Write $M_0=CRT^2$ and split the last sum in part d according as $\tau_4(m)<X$ or $\tau_4(m)\geq X$. For the first part, the <reciprocal fractional-part sum near a rational> gives
$$
\sum_{\substack{m\leq M_0\\\tau_4(m)<X}}
\tau_4(m)\min(R,\|\alpha m\|^{-1})
\ll X(\log q)\left(\frac{M_0R}{q}+M_0+R+q\right).
$$
For the second part, use the supplied second-moment estimate and <truncation of a divisor weight by its second moment>:
$$
\begin{aligned}
\sum_{\substack{m\leq M_0\\\tau_4(m)\geq X}}
\tau_4(m)\min(R,\|\alpha m\|^{-1})
&\leq R\sum_{\substack{m\leq M_0\\\tau_4(m)\geq X}}\tau_4(m)\\
&\ll\frac{RM_0}{X}(\log M_0)^{O(1)}.
\end{aligned}
$$
Substitute $M_0\asymp RT^2$, take fourth roots, and factor out $R^{1/2}T^{1/2}$. The four terms from the bounded-weight estimate become, after harmless enlargement by $(\log q)(\log RT)^{O(1)}$,
$$
\frac{X^{1/4}}{q^{1/4}},qquad
\frac{X^{1/4}}{R^{1/4}},qquad
\frac{X^{1/4}q^{1/4}}{R^{1/2}T^{1/2}},
$$
with the smaller $XR$ term absorbed by $XM_0$. The large-weight part contributes $X^{-1/4}$. Finally, the two diagonal terms from part d contribute $R^{-1/4}$ and $T^{-1/4}$; since $X\geq1$, the first is absorbed by $X^{1/4}R^{-1/4}$. Therefore
$$
\boxed{
|S|\ll\|b\|_2\|c\|_2(\log q)(\log RT)^{O(1)}R^{1/2}T^{1/2}
\left(
\frac1{X^{1/4}}+\frac{X^{1/4}}{q^{1/4}}
+\frac{X^{1/4}}{R^{1/4}}
+\frac{X^{1/4}q^{1/4}}{R^{1/2}T^{1/2}}
+\frac1{T^{1/4}}
\right).}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 117 / 3 / e / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-117/3/e/solution