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.



On Heath-Brown's "Prime twins and Siegel zeros"
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.


