d,k)}~(\text{mod}~dk)$ be the pairwise incongruent solutions to the system $WVx\equiv -b~(\text{mod}~d)$ , $WVx\equiv -(b-1)~(\text{mod}~k)$ (if there are none, the sum (5.7) is empty). Since $dk=[d,k]\mid q$ , after writing $n=x_{j}+dkt$ for some $1\leqslant j\leqslant R(d,k)$ and $1\leqslant t\leqslant q/dk$ , (5.7) transforms into

(5.8) $$\begin{eqnarray}\displaystyle \mathop{\sum }_{j=1}^{R(d,k)}\mathop{\sum }_{\substack{ n~(\text{mod}~q) \\ n\equiv x_{j}~(\text{mod}~dk)}}e\biggl(\frac{an}{q}\biggr) & =\mathop{\sum }_{j=1}^{R(d,k)}e\biggl(\frac{ax_{j}}{q}\biggr)\mathop{\sum }_{t~(\text{mod}~q/dk)}e\biggl(\frac{at}{q/dk}\biggr).\qquad & \displaystyle\end{eqnarray}$$

The inner sum is non-zero only when $dk=q$ , in which case it is $1$ . Taking these considerations into account, (5.6) has absolute value at most

(5.9) $$\begin{eqnarray}\displaystyle \mathop{\sum }_{\substack{ d\mid q \\ k\mid s(q) \\ dk=q}}R(d,k)|\unicode[STIX]{x1D707}(d)||\unicode[STIX]{x1D707}(k)|. & & \displaystyle\end{eqnarray}$$

We estimate this differently depending on whether $(q,W)>1$ or $(q,W)=1$ . In the former case, there is some prime $p$ such that $p\mid q$ , $p\mid W$ , so $dk=q$ tells us that $p$ divides either $d$ or $k$ . If $p\mid d$ , then supposing that $R(d,k)\neq 0$ , the congruence $WVx\equiv -b~(\text{mod}~p)$ must be solvable. It however is not solvable, since $p\nmid b$ for $p\mid W$ by the amenability of $Wn+b$ . If $p\mid k$ , then $k\mid s(q)$ implies that $p\equiv -1~(\text{mod}~4)$ , $p\neq 3$ . If $R(d,k)\neq 0$ , the congruence $WVx\equiv -(b-1)~(\text{mod}~p)$ has a solution, but $p\nmid b-1$ by amenability, so we have a contradiction. We deduce that all the summands in (5.9) vanish for $(q,W)>1$ .

Then let $(q,W)=1$ . As $d,k\mid q$ in (5.9), we also have $(d,W)=(k,W)=1$ and $(d,V)=(k,V)=1$ . Now clearly both of the congruences $WVx\equiv -b~(\text{mod}~d)$ , $WVx\equiv -(b-1)~(\text{mod}~k)$ have a unique solution, so if the two congruences are thought of as a simultaneous equation, it has at most one solution $(\text{mod}~dk)$ . Therefore, $R(d,k)\leqslant 1$ , which leads to (5.9) being at most

$$\begin{eqnarray}\displaystyle \mathop{\sum }_{dk=q}1\leqslant \unicode[STIX]{x1D70F}(q), & & \displaystyle \nonumber\end{eqnarray}$$

as asserted. ◻

Proof of Lemma 5.3.

This is similar to the argument on [Reference Matomäki and Shao15, p. 21]. We can find unique $q^{\prime }$ and $Q^{\prime }$ such that $Q=qq^{\prime }Q^{\prime }$ and $(q,Q^{\prime })=1$ and all the prime divisors of $q^{\prime }$ divide $q$ . Writing $c_{0}=c_{1}q+c_{2}Q^{\prime }$ , $c_{0}$ runs through each residue class $(\text{mod}~Q)$ exactly once as $c_{1}$ runs through residue classes $(\text{mod}~q^{\prime }Q^{\prime })$ and $c_{2}$ runs independently through residue classes $(\text{mod}~q)$ . Now the left-hand side of (5.4) (without absolute values) becomes

(5.10) $$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D6F4}:=\mathop{\sum }_{\substack{ c_{1}~(\text{mod}~q^{\prime }Q^{\prime }) \\ (Wqc_{1}+b,Q^{\prime })=1 \\ (Wqc_{1}+b-1,s(Q^{\prime }))=1}}\mathop{\sum }_{\substack{ c_{2}~(\text{mod}~q) \\ (WQ^{\prime }c_{2}+b,q)=1 \\ (WQ^{\prime }c_{2}+b-1,s(q))=1}}e\biggl(\frac{aQ^{\prime }}{q}c_{2}\biggr). & & \displaystyle\end{eqnarray}$$

Since $(aQ^{\prime },q)=1$ , the inner sum is exactly of the form appearing in Lemma 5.4. Therefore,

$$\begin{eqnarray}\displaystyle |\unicode[STIX]{x1D6F4}|\leqslant \mathop{\sum }_{\substack{ c_{1}~(\text{mod}~q^{\prime }Q^{\prime }) \\ (Wqc_{1}+b,Q^{\prime })=1 \\ (Wqc_{1}+b-1,s(Q^{\prime }))=1}}\unicode[STIX]{x1D70F}(q)\cdot 1_{q>w}. & & \displaystyle \nonumber\end{eqnarray}$$

Since $w\geqslant 10^{10^{10}}$ , estimating the divisor function crudely yields

$$\begin{eqnarray}\displaystyle |\unicode[STIX]{x1D6F4}|\leqslant 1_{q>w}\cdot q^{0.1}\mathop{\sum }_{\substack{ c_{1}~(\text{mod}~q^{\prime }Q^{\prime }) \\ (Wqc_{1}+b,Q^{\prime })=1 \\ (Wqc_{1}+b-1,s(Q^{\prime }))=1}}1 & = & \displaystyle 1_{q>w}\cdot q^{\prime }q^{0.1}\mathop{\sum }_{\substack{ c_{1}~(\text{mod}~Q^{\prime }) \\ (Wqc_{1}+b,Q^{\prime })=1 \\ (Wqc_{1}+b-1,s(Q^{\prime }))=1}}1\nonumber\\ \displaystyle & = & \displaystyle 1_{q>w}\cdot q^{\prime }q^{0.1}Q^{\prime }\mathop{\prod }_{\substack{ p\mid Q^{\prime } \\ p>w}}\biggl(1-\frac{\unicode[STIX]{x1D714}(p)}{p}\biggr),\nonumber\end{eqnarray}$$

where $\unicode[STIX]{x1D714}(p)\in \{1,2\}$ and $\unicode[STIX]{x1D714}(p)=2$ precisely when $p\equiv -1~(\text{mod}~4)$ . The previous expression is, for $q>w\geqslant 10^{10^{10}}$ ,

$$\begin{eqnarray}\displaystyle & {\leqslant} & \displaystyle q^{\prime }q^{0.2}\mathop{\prod }_{\substack{ p\mid q \\ p>w}}\biggl(1-\frac{\unicode[STIX]{x1D714}(p)}{p}\biggr)\cdot Q^{\prime }\mathop{\prod }_{\substack{ p\mid Q^{\prime } \\ p>w}}\biggl(1-\frac{\unicode[STIX]{x1D714}(p)}{p}\biggr)\nonumber\\ \displaystyle & = & \displaystyle \frac{Q}{q^{0.8}}\mathop{\prod }_{\substack{ p\mid Q \\ p>w}}\biggl(1-\frac{\unicode[STIX]{x1D714}(p)}{p}\biggr)\leqslant \frac{|{\mathcal{Q}}|}{w^{1/2}},\nonumber\end{eqnarray}$$

where the last step comes from (5.2). ◻

Hypothesis 6.4. Let $L(n)=Kn+b$ be an amenable linear function with $K\ll (\log N)^{O(1)}$ . Let $(\unicode[STIX]{x1D714}_{n})_{n\sim N}$ be a non-negative sequence of real numbers and let $\unicode[STIX]{x1D6FF}=(b-1,K)$ . Let $\unicode[STIX]{x1D700}>0$ be any small number. Let ${\textstyle \frac{1}{3}}<\unicode[STIX]{x1D70C}_{2}<\unicode[STIX]{x1D70C}_{1}<{\textstyle \frac{1}{2}}-\unicode[STIX]{x1D700}$ , $\unicode[STIX]{x1D70E}\in (3,4)$ . Then, for any sequence $(g(\ell ))_{\ell \leqslant N^{0.9}}$ of convolution type (with parameter  $\unicode[STIX]{x1D70E}$ ),

