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With a graduate student, I'm going through the paper (Proc. London Math. Soc. (3) 47 (1983), no. 2, 193–224.) We are quite baffled by what follows Lemma 3 on p. 198.

Here's the background and notation. We have two linear forms $l_i(n)=\alpha_in+\beta_i$, $i=1,2$ with $(\alpha_i,\beta_i)=1$, $2|\alpha_i$, as well as several other side conditions ((1.6)-(1.9) p. 194)) which imply that $(l_1(n),l_2(n))=1$.

We have a quadratic character $\chi$ modulo $q$, with Siegel zero $\beta_0$. $\eta=((1-\beta_0)\log q)^{-1}$, so $3\le \eta\ll q$ is known. Let $L=\log q$. We take $$ q^{250}\le x\le q^{500}, $$ and $$ 2\le z\le q, $$ and let $z_0=\log q/\log z\ge 1$. Following the statement of Lemma 3 p. 198 are two consecutive estimates: $$ \sum_{p\ge z}\sum_{x<n\le 2x, p^2|l_i}L^2 2^{\omega(l_1)+\omega(l_2)}\ll xL^2\exp(A L/\log L)z^{-1}, $$ and subsequently $$ xL^2\exp(A L/\log L)z^{-1}\ll xz_0^{-1}, $$ if $z_0\le A\log\log\eta$. Here $A$ is a positive constant which need not be the same at each occurrence.

In the first estimate, we need to read $p^2|l_i$ as $p^2$ divides either $l_1(n)$ or $l_2(n)$ since we already know these are relatively prime. Regardless, I'm completely stumped.

In the second, it would suffice to show $$ L^3\exp(AL/\log L)\ll z\log z. $$ We know $\log\log\eta\ll\log L$ so Heath-Brown's side condition is that $L/\log L\ll z$, but again I am baffled.

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    $\begingroup$ If I were you, I would ask Heath-Brown. $\endgroup$
    – GH from MO
    Aug 17, 2015 at 17:01
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    $\begingroup$ @GHfromMO No one likes to think about work they did 33 years ago. In any case, he (very politely) declined to answer a different question about work that's only 20 years old, so I'm not inclined to bother him again. $\endgroup$
    – Stopple
    Aug 17, 2015 at 17:14

1 Answer 1

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In the first step, I think you just estimate the $2^{\omega(l_1)+\omega(l_2)}$ as a divisor function, so bounded by $\exp(A\log n/\log\log n)$, with $n\sim x$ and $\log x\asymp\log q$ then converting this estimate to $L$.

The remaining part of the sum is then "obvious" (I guess), as $p^2$ divides $l_i$ about $1/p^2$ of the time, so you just sum $1/p^2$ for $p\ge z$ and get $\ll 1/z$.

For step 2, after unwinding and taking logs, you want to show essentially that $$2\log L+\frac{A_1L}{\log L}=2\log\log q+\frac{A_1\log q}{\log\log q}\le \log(z/z_0)+O(1),$$ and the first and last terms can essentially be neglected.

Now note $z_0=\frac{\log q}{\log z}\le A_2\log\log q$, so $\log z\ge\frac{\log q}{A_2\log\log q}$, and $\log(z_0)\ll\log\log\log q$.

Just take $A_2$ a bit smaller than the reciprocal of $A_1$ and it is done.

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    $\begingroup$ The overall point is that $z$ is large (by $z_0$ being small), say $\gg q^{C/\log\log q}$, and so the $p\ge z$ chopping dominates the divisor function. There's an extra couple of harmless logs thrown in. $\endgroup$
    – kantelope
    Aug 18, 2015 at 4:51

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