I have trouble to conclude what I have to prove for this exercice.
Let $$ c(a) = \frac{\varphi(a)}{a} \prod_{ p \not\mid a} \left( 1 + \frac{1}{p(p-1)} \right) $$
Prove that $$ \sum_{a < p \leq x } \tau(p-a) = c(a) x + O\left( x \frac{ \log \log x}{\log x} \right) $$
Hint: Use Bombieri-Vinogradov theorem and also that for $(a,q)=1$ we have $ \pi(x;q,a) \ll \frac{x}{\varphi(q) \log (x/q) } $ and also $$ \sum_{n \leq x; (n,a)=1} \frac{n}{\varphi(n)} = c(a) x + O(\tau(a)\log x) $$
What I did:
Since $ \tau(n) = 2 \sum_{ d \mid n , d < n^{1/2}} 1 + s(n) $ where $s(n)=1$ if is a perfect square and 0 otherwise we have $$ \sum_{a < p \leq x } \tau(p-a) = 2 \sum_{d < x^{1/2}, (d,a) = 1} ( \pi(x;d,a) - \pi(a+d^2; d, a) ) + O(x^{1/2}) = \star $$ Then using the hint we have that $$ \sum_{d < x^{1/2}, (d,a) =1} \pi(a+d^2; d, a) \ll \sum_{d < x^{1/2}, (d,a) =1} \frac{ a +d^2}{\varphi(d) \log ( (a+d^2)/d)} \ll \frac{ a+ x}{ \log( (a+x)/x^{1/2}) } $$ $$ \ll x/ \log x $$ hence, by Bombieri-Vinogradov Theorem we have $$ \star = 2 \operatorname{Li}(x) \left( \sum_{d \leq x^{1/2} (\log x)^{-B}; (d,a)=1} \frac{1}{\varphi(d)} \right) + 2\sum_{x^{1/2} (\log x)^{-B} < d < x^{1/2}, (d,a)=1} \pi(x;d,a) + O(x/\log x) $$
$$ = 2 \operatorname{Li}(x) \left( \sum_{d \leq x^{1/2} (\log x)^{-B}; (d,a)=1} \frac{1}{\varphi(d)} \right) + O \left( \frac{x}{\log x} + \sum_{x^{1/2} (\log x)^{-B} < d < x^{1/2}, (d,a)=1} \pi(x;d,a) \right) $$
Now in some way i want to show that $$2 \operatorname{Li}(x) \left( \sum_{d \leq x^{1/2} (\log x)^{-B}; (d,a)=1} \frac{1}{\varphi(d)} \right) = c(a) x $$ and that $$ O \left( \frac{x}{\log x} + \sum_{x^{1/2} (\log x)^{-B} < d < x^{1/2}, (d,a)=1} \pi(x;d,a) \right) = O\left( x \frac{ \log \log x}{\log x} \right) $$
But i don't know how. Somebody can help me?
For the reciprocal sum of $1/\varphi(d)$, partial summation shall give
$$ \sum_{\substack{d\le x\\(d,a)=1}}{1\over\varphi(d)}=c(a)\log x+O(1), $$
which indicates that
\begin{aligned} 2\operatorname{Li}(x)\sum_{\substack{d\le x^{1/2}(\log x)^{-B}\\(d,a)=1}}{1\over\varphi(d)} &=c(a){2x\over\log x}\log(x^{1/2}\log^{-B}x)+O\left(x\over\log x\right) \\ &=c(a)x+O\left(x\log\log x\over\log x\right). \end{aligned}
For the second part, apply the Brun-Titchmarsh inequality (i.e. $\pi(x;q,a)\ll{x\over\varphi(q)\log(x/q)}$), so that
\begin{aligned} \sum_{\substack{x^{1/2}(\log x)^{-B}<d\le x^{1/2}\\(d,a)=1}}\pi(x;d,a) &\ll{x\over\log x}\sum_{x^{1/2}(\log x)^{-B}<d\le x^{1/2}}{1\over\varphi(d)} \\ &\ll{x\over\log x}\log{x^{1/2}\over x^{1/2}(\log x)^{-B}}\ll{x\log\log x\over\log x}. \end{aligned}