$$\begin{eqnarray}\displaystyle & & \displaystyle \mathop{\sum }_{\substack{ d\leqslant N^{\unicode[STIX]{x1D70C}_{1}} \\ (d,K)=1}}\unicode[STIX]{x1D706}_{d}^{+,\text{LIN}}\mathop{\sum }_{\substack{ \ell \leqslant N^{0.9} \\ (\ell ,K)=\unicode[STIX]{x1D6FF} \\ (\ell ,d)=1}}g(\ell )\bigg(\mathop{\sum }_{\substack{ n\sim N \\ L(n)=\ell p+1 \\ L(n)\equiv 0~(\text{mod}~d)}}\unicode[STIX]{x1D714}_{n}-\frac{1}{\unicode[STIX]{x1D711}(d)}\frac{K}{\unicode[STIX]{x1D711}(K/\unicode[STIX]{x1D6FF})}\mathop{\sum }_{n\sim N}\frac{\unicode[STIX]{x1D714}_{n}}{\ell \log (Kn/\ell )}\bigg)\nonumber\\ \displaystyle & & \displaystyle \quad \ll \frac{\mathop{\sum }_{n\sim N}\unicode[STIX]{x1D714}_{n}}{(\log N)^{100}},\nonumber\\ \displaystyle & & \displaystyle \mathop{\sum }_{\substack{ d\leqslant N^{\unicode[STIX]{x1D70C}_{2}} \\ (d,K)=1}}\unicode[STIX]{x1D706}_{d}^{-,\text{SEM}}\bigg(\mathop{\sum }_{\substack{ n\sim N \\ L(n)\in \mathbb{P} \\ L(n)\equiv 1~(\text{mod}~d)}}\unicode[STIX]{x1D714}_{n}-\frac{1}{\unicode[STIX]{x1D711}(d)}\frac{K}{\unicode[STIX]{x1D711}(K)}\mathop{\sum }_{n\sim N}\frac{\unicode[STIX]{x1D714}_{n}}{\log (Kn)}\bigg)\nonumber\\ \displaystyle & & \displaystyle \quad \ll \frac{\mathop{\sum }_{n\sim N}\unicode[STIX]{x1D714}_{n}}{(\log N)^{100}},\nonumber\end{eqnarray}$$

where $\unicode[STIX]{x1D706}_{d}^{+,\text{LIN}}$ are the upper bound linear sieve weights with sifting parameter $z_{1}=N^{1/5}$ and $\unicode[STIX]{x1D706}_{d}^{-,\text{SEM}}$ are the lower bound semilinear sieve weights with sifting parameter $z_{2}=N^{1/\unicode[STIX]{x1D70E}}$ (the weights $\unicode[STIX]{x1D706}_{d}^{\pm ,\text{SEM}}$ were defined in Theorem 1.5, and the weights $\unicode[STIX]{x1D706}_{d}^{\pm ,\text{LIN}}$ are defined analogously by replacing $\unicode[STIX]{x1D6FD}=1$ by $\unicode[STIX]{x1D6FD}=2$ in that definition).

Proof. Put

$$\begin{eqnarray}\displaystyle {\mathcal{A}} & = & \displaystyle \{L(n)-1:n\sim N,L(n)\in \mathbb{P}\},\nonumber\\ \displaystyle {\mathcal{P}}_{4,-1} & = & \displaystyle \{p\in \mathbb{P}:p\equiv -1~(\text{mod}~4),p\neq 3\},\nonumber\\ \displaystyle P(z) & = & \displaystyle \mathop{\prod }_{\substack{ p<z \\ p\in {\mathcal{P}}_{4,-1}}}p,\nonumber\\ \displaystyle {\mathcal{P}}_{4,1}^{\ast } & = & \displaystyle \{n\geqslant 1:p\mid n\Rightarrow p\equiv 1~(\text{mod}~4)\}.\nonumber\end{eqnarray}$$

If we weight the elements of ${\mathcal{A}}$ by $\unicode[STIX]{x1D708}_{n}=\unicode[STIX]{x1D714}_{(L^{-1}(n+1))}$ , where $L^{-1}$ is the inverse function of $L$ , the sifting function is

$$\begin{eqnarray}\displaystyle S({\mathcal{A}},{\mathcal{P}}_{4,-1},z)=\mathop{\sum }_{\substack{ n\sim N \\ L(n)\in \mathbb{P} \\ (L(n)-1,P(z))=1}}\unicode[STIX]{x1D714}_{n}. & & \displaystyle \nonumber\end{eqnarray}$$

Note that $L(n)-1\equiv 2^{\unicode[STIX]{x1D6FD}}~(\text{mod}~2^{\unicode[STIX]{x1D6FD}+2})$ for some $\unicode[STIX]{x1D6FD}\geqslant 1$ by the definition of amenability, so that $L(n)-1$ has an even number of prime factors that are $\equiv -1~(\text{mod}~4)$ (counted with multiplicity). We have

(6.2) $$\begin{eqnarray}\displaystyle \mathop{\sum }_{\substack{ n\sim N \\ L(n)\in \mathbb{P} \\ L(n)-1\in {\mathcal{S}}}}\unicode[STIX]{x1D714}_{n}=S({\mathcal{A}},{\mathcal{P}}_{4,-1},(3KN)^{1/2}), & & \displaystyle\end{eqnarray}$$

since the right-hand side counts with weight $\unicode[STIX]{x1D714}_{n}$ the numbers $L(n)-1=2^{\unicode[STIX]{x1D6FC}_{1}}3^{\unicode[STIX]{x1D6FC}_{2}}k\in {\mathcal{A}}$ with $k\in {\mathcal{P}}_{4,1}^{\ast }$ , and we claim that these numbers are precisely the numbers in ${\mathcal{S}}\cap {\mathcal{A}}$ . We have $2^{\unicode[STIX]{x1D6FC}_{1}}3^{\unicode[STIX]{x1D6FC}_{2}}k=L(n)-1$ , so by amenability $\unicode[STIX]{x1D6FC}_{2}\equiv 0~(\text{mod}~2)$ . It is a fact in elementary number theory that for $k\in {\mathcal{P}}_{4,1}^{\ast }$ , both $k$ and $2k$ can be expressed in the form $a^{2}+b^{2}$ with $(a,b)=1$ , and additionally no number of the form $2^{\unicode[STIX]{x1D6FC}_{1}}3^{\unicode[STIX]{x1D6FC}_{2}}k$ with $(k,6)=1$ and $\unicode[STIX]{x1D6FC}_{2}$ odd or $k\not \in {\mathcal{P}}_{4,1}^{\ast }$ is of the form $x^{2}+y^{2}$ with $(x,y)\mid 6^{\infty }$ . Hence, both sides of (6.2) indeed count the same integers.

Buchstab’s identity reveals that

$$\begin{eqnarray}\displaystyle S({\mathcal{A}},{\mathcal{P}}_{4,-1},(3KN)^{1/2})=S({\mathcal{A}},{\mathcal{P}}_{4,-1},N^{1/\unicode[STIX]{x1D70E}})-\mathop{\sum }_{\substack{ n\sim N \\ L(n)\in \mathbb{P}}}\mathop{\sum }_{\substack{ p_{2}\mid L(n)-1 \\ N^{1/\unicode[STIX]{x1D70E}}\leqslant p_{2}<(3KN)^{1/2} \\ (L(n)-1,P(p_{2}))=1 \\ p_{2}\in {\mathcal{P}}_{4,-1}}}\unicode[STIX]{x1D714}_{n}. & & \displaystyle \nonumber\end{eqnarray}$$

The condition $p_{2}\mid L(n)-1\equiv 2^{\unicode[STIX]{x1D6FD}}~(\text{mod}~2^{\unicode[STIX]{x1D6FD}+2})$ implies that $L(n)-1$ has either exactly two prime divisors from ${\mathcal{P}}_{4,-1}$ or at least four such prime divisors (with multiplicities). The second case is impossible, since all the prime divisors of $L(n)-1$ that are from ${\mathcal{P}}_{4,-1}$ are ${\geqslant}p_{2}$ and $p_{2}^{4}\geqslant N^{4/\unicode[STIX]{x1D70E}}>L(2N)-1.$ This means that we may write $L(n)-1=p_{1}p_{2}m^{\prime },$ $p_{1}\geqslant p_{2},$ $p_{1}\in {\mathcal{P}}_{4,-1},$ with $m^{\prime }$ having no prime divisors from ${\mathcal{P}}_{4,-1}$ . Now $\unicode[STIX]{x1D6FF}\mid L(n)-1=Kn+b-1$ with $\unicode[STIX]{x1D6FF}=(b-1,K)$ and, since $p_{1}\geqslant p_{2}\geqslant N^{1/\unicode[STIX]{x1D70E}}>K$ , we have $\unicode[STIX]{x1D6FF}\mid m^{\prime }$ . Hence, we may write $m^{\prime }=\unicode[STIX]{x1D6FF}m$ , where $m\in {\mathcal{P}}_{4,1}^{\ast }$ (we have $3\nmid m$ , since $K$ is divisible by a larger power of $3$ than $b-1$ is, by the definition of amenability. Similarly, $2\nmid m$ ). We claim that $(m,K/\unicode[STIX]{x1D6FF})=1$ . Indeed, if $p\mid m$ and $p\mid K/\unicode[STIX]{x1D6FF}$ , we must have $p\mid (b-1)/\unicode[STIX]{x1D6FF}$ , which is a contradiction to $(K,b-1)=\unicode[STIX]{x1D6FF}$ . Now we have

(6.3) $$\begin{eqnarray}\displaystyle S({\mathcal{A}},{\mathcal{P}}_{4,-1},(3KN)^{1/2})=S-T. & & \displaystyle\end{eqnarray}$$

