The [Bombieri-Vinogradov Theorem][1] states that given $A>0$, there exists $B>0$ such that for $Q=\sqrt{x}\left(\log x\right)^{-B},$  we have $$\sum_{q\leq Q}\max_{y\leq x}\max_{\begin{array}{c}
a\text{ mod q}\\
(a,q)=1
\end{array}}\left|\psi(y;q,a)-\frac{y}{\phi(q)}\right|\ll \frac{x}{(\log x)^A}.$$ 

I was wondering what happens if a restriction is put on the $q$ so that they are all divisible by some smaller integer $k$. Are there any non-trivial bounds on the average over $q$ divisible by $k$? Specifically, suppose that $k\leq Q^{1-\epsilon},$ and that there is no [Siegel zero][2]${}^{++}$ for any $\chi$ modulo $k$.  Is it true that $$\sum_{\begin{array}{c}
q\leq Q\\
k|q
\end{array}}\max_{y\leq x}\max_{\begin{array}{c}
a\text{ mod q}\\
(a,q)=1
\end{array}}\left|\psi(y;q,a)-\frac{y}{\phi(q)}\right|\ll_{\epsilon}\frac{1}{k} \frac{x}{(\log x)^A}.$$ 

Any references would be greatly appreciated.

Thanks for your help,

${}^{++}$ **Edit:** As mentioned by Terence Tao in the comments, there is an issue regarding Siegel zeros modulo $k$.  The original result I was asking about would give stronger bounds on the location of a possible Siegel zero modulo $k$, and for that reason it is out of reach.  

I have added a brief heuristic for why we have to assume that there are no Siegel zeros modulo $k$ for anyone who is interested. (I wrote this mainly for my own understanding)
  
>**Heuristic:** Suppose that $k$ is a small power of $x$.  If there is an exceptional zero $\beta$ for a quadratic character $\chi$  modulo $k$ , then for every $q$ such that $k|q,$ the induced character modulo $q,$ $\chi^{\star},$ will have the same exceptional zero, and so we expect that $$\psi(x;q,a)\approx\frac{x}{\phi(q)}-\frac{\chi^{\star}(a)}{\phi(q)}\frac{x^{\beta}}{\beta}+small$$ for each $q.$  This leads us to expect that $$\sum_{\begin{array}{c}
q\leq Q\\
k|q
\end{array}}\max_{y\leq x}\max_{a\text{ mod }q}\left|\psi\left(y;q,a\right)-\frac{y}{\phi(q)}\right|\approx x^{\beta}\sum_{\begin{array}{c}
q\leq Q\\
k|q
\end{array}}\frac{1}{\phi(q)}\approx\frac{x^{\beta}\log x}{k},$$ however, the desired upper bound is too strong, as $$x^{\beta}\log x\ll \frac{x}{\left(\log x\right)^{B}},$$  implies that for some $C>0$, $$\beta\leq1-C\frac{\log\log x}{\log x},$$ which is stronger than the long standing bound by Siegel $$\beta\leq1-C(\epsilon)x^{-\epsilon}.$$  In the above, we have used $x$ interchangeably with $k$ for the bounds since $k$ is a small power of $x$.


  [1]: http://en.wikipedia.org/wiki/Bombieri%E2%80%93Vinogradov_theorem
  [2]: http://en.wikipedia.org/wiki/Siegel_zero