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

With ${\mu }_{\le U}(d)=\mu (d)1_{d\le U}$ and analogous notation for ${\Lambda }_{\le V}$, Vaughan identity isThus, when $UV\le N$,The first term is short, the next two are Type I sums, and the last becomes a Type II bilinear sum after grouping variables and applying a dyadic decomposition.

Write $\alpha =a/q+\beta$, where $|\beta |\le q^{-2}$. Split the $m$-interval into $O(M/q+1)$ consecutive blocks of length at most $q/2$. If distinct integers $m,n$ lie in one block, then $0<|m-n|<q/2$. Since $(a,q)=1$,whereas $|\beta (m-n)|\le 1/(2q)$. Hence $\Vert \alpha m-\alpha n\Vert \ge 1/(2q)$.

The points $\alpha m$ in one block are therefore $1/(2q)$-separated modulo one. Order them by distance from the nearest integer. Apart from a bounded number of endpoints, the $j$th closest point has distance $\gg j/q$, and consequentlyMultiplying by $O(M/q+1)$ givesThis is the reciprocal fractional-part sum near a rational estimate.

LetApplying the Cauchy-Schwarz inequality in $t$ givesTake absolute values and apply Cauchy-Schwarz to $(r_1,r_2)$. Since ${\sum }_{r_1,r_2}|c_{r_1}{c}_{r_2}{|}^2=\Vert c{\Vert }_2^4$, expanding the remaining square givesTaking fourth roots proves the claimed bilinear quadratic exponential sum fourth-moment bound.

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.

Write $M_0=CR{T}^2$ and split the last sum in part d according as ${\tau }_4(m)<X$ or ${\tau }_4(m)\ge X$. For the first part, the reciprocal fractional-part sum near a rational givesFor the second part, use the supplied second-moment estimate and truncation of a divisor weight by its second moment:Substitute $M_0\asymp R{T}^2$, 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 $(\log q)(\log RT{)}^{O(1)}$,with the smaller $XR$ term absorbed by $X{M}_0$. 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 $X\ge 1$, the first is absorbed by $X^{1/4}{R}^{-1/4}$. Therefore

Assume $RT=N$. To beat the trivial bound $\Vert b{\Vert }_2\Vert c{\Vert }_2{N}^{1/2}$ by $(\log N{)}^{-A}$, it is enough to make every term in the parentheses of part e smaller than a sufficiently larger negative power of $\log N$, allowing for the prefactor $(\log q)(\log N{)}^{O(1)}$.

Choose the splitting parameter $X=(\log N{)}^B$ with $B$ large in terms of $A$. It then suffices, for a still larger constant $C=C(A,B)$, thatIndeed, these four conditions control respectively the last, third, second, and fourth terms, while the choice of $X$ controls the first. Equivalently, away from polylogarithmic neighborhoods of the endpoints, the estimate gives a logarithmic saving wheneverare all sufficiently large powers of $\log N$, with $R$ also larger than the chosen divisor cutoff $X$ by such a power. This is the Type II range used after Vaughan identity.