Here

$$\begin{eqnarray}\displaystyle S & = & \displaystyle S({\mathcal{A}},{\mathcal{P}}_{4,-1},N^{1/\unicode[STIX]{x1D70E}}),\nonumber\\ \displaystyle T & = & \displaystyle \mathop{\sum }_{\substack{ n\sim N \\ L(n)\in \mathbb{P}}}\mathop{\sum }_{\substack{ L(n)-1=\unicode[STIX]{x1D6FF}p_{1}p_{2}m \\ p_{1},p_{2}\in {\mathcal{P}}_{4,-1} \\ N^{1/\unicode[STIX]{x1D70E}}\leqslant p_{2}\leqslant p_{1} \\ m\in {\mathcal{P}}_{4,1}^{\ast }}}\unicode[STIX]{x1D714}_{n}\leqslant \mathop{\sum }_{\ell \in {\mathcal{L}}}S({\mathcal{M}}(\ell ),{\mathcal{P}}(\ell ),N^{1/6})\nonumber\end{eqnarray}$$

with

$$\begin{eqnarray}\displaystyle {\mathcal{L}} & = & \displaystyle \biggl\{\unicode[STIX]{x1D6FF}p_{2}m:N^{1/\unicode[STIX]{x1D70E}}\leqslant p_{2}\leqslant (3KNm^{-1})^{1/2},\nonumber\\ \displaystyle & & \displaystyle \quad p_{2}\in {\mathcal{P}}_{4,-1},m\in {\mathcal{P}}_{4,1}^{\ast },\biggl(m,\frac{K}{\unicode[STIX]{x1D6FF}}\biggr)=1\biggr\},\nonumber\\ \displaystyle {\mathcal{M}}(\ell ) & = & \displaystyle \{L(n):L(n)=\ell p+1:n\sim N,p\in \mathbb{P}\},\nonumber\\ \displaystyle {\mathcal{P}}(\ell ) & = & \displaystyle \{p\in \mathbb{P}:(p,2\ell )=1\},\qquad Q(z)=\mathop{\prod }_{\substack{ p<z \\ p\in {\mathcal{P}}(\ell )}}p\nonumber\end{eqnarray}$$

and $M(\ell )$ has been assigned the weights $\unicode[STIX]{x1D708}_{n}=\unicode[STIX]{x1D714}_{L^{-1}(n)}$ , so that

$$\begin{eqnarray}\displaystyle S({\mathcal{M}}(\ell ),{\mathcal{P}}(\ell ),z)=\mathop{\sum }_{\substack{ n\sim N \\ L(n)=\ell p+1 \\ (L(n),Q(z))=1}}\unicode[STIX]{x1D714}_{n}. & & \displaystyle \nonumber\end{eqnarray}$$

We carry out bounding $S$ from below and bounding $T$ from above separately.

Bounding  $S$ . For $d\mid P(z)$ , $(d,K)=1$ , let

$$\begin{eqnarray}\displaystyle r({\mathcal{A}},d)=\mathop{\sum }_{\substack{ n\sim N \\ L(n)\in \mathbb{P} \\ L(n)-1\equiv 0~(\text{mod}~d)}}\unicode[STIX]{x1D714}_{n}-\frac{1}{\unicode[STIX]{x1D711}(d)}\frac{K}{\unicode[STIX]{x1D711}(K)}\mathop{\sum }_{n\sim N}\frac{\unicode[STIX]{x1D714}_{n}}{\log (Kn)} & & \displaystyle \nonumber\end{eqnarray}$$

and, for $(d,K)>1$ , let $r({\mathcal{A}},d)=0$ (since if $p\mid d$ , $p\mid K$ and $p\in {\mathcal{P}}_{4,-1}$ , then $p$ does not divide any element of ${\mathcal{A}}$ by the amenability of $L(n)$ ). Let $\unicode[STIX]{x1D70E}\in (3,4)$ be as in Hypothesis 6.4. The semilinear sieve [Reference Friedlander and Iwaniec2, Theorem 11.13], with $\unicode[STIX]{x1D6FD}=1$ , sifting parameter $z=N^{1/\unicode[STIX]{x1D70E}}$ and level $D=z^{s}$ , $1\leqslant s\leqslant 2$ , gives

(6.4) $$\begin{eqnarray}\displaystyle S({\mathcal{A}},{\mathcal{P}}_{4,-1},N^{1/\unicode[STIX]{x1D70E}}) & {\geqslant} & \displaystyle \frac{K}{\unicode[STIX]{x1D711}(K)}\mathop{\sum }_{n\sim N}\frac{\unicode[STIX]{x1D714}_{n}}{\log (Kn)}V_{K}^{\text{SEM}}(N^{1/\unicode[STIX]{x1D70E}})\nonumber\\ \displaystyle & & \displaystyle \times \,(f(s)+O((\log N)^{-0.1}))\nonumber\\ \displaystyle & & \displaystyle +\,\mathop{\sum }_{d\leqslant N^{s/\unicode[STIX]{x1D70E}}}\unicode[STIX]{x1D706}_{d}^{-,\text{SEM}}r({\mathcal{A}},d),\end{eqnarray}$$

where $\unicode[STIX]{x1D706}_{d}^{-,\text{SEM}}$ are the lower bound semilinear weights with sifting parameter $z=N^{1/\unicode[STIX]{x1D70E}}$ and we have introduced the quantities

$$\begin{eqnarray}\displaystyle f(s)=\sqrt{\frac{\text{e}^{\unicode[STIX]{x1D6FE}}}{\unicode[STIX]{x1D70B}s}}\int _{1}^{s}\,\frac{dt}{\sqrt{t(t-1)}}\quad \text{and}\quad V_{K}^{\text{SEM}}(z)=\mathop{\prod }_{\substack{ p<z \\ p\equiv -1~(\text{mod}~4) \\ p\nmid K}}\biggl(1-\frac{1}{\unicode[STIX]{x1D711}(p)}\biggr). & & \displaystyle \nonumber\end{eqnarray}$$

We take $s=\unicode[STIX]{x1D70C}_{2}\unicode[STIX]{x1D70E}\in [1,2]$ , where $\unicode[STIX]{x1D70C}_{2}$ is as in Hypothesis 6.4. Now Hypothesis 6.4 permits replacing the last sum in (6.4) with an error of $\ll \sum _{n\sim N}\unicode[STIX]{x1D714}_{n}/(\log N)^{100}$ (since the terms of that sum in (6.4) vanish unless $(d,K)=1$ ). Moreover, the term $V_{K}^{\text{SEM}}(N^{1/\unicode[STIX]{x1D70E}})$ can be computed asymptotically using [Reference Wu24, Proposition 1], which implies that

$$\begin{eqnarray}\displaystyle V_{K}^{\text{SEM}}(z)=(1+o(1))\mathop{\prod }_{\substack{ p\mid K \\ p\equiv -1~(\text{mod}~4)}}\biggl(1-\frac{1}{p-1}\biggr)^{-1}\cdot 2AC_{4,-1}\cdot \biggl(\frac{\unicode[STIX]{x1D70B}\text{e}^{-\unicode[STIX]{x1D6FE}}}{\log z}\biggr)^{1/2}, & & \displaystyle \nonumber\end{eqnarray}$$

where

$$\begin{eqnarray}\displaystyle A=\frac{1}{2\sqrt{2}}\mathop{\prod }_{p\equiv -1~(\text{mod}~4)}\biggl(1-\frac{1}{p^{2}}\biggr)^{1/2} & & \displaystyle \nonumber\end{eqnarray}$$

and

$$\begin{eqnarray}\displaystyle C_{4,i}=\mathop{\prod }_{p\equiv i~(\text{mod}~4)}\biggl(1-\frac{1}{(p-1)^{2}}\biggr) & & \displaystyle \nonumber\end{eqnarray}$$

for $i\in \{-1,1\}$ . Therefore, we end up with the bound

(6.5) $$\begin{eqnarray}\displaystyle S & {\geqslant} & \displaystyle \frac{4AC_{4,-1}+o(1)}{(\log N)^{1/2}}\cdot I_{1}(\unicode[STIX]{x1D70C}_{2},\unicode[STIX]{x1D70E})\frac{K}{\unicode[STIX]{x1D711}(K)}\mathop{\prod }_{\substack{ p\mid K \\ p\equiv -1~(\text{mod}~4)}}\biggl(1-\frac{1}{p-1}\biggr)^{-1}\nonumber\\ \displaystyle & & \displaystyle \times \,\mathop{\sum }_{n\sim N}\frac{\unicode[STIX]{x1D714}_{n}}{\log (Kn)}\nonumber\\ \displaystyle & = & \displaystyle \frac{4AC_{4,-1}+o(1)}{(\log N)^{3/2}}\cdot I_{1}(\unicode[STIX]{x1D70C}_{2},\unicode[STIX]{x1D70E})\frac{K}{\unicode[STIX]{x1D711}(K)}\mathop{\prod }_{\substack{ p\mid K \\ p\equiv -1~(\text{mod}~4)}}\biggl(1-\frac{1}{p-1}\biggr)^{-1}\nonumber\\ \displaystyle & & \displaystyle \times \,\mathop{\sum }_{n\sim N}\unicode[STIX]{x1D714}_{n},\end{eqnarray}$$

where

$$\begin{eqnarray}\displaystyle I_{1}(\unicode[STIX]{x1D70C}_{2},\unicode[STIX]{x1D70E})=\frac{1}{2\sqrt{\unicode[STIX]{x1D70C}_{2}}}\int _{1}^{\unicode[STIX]{x1D70C}_{2}\unicode[STIX]{x1D70E}}\,\frac{dt}{\sqrt{t(t-1)}}. & & \displaystyle \nonumber\end{eqnarray}$$

