 Solution

In the fourth moment from part c, putThen the phase is $e(\alpha s_1{s}_2{u}_1{u}_2)$. The contribution with $s_1=0$ is $O(R{T}^2)$, because $r_1=r_2$; the contribution with $u_1=0$ is $O(T{R}^2)$. Their fourth roots are respectively $R^{1/4}{T}^{1/2}$ and $R^{1/2}{T}^{1/4}$.

Off the diagonals, fix $s_2,u_1,u_2$. The variable $s_1$ ranges over an interval of length $O(R)$, subject only to harmless parity and range restrictions. The exponential geometric sum bound givesPut $m=|s_2{u}_1{u}_2|$. We have $m\le CR{T}^2$, and the number of representations of a fixed $m$ by the three factors, including signs and range restrictions, is $O({\tau }_4(m))$. Hence the fourth moment isUsing $(A+B+C{)}^{1/4}\le A^{1/4}+B^{1/4}+C^{1/4}$ in part c provesThis is the factorized fourth moment for a bilinear quadratic exponential sum.