Bilinear quadratic exponential sum fourth-moment bound

Two applications of the Cauchy-Schwarz inequality giveThe fourth moment exposes differences of squares that can be factored and estimated by one-dimensional geometric sums.

Factoring $r_1^2-r_2^2=(r_1-r_2)(r_1+r_2)$ and similarly in $t$ separates diagonal contributions from a geometric sum. Grouping the three fixed factors by their product $m$ bounds the off-diagonal fourth moment by

If ${\sum }_{n\le N}{\tau }_k(n{)}^2\ll N(\log N{)}^{O(1)}$, then for any $X\ge 1$,This separates a divisor-weighted exponential sum into a bounded-weight part and a sparse large-weight part.