Bounding  $T$ . Write, for $d\mid Q(z)$ , $(d,K)=1$ , $(\ell ,d)=1$ and $(\ell ,K)=\unicode[STIX]{x1D6FF}$ ,

$$\begin{eqnarray}\displaystyle r({\mathcal{M}}(\ell ),d)=\mathop{\sum }_{\substack{ n\sim N \\ L(n)-1=\ell p \\ L(n)\equiv 0~(\text{mod}~d)}}\unicode[STIX]{x1D714}_{n}-\frac{1}{\unicode[STIX]{x1D711}(d)}\frac{K}{\unicode[STIX]{x1D711}(K/\unicode[STIX]{x1D6FF})}\mathop{\sum }_{n\sim N}\frac{\unicode[STIX]{x1D714}_{n}}{\ell \log \!(Kn/\ell )}. & & \displaystyle \nonumber\end{eqnarray}$$

For all other $d$ such that $d\mid Q(z)$ , let $r({\mathcal{M}}(\ell ),d)=0$ (since if $(d,K)>1$ , then $L(n)-1=\ell p,$ $L(n)\equiv 0~(\text{mod}~d)$ is impossible). With these notations, for $1\leqslant s\leqslant 3$ the linear sieve [Reference Friedlander and Iwaniec2, Theorem 11.13] with $\unicode[STIX]{x1D6FD}=2$ provides the bound

(6.6) $$\begin{eqnarray}\displaystyle S({\mathcal{M}}(\ell ),{\mathcal{P}}(\ell ),N^{1/6}) & {\leqslant} & \displaystyle \frac{(1+o(1))K}{\unicode[STIX]{x1D711}(K/\unicode[STIX]{x1D6FF})}\mathop{\sum }_{n\sim N}\frac{\unicode[STIX]{x1D714}_{n}}{\ell \log \!(Kn/\ell )}V_{K}^{\text{LIN}}(N^{1/5},\ell )F(s)\nonumber\\ \displaystyle & & \displaystyle +\,\mathop{\sum }_{d\leqslant N^{s/5}}\unicode[STIX]{x1D706}_{d}^{+,\text{LIN}}r({\mathcal{M}}(\ell ),d),\end{eqnarray}$$

where  $\unicode[STIX]{x1D706}_{d}^{+,\text{LIN}}$ are the upper bound linear sieve coefficients with sifting parameter $z=N^{1/5}$ , $F(s)=2\text{e}^{\unicode[STIX]{x1D6FE}}/s$ and

$$\begin{eqnarray}\displaystyle V_{K}^{\text{LIN}}(z,\ell )=\mathop{\prod }_{\substack{ p\in {\mathcal{P}}(\ell ) \\ p<z \\ p\nmid K}}\!\biggl(1-\frac{1}{\unicode[STIX]{x1D711}(p)}\biggr)=\mathop{\prod }_{2<p<z}\!\biggl(1-\frac{1}{p-1}\biggr)\mathop{\prod }_{\substack{ p\mid K\ell \\ 2<p<z}}\!\biggl(1-\frac{1}{p-1}\biggr)^{-1}. & & \displaystyle \nonumber\end{eqnarray}$$

Applying [Reference Wu24, formula (4.6)], we get the asymptotic

(6.7) $$\begin{eqnarray}\displaystyle \begin{array}{@{}rcl@{}}V_{K}^{\text{LIN}}(z,\ell )\, & =\, & (1+o(1))\frac{2C_{4,1}C_{4,-1}\text{e}^{-\unicode[STIX]{x1D6FE}}\mathfrak{f}(K\ell )}{\log z},\\ \, & \, & \text{where}~\mathfrak{f}(d)=\mathop{\prod }_{\substack{ p\mid d \\ p>2}}\biggl(1-\frac{1}{p-1}\biggr)^{-1}.\end{array} & & \displaystyle\end{eqnarray}$$

We take $s=5\unicode[STIX]{x1D70C}_{1}\in [1,3]$ in the linear sieve. Then we have

(6.8) $$\begin{eqnarray}\displaystyle \mathop{\sum }_{\ell \in {\mathcal{L}}}\mathop{\sum }_{d\leqslant N^{\unicode[STIX]{x1D70C}_{1}}}\unicode[STIX]{x1D706}_{d}^{+,\text{LIN}}r({\mathcal{M}}(\ell ),d)=\mathop{\sum }_{\substack{ d\leqslant N^{\unicode[STIX]{x1D70C}_{1}} \\ (d,K)=1}}\unicode[STIX]{x1D706}_{d}^{+,\text{LIN}}\mathop{\sum }_{\substack{ \ell \leqslant N^{3/4+\unicode[STIX]{x1D700}} \\ (\ell ,d)=1 \\ (\ell ,K)=\unicode[STIX]{x1D6FF}}}1_{{\mathcal{L}}}(\ell )r({\mathcal{M}}(\ell ),d),\qquad \; & & \displaystyle\end{eqnarray}$$

since $1_{{\mathcal{L}}}(\ell )$ is supported on $\ell \leqslant 3K^{2}N^{1-1/\unicode[STIX]{x1D70E}}\leqslant N^{3/4+\unicode[STIX]{x1D700}}$ . Concerning the error sum in (6.6), observe that

$$\begin{eqnarray}\displaystyle 1_{{\mathcal{L}}}(\ell )=\mathop{\sum }_{\substack{ \ell =k\cdot \unicode[STIX]{x1D6FF}m \\ N^{1/\unicode[STIX]{x1D70E}}\leqslant k\leqslant (3KN)^{1/2} \\ k\leqslant (3KN/m)^{1/2}}}1_{{\mathcal{P}}_{4,-1}}(k)1_{{\mathcal{P}}_{4,1}^{\ast }}(m)1_{(m,K/\unicode[STIX]{x1D6FF})=1}, & & \displaystyle \nonumber\end{eqnarray}$$

so $1_{{\mathcal{L}}}(\ell )$ is of convolution type (for the value of $\unicode[STIX]{x1D70E}$ we are considering), except for the cross condition $k\leqslant (3KN/m)^{1/2}$ . We use Perron’s formula in the form

$$\begin{eqnarray}\displaystyle 1_{(1,\infty )}(y) & = & \displaystyle \frac{1}{\unicode[STIX]{x1D70B}}\int _{-N^{4}}^{N^{4}}\frac{\sin (t\log y)}{t}\,dt+O\biggl(\frac{1}{N^{4}|\!\log y|}\biggr)\nonumber\\ \displaystyle & = & \displaystyle \frac{2}{\unicode[STIX]{x1D70B}}\int _{N^{-5}}^{N^{4}}\frac{\sin (t\log y)}{t}\,dt+O\biggl(\frac{1}{N^{4}|\!\log y|}+\frac{|\!\log y|}{N^{5}}\biggr)\nonumber\end{eqnarray}$$

for $N^{-3}<y\leqslant N^{3},y\neq 1$ to dispose of the cross condition. We choose $y=3KN/k^{2}m$ , which satisfies $|y-1|\geqslant 1/3KN^{2}$ after altering $N$ by ${\leqslant}1$ if necessary, so that the error term in Perron’s formula becomes $O(K/N^{2})$ . According to the addition formula for sine, we have

$$\begin{eqnarray}\displaystyle \sin (t\log y) & = & \displaystyle \sin (t\log (3KN)-t\log k^{2})\cos (t\log m)\nonumber\\ \displaystyle & & \displaystyle -\,\cos (t\log (3KN)-t\log k^{2})\sin (t\log m),\nonumber\end{eqnarray}$$

which permits us to separate the variables $k$ and $m$ . Then we have

$$\begin{eqnarray}\displaystyle 1_{{\mathcal{L}}}(\ell ) & = & \displaystyle \frac{2}{\unicode[STIX]{x1D70B}}\int _{N^{-4}}^{N^{3}}\frac{1}{t}\mathop{\sum }_{\substack{ \ell =k\cdot \unicode[STIX]{x1D6FF}m \\ N^{1/\unicode[STIX]{x1D70E}}\leqslant k\leqslant (3KN)^{1/2}}}(\unicode[STIX]{x1D6FC}_{k}^{(1)}(t)\unicode[STIX]{x1D6FD}_{m}^{(1)}(t)-\unicode[STIX]{x1D6FC}_{k}^{(2)}(t)\unicode[STIX]{x1D6FD}_{m}^{(2)}(t))\,dt\nonumber\\ \displaystyle & & \displaystyle +\,O\biggl(\frac{1}{N^{2-\unicode[STIX]{x1D700}}}\biggr),\nonumber\end{eqnarray}$$

where $|\unicode[STIX]{x1D6FC}_{k}^{(j)}(t)|,|\unicode[STIX]{x1D6FD}_{m}^{(j)}(t)|\leqslant 1$ and $t\mapsto \unicode[STIX]{x1D6FC}_{k}^{(j)}(t)$ and $t\mapsto \unicode[STIX]{x1D6FD}_{m}^{(j)}(t)$ are continuous and $\unicode[STIX]{x1D6FC}_{k}^{(j)}(t)$ is supported on $N^{1/\unicode[STIX]{x1D70E}}\leqslant k\leqslant (3KN)^{1/2}$ . Substituting this to (6.8), Hypothesis 6.4 tells us that

$$\begin{eqnarray}\displaystyle \mathop{\sum }_{\ell \in {\mathcal{L}}}\mathop{\sum }_{d\leqslant N^{\unicode[STIX]{x1D70C}_{1}}}\unicode[STIX]{x1D706}_{d}^{+,\text{LIN}}r({\mathcal{M}}(\ell ),d)\ll \frac{\mathop{\sum }_{n\sim N}\unicode[STIX]{x1D714}_{n}}{(\log N)^{99}}+O(N^{1/2-\unicode[STIX]{x1D700}}). & & \displaystyle \nonumber\end{eqnarray}$$

