= Solution
In the fourth moment from part c, put
$$
s_1=r_1-r_2,\quad s_2=r_1+r_2,\qquad
u_1=t_1-t_2,\quad u_2=t_1+t_2.
$$
Then the phase is $e(\alpha s_1s_2u_1u_2)$. The contribution with $s_1=0$ is $O(RT^2)$, because $r_1=r_2$; the contribution with $u_1=0$ is $O(TR^2)$. 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> gives
$$
\left|\sum_{s_1\in I}e(\alpha s_1s_2u_1u_2)\right|
\ll\min(R,\|\alpha s_2u_1u_2\|^{-1}).
$$
Put $m=|s_2u_1u_2|$. We have $m\leq CRT^2$, 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 is
$$
\ll RT^2+TR^2+\sum_{m\leq CRT^2}\tau_4(m)\min(R,\|\alpha m\|^{-1}).
$$
Using $(A+B+C)^{1/4}\leq A^{1/4}+B^{1/4}+C^{1/4}$ in part c proves
$$
|S|\ll\|b\|_2\|c\|_2\left(
R^{1/4}T^{1/2}+R^{1/2}T^{1/4}
+\left(\sum_{m\leq CRT^2}\tau_4(m)\min(R,\|\alpha m\|^{-1})\right)^{1/4}
\right).
$$
This is the <factorized fourth moment for a bilinear quadratic exponential sum>.
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