We sum (6.6) over $\ell \in {\mathcal{L}}$ and make use of (6.7), after which we have obtained

$$\begin{eqnarray}\displaystyle \mathop{\sum }_{\ell \in {\mathcal{L}}}S({\mathcal{M}}(\ell ),{\mathcal{P}}(\ell ),N^{1/5}) & {\leqslant} & \displaystyle (F(s)+o(1))\nonumber\\ \displaystyle & & \displaystyle \times \,\frac{K}{\unicode[STIX]{x1D711}(K/\unicode[STIX]{x1D6FF})}\mathop{\sum }_{n\sim N}\mathop{\sum }_{\ell \in {\mathcal{L}}}\frac{\unicode[STIX]{x1D714}_{n}}{\ell \log \!(Kn/\ell )}V_{K}^{\text{LIN}}(N^{1/6},\ell )\nonumber\\ \displaystyle & & \displaystyle +\,O\biggl(\frac{\mathop{\sum }_{n\sim N}\unicode[STIX]{x1D714}_{n}}{(\log N)^{99}}\biggr)\nonumber\\ \displaystyle & = & \displaystyle \biggl(\frac{2\text{e}^{\unicode[STIX]{x1D6FE}}}{5\unicode[STIX]{x1D70C}_{1}}+o(1)\biggr)\cdot \frac{K}{\unicode[STIX]{x1D711}(K/\unicode[STIX]{x1D6FF})}\mathop{\sum }_{\ell \in {\mathcal{L}}}\frac{\mathfrak{f}(K\ell )}{\ell \log \!(KN/\ell )}\nonumber\\ \displaystyle & & \displaystyle \times \,\mathop{\sum }_{n\sim N}\unicode[STIX]{x1D714}_{n}\cdot \frac{2C_{4,1}C_{4,-1}\text{e}^{-\unicode[STIX]{x1D6FE}}}{1/5\log N}+O\biggl(\frac{\mathop{\sum }_{n\sim N}\unicode[STIX]{x1D714}_{n}}{(\log N)^{99}}\biggr).\nonumber\end{eqnarray}$$

We analyze the sum over ${\mathcal{L}}$ in the above formula. Denoting ${\mathcal{L}}^{\prime }=\{\ell /\unicode[STIX]{x1D6FF}:\ell \in {\mathcal{L}}\}$ , it is

$$\begin{eqnarray}\displaystyle \mathop{\sum }_{\ell \in {\mathcal{L}}}\frac{\mathfrak{f}(K\ell )}{\ell \log \!(KN/\ell )}=\biggl(\frac{1}{\unicode[STIX]{x1D6FF}}+o(1)\biggr)\mathop{\sum }_{\ell ^{\prime }\in {\mathcal{L}}^{\prime }}\frac{\mathfrak{f}(K\ell ^{\prime })}{\ell ^{\prime }\log \!(KN/\ell ^{\prime })}1_{(\ell ^{\prime },K/\unicode[STIX]{x1D6FF})=1}, & & \displaystyle \nonumber\end{eqnarray}$$

since $\unicode[STIX]{x1D6FF}\mid K$ . The previous sum can be written as

(6.9) $$\begin{eqnarray}\displaystyle & & \displaystyle (1+o(1))\mathop{\sum }_{m\leqslant N^{1-2/\unicode[STIX]{x1D70E}+\unicode[STIX]{x1D700}}}\frac{u(m)\mathfrak{f}(Km)1_{(m,K/\unicode[STIX]{x1D6FF})=1}}{m}\nonumber\\ \displaystyle & & \displaystyle \quad \times \,\mathop{\sum }_{\substack{ N^{1/\unicode[STIX]{x1D70E}}\leqslant p\leqslant (3KN/m)^{1/2} \\ p\equiv -1~(\text{mod}~4)}}\frac{1}{p\log \!(N/pm)},\end{eqnarray}$$

where $u(m)$ is the characteristic function of ${\mathcal{P}}_{4,1}^{\ast }$ . To evaluate this sum, we study the sum

(6.10) $$\begin{eqnarray}\displaystyle \mathop{\sum }_{m\leqslant x}u(m)\mathfrak{f}(Km)1_{(m,K/\unicode[STIX]{x1D6FF})=1}. & & \displaystyle\end{eqnarray}$$

The sum can be written as

$$\begin{eqnarray}\displaystyle \mathfrak{f}(K)\mathop{\sum }_{m\leqslant x}u(m)\mathfrak{f}(\unicode[STIX]{x1D713}_{K}(m))1_{(m,K/\unicode[STIX]{x1D6FF})=1}\quad \text{where}~\unicode[STIX]{x1D713}_{K}(m)=\mathop{\prod }_{\substack{ p\mid m \\ p\nmid K}}p & & \displaystyle \nonumber\end{eqnarray}$$

and the advantage is that $\mathfrak{f}(\unicode[STIX]{x1D713}_{K}(m))$ is a multiplicative function. By Wirsing’s theorem [Reference Wirsing23, Satz 1] applied to the non-negative multiplicative function $h(m)=u(m)\mathfrak{f}(\unicode[STIX]{x1D713}_{K}(m))1_{(m,K/\unicode[STIX]{x1D6FF})=1}$ (which is bounded by $2$ at prime powers and fulfills $\sum _{p\leqslant x}h(p)\log p=({\textstyle \frac{1}{2}}+o(1))x$ ), we see that (6.10) equals

$$\begin{eqnarray}\displaystyle & & \displaystyle (\mathfrak{f}(K)+o(1))\frac{\text{e}^{-\unicode[STIX]{x1D6FE}/2}}{\sqrt{\unicode[STIX]{x1D70B}}}\frac{x}{\log x}\mathop{\prod }_{\substack{ p\leqslant x \\ p\nmid K/\unicode[STIX]{x1D6FF} \\ p\equiv 1~(\text{mod}~4)}}\biggl(1+\frac{h(p)}{p}+\frac{h(p^{2})}{p^{2}}+\cdots \,\biggr)\nonumber\\ \displaystyle & & \displaystyle \quad =(\mathfrak{f}(K)+o(1))\frac{\text{e}^{-\unicode[STIX]{x1D6FE}/2}}{\sqrt{\unicode[STIX]{x1D70B}}}\frac{x}{\log x}\mathop{\prod }_{\substack{ p\leqslant x \\ p\nmid K \\ p\equiv 1~(\text{mod}~4)}}\biggl(1+\frac{1}{p-2}\biggr)\mathop{\prod }_{\substack{ p\mid K \\ p\nmid K/\unicode[STIX]{x1D6FF} \\ p\equiv 1~(\text{mod}~4)}}\biggl(1-\frac{1}{p}\biggr)^{-1}.\nonumber\end{eqnarray}$$

Applying Wirsing’s theorem reversely, this is

$$\begin{eqnarray}\displaystyle (\mathfrak{f}(K)+o(1))\mathop{\sum }_{m\leqslant x}u(m)\mathfrak{f}(m)\cdot \!\!\mathop{\prod }_{\substack{ p\mid K \\ p\equiv 1~(\text{mod}~4)}}\!\biggl(1+\frac{1}{p-2}\biggr)^{-1}\!\!\mathop{\prod }_{\substack{ p\mid K \\ p\nmid K/\unicode[STIX]{x1D6FF} \\ p\equiv 1~(\text{mod}~4)}}\!\biggl(1-\frac{1}{p}\biggr)^{-1}. & & \displaystyle \nonumber\end{eqnarray}$$

By [Reference Wu24, Lemma 3], we have

$$\begin{eqnarray}\displaystyle \mathop{\sum }_{m\leqslant x}u(m)\mathfrak{f}(m)=(1+o(1))\frac{A}{C_{4,1}}\frac{x}{(\log x)^{1/2}}. & & \displaystyle \nonumber\end{eqnarray}$$

Now, using the same argument as in the proof of [Reference Matomäki12, Lemma 5], we compute that (6.9) equals

$$\begin{eqnarray}\displaystyle & & \displaystyle \frac{A+o(1)}{C_{4,1}(\log N)^{1/2}}\cdot \frac{1}{2}\int _{2}^{\unicode[STIX]{x1D70E}}\frac{\log (t-1)}{t(1-t/\unicode[STIX]{x1D70E})^{1/2}}\,dt\nonumber\\ \displaystyle & & \displaystyle \quad \times \,\frac{\mathfrak{f}(K)}{\unicode[STIX]{x1D6FF}}\mathop{\prod }_{\substack{ p\mid K \\ p\equiv 1~(\text{mod}~4)}}\biggl(1+\frac{1}{p-2}\biggr)^{-1}\mathop{\prod }_{\substack{ p\mid K \\ p\nmid K/\unicode[STIX]{x1D6FF} \\ p\equiv 1~(\text{mod}~4)}}\biggl(1-\frac{1}{p}\biggr)^{-1}.\nonumber\end{eqnarray}$$

Concluding the proof. Now we have

(6.11) $$\begin{eqnarray}\displaystyle T & {\leqslant} & \displaystyle \frac{4AC_{4,-1}+o(1)}{(\log N)^{3/2}}\frac{I_{2}(\unicode[STIX]{x1D70C}_{1},\unicode[STIX]{x1D70E})K\mathfrak{f}(K)}{\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D711}(K/\unicode[STIX]{x1D6FF})}\nonumber\\ \displaystyle & & \displaystyle \times \,\mathop{\prod }_{\substack{ p\mid K \\ p\equiv 1~(\text{mod}~4)}}\biggl(1+\frac{1}{p-2}\biggr)^{-1}\mathop{\prod }_{\substack{ p\mid K \\ p\nmid K/\unicode[STIX]{x1D6FF} \\ p\equiv 1~(\text{mod}~4)}}\biggl(1-\frac{1}{p}\biggr)^{-1}\mathop{\sum }_{n\sim N}\unicode[STIX]{x1D714}_{n},\quad \qquad\end{eqnarray}$$

where

$$\begin{eqnarray}\displaystyle I_{2}(\unicode[STIX]{x1D70C}_{1},\unicode[STIX]{x1D70E})=\frac{1}{2\unicode[STIX]{x1D70C}_{1}}\int _{2}^{\unicode[STIX]{x1D70E}}\frac{\log (t-1)}{t(1-t/\unicode[STIX]{x1D70E})^{1/2}}\,dt. & & \displaystyle \nonumber\end{eqnarray}$$

We claim that the local factors in (6.5) and (6.11) are identical, or in other words that

(6.12) $$\begin{eqnarray}\displaystyle & & \displaystyle \mathop{\prod }_{p\mid K}\biggl(1-\frac{1}{p}\biggr)^{-1}\mathop{\prod }_{\substack{ p\mid K \\ p\equiv -1~(\text{mod}~4)}}\biggl(1-\frac{1}{p-1}\biggr)^{-1}\nonumber\\ \displaystyle & & \displaystyle \quad =\mathop{\prod }_{p\mid K/\unicode[STIX]{x1D6FF}}\biggl(1-\frac{1}{p}\biggr)^{-1}\mathop{\prod }_{\substack{ p\mid K \\ p>2}}\biggl(1-\frac{1}{p-1}\biggr)^{-1}\nonumber\\ \displaystyle & & \displaystyle \qquad \times \,\mathop{\prod }_{\substack{ p\mid K \\ p\equiv 1~(\text{mod}~4)}}\biggl(1+\frac{1}{p-2}\biggr)^{-1}\mathop{\prod }_{\substack{ p\mid K \\ p\nmid K/\unicode[STIX]{x1D6FF} \\ p\equiv 1~(\text{mod}~4)}}\biggl(1-\frac{1}{p}\biggr)^{-1}.\end{eqnarray}$$

By the identity $(1+1/(p-2))^{-1}=1-1/(p-1)$ , (6.12) is equivalent to

$$\begin{eqnarray}\displaystyle \mathop{\prod }_{p\mid K}\biggl(1-\frac{1}{p}\biggr)^{-1}=\mathop{\prod }_{p\mid K/\unicode[STIX]{x1D6FF}}\biggl(1-\frac{1}{p}\biggr)^{-1}\mathop{\prod }_{\substack{ p\mid K \\ p\nmid K/\unicode[STIX]{x1D6FF} \\ p\equiv 1~(\text{mod}~4)}}\biggl(1-\frac{1}{p}\biggr)^{-1}, & & \displaystyle \nonumber\end{eqnarray}$$

which in turn is equivalent to the non-existence of a prime $p\not \equiv 1~(\text{mod}~4)$ for which $p\mid K$ , $p\nmid K/\unicode[STIX]{x1D6FF}$ . If $p\geqslant 5$ were such a prime, we would have $p\mid \unicode[STIX]{x1D6FF}$ , so $p\mid b-1$ , which contradicts the definition of amenability. We also cannot have $p=2$ or $p=3$ , since $2\mid K/\unicode[STIX]{x1D6FF}$ and $3\mid K/\unicode[STIX]{x1D6FF}$ for $\unicode[STIX]{x1D6FF}=(b-1,K)$ by amenability.

Thus, no such $p$ exists and (6.12) holds. Furthermore, it is clear that (6.12) is at least $0.01\mathfrak{S}(L)$ . Consequently,

$$\begin{eqnarray}\displaystyle S-T & {\geqslant} & \displaystyle (0.01+o(1))4AC_{4,-1}\mathfrak{S}(L)(I_{1}(\unicode[STIX]{x1D70C}_{2},\unicode[STIX]{x1D70E})-I_{2}(\unicode[STIX]{x1D70C}_{1},\unicode[STIX]{x1D70E}))\nonumber\\ \displaystyle & & \displaystyle \times \,\frac{\mathop{\sum }_{n\sim N}\unicode[STIX]{x1D714}_{n}}{(\log N)^{3/2}}+O(N^{1/2}).\nonumber\end{eqnarray}$$

Owing to the fact that $\text{H}(\unicode[STIX]{x1D70C}_{1},\unicode[STIX]{x1D70C}_{2},\unicode[STIX]{x1D70E})$ is assumed to be true, we have $I_{1}(\unicode[STIX]{x1D70C}_{2},\unicode[STIX]{x1D70E})-I_{2}(\unicode[STIX]{x1D70C}_{1},\unicode[STIX]{x1D70E})\geqslant 10^{-10}$ , and this completes the proof of Theorem 6.5 in view of (6.2) and (6.3).◻

Proof. This is [Reference Matomäki and Shao15, Lemma 3.2]. ◻

Proof. We follow the proof of [Reference Matomäki and Shao15, Lemma 8.3]. It suffices to consider the sum over $0<r\leqslant R$ . Our task is to estimate

$$\begin{eqnarray}\displaystyle S_{r}=\mathop{\sum }_{d\leqslant N^{\unicode[STIX]{x1D70C}}}\max _{(c,dv)=1}\bigg|\mathop{\sum }_{\substack{ N\leqslant mn\leqslant t \\ mn\equiv c~(\text{mod}~dv) \\ m\leqslant M}}\unicode[STIX]{x1D6FC}_{m}e(\unicode[STIX]{x1D709}rmn)\bigg| & & \displaystyle \nonumber\end{eqnarray}$$

for $r\leqslant R$ . The inner sum in the definition of $S_{r}$ is a geometric sum in the variable $n$ , so evaluating it provides the bound

$$\begin{eqnarray}\displaystyle S_{r}\ll \mathop{\sum }_{d\leqslant N^{\unicode[STIX]{x1D70C}}}\mathop{\sum }_{m\leqslant M}|\unicode[STIX]{x1D6FC}_{m}|\min \biggl\{\frac{RN}{rm\,dv},\frac{1}{\Vert r\unicode[STIX]{x1D709}m\,dv\Vert }\biggr\}. & & \displaystyle \nonumber\end{eqnarray}$$

Observe that $|v\unicode[STIX]{x1D709}-av/q|\leqslant 1/q^{2}$ . Based on this, writing $d^{\prime }=rm\,d$ and using a standard bound for sums over fractional parts [Reference Matomäki and Shao15, Lemma B.3] (taking $x=RN/v$ in that lemma), we get

$$\begin{eqnarray}\displaystyle \mathop{\sum }_{r\leqslant R}S_{r} & \ll & \displaystyle \mathop{\sum }_{d^{\prime }\leqslant RMN^{\unicode[STIX]{x1D70C}}}\unicode[STIX]{x1D70F}(d^{\prime })^{5}\min \biggl\{\frac{RN}{d^{\prime }v},\frac{1}{\Vert v\unicode[STIX]{x1D709}d^{\prime }\Vert }\biggr\}(\log N)\nonumber\\ \displaystyle & \ll & \displaystyle \biggl(\frac{RN}{vq_{v}^{1/2}}+\biggl(\frac{RN\cdot RMN^{\unicode[STIX]{x1D70C}}}{v}\biggr)^{1/2}+\biggl(\frac{RN}{v}q_{v}\biggr)^{1/2}\biggr)(\log N)^{1000}\nonumber\\ \displaystyle & \ll & \displaystyle \biggl(\frac{RN}{v}\biggr)^{1/2}\biggl(RMN^{\unicode[STIX]{x1D70C}}+\frac{RN}{vq_{v}}+q_{v}\biggr)^{1/2}(\log N)^{1000},\nonumber\end{eqnarray}$$

as wanted. ◻

Proof. We follow the proof of [Reference Matomäki and Shao15, Lemma 8.4], which in turn is based on an argument of Mikawa [Reference Mikawa16]. It suffices to consider the case $r>0$ . We will first prove the lemma in the case $F_{1}=\min \{F_{1},F_{2}\}$ . Let us write

$$\begin{eqnarray}\displaystyle I_{r}=\mathop{\sum }_{d_{1}\sim \unicode[STIX]{x1D6E5}_{1}}\mathop{\sum }_{\substack{ d_{2}\sim \unicode[STIX]{x1D6E5}_{2} \\ (d_{2},c^{\prime }d_{1}v)=1}}\max _{(c,d_{1}v)=1}\bigg|\mathop{\sum }_{\substack{ N\leqslant mn\leqslant t \\ mn\equiv c~(\text{mod}~d_{1}v) \\ mn\equiv c^{\prime }~(\text{mod}~d_{2}) \\ m\sim M}}\unicode[STIX]{x1D6FC}_{m}\unicode[STIX]{x1D6FD}_{n}e(\unicode[STIX]{x1D709}rmn)\bigg|, & & \displaystyle \nonumber\end{eqnarray}$$

so that $\sum _{r\leqslant R}I_{r}$ is what we are interested in. Since $\unicode[STIX]{x1D6E5}_{1}\unicode[STIX]{x1D6E5}_{2}^{2}\leqslant M/v$ , a formula on [Reference Matomäki and Shao15, p. 37] tells us (with $x=N,$ $D=\unicode[STIX]{x1D6E5}_{1}$ , $\unicode[STIX]{x1D6FC}=r\unicode[STIX]{x1D709}$ ) that

$$\begin{eqnarray}\displaystyle I_{r}^{2} & \ll & \displaystyle N(\log N)^{100}\biggl(\unicode[STIX]{x1D6E5}_{1}\mathop{\sum }_{d_{1}\sim \unicode[STIX]{x1D6E5}_{1}}\mathop{\sum }_{0<|j|\leqslant 8\unicode[STIX]{x1D6E5}_{2}^{2}N/\unicode[STIX]{x1D6E5}_{1}Mv}\unicode[STIX]{x1D70F}_{3}(j)\nonumber\\ \displaystyle & & \displaystyle \times \,\min \biggl\{\frac{RN}{r(d_{1}v)^{2}|j|},\frac{1}{\Vert r\unicode[STIX]{x1D709}(d_{1}v)^{2}|j|\Vert }\biggr\}+\frac{\unicode[STIX]{x1D6E5}_{1}M}{v}\biggr)\nonumber\end{eqnarray}$$

(since the term $(x^{2}/Q^{2})(\log x)^{-C+10}$ present in that formula of [Reference Matomäki and Shao15] can be replaced with $(DMx/Q)(\log x)^{100}$ without changing anything in the proof). Using the Cauchy–Schwarz inequality, we obtain

(8.1) $$\begin{eqnarray}\displaystyle \frac{1}{(\log N)^{200}}\mathop{\sum }_{r\leqslant R}I_{r} & {\leqslant} & \displaystyle \frac{1}{(\log N)^{200}}R^{1/2}\biggl(\mathop{\sum }_{r\leqslant R}I_{r}^{2}\biggr)^{1/2}\nonumber\\ \displaystyle & {\leqslant} & \displaystyle (RN)^{1/2}\biggl(\unicode[STIX]{x1D6E5}_{1}\mathop{\sum }_{d_{1}\sim \unicode[STIX]{x1D6E5}_{1}}\mathop{\sum }_{0<|j|\leqslant 8\unicode[STIX]{x1D6E5}_{2}^{2}N/\unicode[STIX]{x1D6E5}_{1}Mv}\mathop{\sum }_{r\leqslant R}\unicode[STIX]{x1D70F}_{3}(j)\nonumber\\ \displaystyle & & \displaystyle \times \,\min \biggl\{\frac{RN}{r(d_{1}v)^{2}|j|},\frac{1}{\Vert r\unicode[STIX]{x1D709}(d_{1}v)^{2}|j|\Vert }\biggr\}+\frac{\unicode[STIX]{x1D6E5}_{1}RM}{v}\biggr)^{1/2}\nonumber\\ \displaystyle & \ll & \displaystyle (RN)^{1/2}\biggl(\unicode[STIX]{x1D6E5}_{1}\mathop{\sum }_{d_{1}\sim \unicode[STIX]{x1D6E5}_{1}}\mathop{\sum }_{1\leqslant \ell \leqslant 8\unicode[STIX]{x1D6E5}_{2}^{2}RN/\unicode[STIX]{x1D6E5}_{1}Mv}\unicode[STIX]{x1D70F}_{4}(\ell )\nonumber\\ \displaystyle & & \displaystyle \times \,\min \biggl\{\frac{RN}{(d_{1}v)^{2}\ell },\frac{1}{\Vert v^{2}\unicode[STIX]{x1D709}d_{1}^{2}\ell \Vert }\biggr\}+\frac{\unicode[STIX]{x1D6E5}_{1}RM}{v}\biggr)^{1/2},\end{eqnarray}$$

after writing $\ell =rj$ . When it comes to the sum above, we can estimate it using the lemma on [Reference Mikawa16, p. 6] (with $\unicode[STIX]{x1D70F}_{3}(\cdot )$ replaced by $\unicode[STIX]{x1D70F}_{4}(\cdot )$ ), stating that

(8.2) $$\begin{eqnarray}\displaystyle & & \displaystyle \unicode[STIX]{x1D6E5}_{1}\mathop{\sum }_{d_{1}\sim \unicode[STIX]{x1D6E5}_{1}}\mathop{\sum }_{\ell \sim J}\unicode[STIX]{x1D70F}_{4}(\ell )\min \biggl\{\frac{x}{d_{1}^{2}\ell },\frac{1}{\Vert \unicode[STIX]{x1D709}^{\prime }d_{1}^{2}\ell \Vert }\biggr\}\nonumber\\ \displaystyle & & \displaystyle \quad \ll \biggl(\unicode[STIX]{x1D6E5}_{1}^{2}J+x^{3/4}\biggl(q^{\prime }+\frac{x}{q^{\prime }}+\frac{x}{\unicode[STIX]{x1D6E5}_{1}}\biggr)^{1/4}\biggr)(\log x)^{100}\end{eqnarray}$$

for $1\leqslant J\leqslant 10x$ and any real number $\unicode[STIX]{x1D709}^{\prime }$ satisfying $|\unicode[STIX]{x1D709}^{\prime }-a^{\prime }/q^{\prime }|\leqslant 1/{q^{\prime }}^{2}$ for some coprime $a^{\prime }$ and $q^{\prime }\leqslant x$ . In the case $q^{\prime }>x$ , (8.2) continues to hold, by trivial estimates. We substitute (8.2) with $x=RN/v^{2}$ , $\unicode[STIX]{x1D709}^{\prime }=v^{2}\unicode[STIX]{x1D709}$ and $J\leqslant 8\unicode[STIX]{x1D6E5}_{2}^{2}RN/\unicode[STIX]{x1D6E5}_{1}Mv$ into (8.1) (we have $J\leqslant 10(RN/v^{2})$ , since $\unicode[STIX]{x1D6E5}_{1}\unicode[STIX]{x1D6E5}_{2}^{2}\leqslant M/v$ ), making use of our assumption on $\unicode[STIX]{x1D709}$ , which implies that $|v^{2}\unicode[STIX]{x1D709}-(av^{2}/(q,v^{2}))/q_{v}|\leqslant 1/q_{v}^{2}$ . This results in the claimed bound.

Then let $F_{2}=\min \{F_{1},F_{2}\}$ . In this situation, we use the orthogonality of characters to bound the sum in Lemma 8.2 with

(8.3) $$\begin{eqnarray}\displaystyle \mathop{\sum }_{r\leqslant R}\mathop{\sum }_{d_{1}\sim \unicode[STIX]{x1D6E5}_{1}}\mathop{\sum }_{d_{2}\sim \unicode[STIX]{x1D6E5}_{2}}\!\!\max _{\unicode[STIX]{x1D713}~(\text{mod}~d_{1}d_{2})}\bigg|\!\mathop{\sum }_{\substack{ N\leqslant mn\leqslant t \\ mn\equiv c_{v}(d_{1},d_{2})~(\text{mod}~v) \\ m\sim M}}\!\unicode[STIX]{x1D6FC}_{m}\unicode[STIX]{x1D713}(m)\unicode[STIX]{x1D6FD}_{n}\unicode[STIX]{x1D713}(n)e(\unicode[STIX]{x1D709}rmn)\bigg|, & & \displaystyle \nonumber\\ \displaystyle & & \displaystyle\end{eqnarray}$$

where $c_{v}(d_{1},d_{2})$ is a suitably chosen integer coprime to $v$ . Estimating the sums over $d_{1}$ and $d_{2}$ trivially and using the Cauchy–Schwarz inequality and expanding a square, we find that (8.3) is, for some $|\unicode[STIX]{x1D6FD}_{n}^{\prime }|\leqslant \unicode[STIX]{x1D70F}(n)^{2}\log n$ and some $c_{v}$ coprime to $v$ ,

(8.4) $$\begin{eqnarray}\displaystyle & {\leqslant} & \displaystyle \unicode[STIX]{x1D6E5}_{1}\unicode[STIX]{x1D6E5}_{2}(RM)^{1/2}\bigg(\mathop{\sum }_{r\leqslant R}\mathop{\sum }_{m\leqslant M}\bigg|\mathop{\sum }_{\substack{ N/m\leqslant n\leqslant t/m \\ n\equiv c_{v}m^{-1}~(\text{mod}~v)}}\unicode[STIX]{x1D6FD}_{n}^{\prime }e(\unicode[STIX]{x1D709}rmn)\bigg|^{2}\bigg)^{1/2}(\log M)^{100}\nonumber\\ \displaystyle & = & \displaystyle \unicode[STIX]{x1D6E5}_{1}\unicode[STIX]{x1D6E5}_{2}(RM)^{1/2}\biggl(\mathop{\sum }_{r\leqslant R}\mathop{\sum }_{\substack{ N/2M\leqslant n_{i}\leqslant 2N/M \\ n_{1}\equiv n_{2}~(\text{mod}~v) \\ \text{for}~i\in \{1,2\}}}\unicode[STIX]{x1D6FD}_{n_{1}}^{\prime }\overline{\unicode[STIX]{x1D6FD}_{n_{2}}^{\prime }}\nonumber\\ \displaystyle & & \displaystyle \times \,\mathop{\sum }_{\substack{ m\leqslant M \\ N/n_{i}\leqslant m\leqslant t/n_{i} \\ m\equiv c_{v}n_{i}^{-1}~(\text{mod}~v) \\ \text{for}~\in \{1,2\}}}e(\unicode[STIX]{x1D709}rm(n_{1}-n_{2}))\biggr)^{1/2}(\log M)^{100}\nonumber\\ \displaystyle & \ll & \displaystyle \unicode[STIX]{x1D6E5}_{1}\unicode[STIX]{x1D6E5}_{2}(RN)^{1/2}\biggl(RM+\mathop{\sum }_{r\leqslant R}\mathop{\sum }_{\substack{ 1\leqslant n\leqslant 2N/M \\ n\equiv 0~(\text{mod}~v)}}T(n)\nonumber\\ \displaystyle & & \displaystyle \times \,\min \biggl\{\frac{RN}{rnv}+1,\frac{1}{\Vert v\unicode[STIX]{x1D709}rn\Vert }\biggr\}\biggr)^{1/2}(\log M)^{101},\end{eqnarray}$$

where

$$\begin{eqnarray}\displaystyle T(n)=\frac{M}{N}\mathop{\sum }_{\substack{ n=n_{1}-n_{2} \\ n_{1},n_{2}\leqslant 2N/M}}\unicode[STIX]{x1D70F}(n_{1})^{2}\unicode[STIX]{x1D70F}(n_{2})^{2}. & & \displaystyle \nonumber\end{eqnarray}$$

We can write $n=kv$ and $\ell =kr$ to bound (8.4) with

(8.5) $$\begin{eqnarray}\displaystyle & \ll & \displaystyle \unicode[STIX]{x1D6E5}_{1}\unicode[STIX]{x1D6E5}_{2}(RN)^{1/2}\biggl(RM+\mathop{\sum }_{\ell \leqslant 2RN/Mv}U(\ell )\min \biggl\{\frac{RN}{\ell v^{2}}+1,\frac{1}{\Vert v^{2}\unicode[STIX]{x1D709}\ell \Vert }\biggr\}\biggr)^{1/2}\nonumber\\ \displaystyle & & \displaystyle \times \,(\log N)^{101},\end{eqnarray}$$

where

$$\begin{eqnarray}\displaystyle U(\ell )=\mathop{\sum }_{\substack{ \ell =\ell _{1}\ell _{2} \\ \ell _{1}\leqslant 2N/Mv}}T(\ell _{1}v). & & \displaystyle \nonumber\end{eqnarray}$$

We apply [Reference Matomäki and Shao15, Lemma B.3] (with $k=20$ ) to (8.5). The weight function $U(\ell )$ is not a divisor function, but the only property of the weight function needed in that lemma is a second-moment bound. Therefore, (8.5) can be bounded with

(8.6) $$\begin{eqnarray}\displaystyle & \ll & \displaystyle \unicode[STIX]{x1D6E5}_{1}\unicode[STIX]{x1D6E5}_{2}(RN)^{1/2}\bigg(\frac{RN}{q_{v}^{1/2}v^{2}}+\frac{RN}{(v^{2}M)^{1/2}}+RM+\bigg(\frac{RNq_{v}}{v^{2}}\bigg)^{1/2}\bigg)^{1/2}\nonumber\\ \displaystyle & & \displaystyle \times \,(\log N)^{1000},\end{eqnarray}$$

once we prove that

(8.7) $$\begin{eqnarray}\displaystyle \mathop{\sum }_{\ell \leqslant 2RN/Mv}U(\ell )^{2}\ll \frac{RN}{Mv}(\log N)^{100}. & & \displaystyle\end{eqnarray}$$

We calculate

(8.8) $$\begin{eqnarray}\displaystyle & & \displaystyle \mathop{\sum }_{\ell \leqslant 2RN/Mv}\biggl(\mathop{\sum }_{\substack{ \ell =\ell _{1}\ell _{2} \\ \ell _{1}\leqslant 2N/Mv}}T(\ell _{1}v)\biggr)^{2}\nonumber\\ \displaystyle & & \displaystyle \quad \ll \frac{RN}{Mv}\mathop{\sum }_{\substack{ \ell _{1}\leqslant 2N/Mv \\ \ell _{1}^{\prime }\leqslant 2N/Mv}}\frac{T(\ell _{1}v)T(\ell _{1}^{\prime }v)}{[\ell _{1},\ell _{1}^{\prime }]}\nonumber\\ \displaystyle & & \displaystyle \quad \ll \frac{RN}{Mv}\mathop{\sum }_{d\leqslant 2N/Mv}\frac{1}{d}\mathop{\sum }_{\substack{ \ell _{1}\leqslant 2N/dMv \\ \ell _{1}^{\prime }\leqslant 2N/dMv}}\frac{T(\ell _{1}dv)T(\ell _{1}^{\prime }dv)}{\ell _{1}\ell _{1}^{\prime }}\nonumber\\ \displaystyle & & \displaystyle \quad =\frac{RN}{Mv}\mathop{\sum }_{d\leqslant 2N/Mv}\frac{1}{d}\biggl(\mathop{\sum }_{\ell \leqslant 2N/dMv}\frac{T(\ell dv)}{\ell }\biggr)^{2}.\end{eqnarray}$$

We can estimate the sum inside the square using

$$\begin{eqnarray}\displaystyle \mathop{\sum }_{\substack{ n\leqslant 2N/M \\ n\equiv 0~(\text{mod}~c)}}\frac{T(n)}{n} & \ll & \displaystyle \frac{M}{N}\mathop{\sum }_{\substack{ n_{1}\leqslant 2N/M \\ n_{2}\leqslant 2N/M \\ n_{1}\equiv n_{2}~(\text{mod}~c) \\ n_{1}>n_{2}}}\frac{\unicode[STIX]{x1D70F}(n_{1})^{2}\unicode[STIX]{x1D70F}(n_{2})^{2}}{n_{1}-n_{2}}\nonumber\\ \displaystyle & \ll & \displaystyle \frac{M}{Nc}\mathop{\sum }_{1\leqslant a\leqslant c}\mathop{\sum }_{\substack{ n_{1}^{\prime }\leqslant 2N/Mc \\ n_{2}^{\prime }\leqslant 2N/Mc \\ n_{1}^{\prime }>n_{2}^{\prime }}}\frac{\unicode[STIX]{x1D70F}(cn_{1}^{\prime }+a)^{2}\unicode[STIX]{x1D70F}(cn_{2}^{\prime }+a)^{2}}{n_{1}^{\prime }-n_{2}^{\prime }}\nonumber\\ \displaystyle & \ll & \displaystyle \frac{M}{Nc}\mathop{\sum }_{1\leqslant a\leqslant c}\mathop{\sum }_{n\leqslant 2N/Mc}\unicode[STIX]{x1D70F}(cn+a)^{4}\nonumber\\ \displaystyle & \ll & \displaystyle \frac{M}{Nc}\mathop{\sum }_{m\leqslant 2N/M+c}\unicode[STIX]{x1D70F}(m)^{4}\ll \frac{1}{c}(\log N)^{15}\nonumber\end{eqnarray}$$

for $c\leqslant 2N/M$ , where we used Hilbert’s inequality [Reference Montgomery17, Ch. 7] in the third last step. Taking $c=dv$ , and substituting to (8.8), we see that (8.7) holds, as claimed. Therefore, we indeed have the bound (8.6) for (8.5) and that bound can be rewritten as the desired bound $F_{2}$ .◻