Here is a sketch of the argument mentioned in the previous post (which arose from the discussions with Étienne Fouvry, Philippe Michel, Paul Nelson, etc, but presentation mistakes are fully mine…).
Theorem. We have
$latex \frac{1}{Q}\sum_{r\sim R}\sum_{s\sim S}\sum_{a\bmod q,\ P(a)= 0\bmod q}\Bigl|S(a,q)-MT_q\Bigr|=O\Bigl(\frac{X}{Q\log^A X}\Bigr)$
provided
$latex Q^{3/2}S^{1/2}<N^{1-\epsilon},\quad\quad R<S,$
where
(1) we put
$latex S(a,q)=\sum_{m\sim M}\alpha_m\sum_{mn_1n_2n_3\equiv a\bmod{q},\ n_i\sim N_i}1,$
and denote by $latex MT_q$ the expected main term;
(2) the parameters are $latex X=MN=MN_1N_2N_3$, $latex Q=RS$, the modulus is $latex q=rs$, the moduli $latex r$ and $latex s$ are coprime and squarefree, and $latex P\in \mathbf{Z}[X]$ is the usual polynomial associated to an admissible tuple.
If we take $latex R=Q^{1/2-\epsilon}$ and $latex S=Q^{1/2+\epsilon}$, we get a non-trivial result for $latex Q$ as large as $latex N^{4/7}X^{-\epsilon}$.
(In fact, the special shape of $latex P$ will play no role in this argument, and any non-constant polynomial will work just as well.)
More precisely, I will give a proof which is — except for its terseness — essentially complete for $latex r$ and $latex s$ of special type, and we anticipate only technical adjustments to cover the general case — we will write this down carefully of course.
Before starting, a natural question may come to mind: given that qui peut le plus, peut le moins, can one give an analogue result for the usual divisor function? Recall that, for the latter, the (individual) exponent of distribution has been known to be at least $latex 2/3$ for a long time (by work of Linnik and Selberg independently, both using the Weil bound for Kloosterman sums.) This exponent has not been improved, even on average over $latex q$ (although Fouvry succeeded, on average over $latex q$, in covering the range $latex X^{2/3-\epsilon}\leq q\leq X^{1-\epsilon}$) despite much effort. However, Fouvry and Iwaniec (with an Appendix by Katz to treat yet another complete exponential sum over finite fields) proved already twenty years ago that one could improve it to $latex 2/3+1/48$ if one averages for a fixed $latex r \leq X^{3/8}$ over special moduli $latex q=rs$ with $latex rs^2\leq X^{1-\epsilon}$ — this gives in particular a nice earlier illustration of the usefulness of factorable moduli for this type of questions.
So, to work. For fixed $latex r$ and $latex s$, we begin by applying the Poisson summation formula to the three “smooth” variables $latex n_i$ (the smoothing is hidden in the notation $latex n_i\sim N_i$); the simultaneous zero frequencies $latex (h_1,h_2,h_3)=(0,0,0)$ give the main term, as they should, and the other degenerate cases are easier to handle than the contribution of the non-zero $latex h_i,$ so the main secondary term for a given $latex q$ and $latex a $ is given by
$latex S_2=\frac{N}{q^2}\sum_m \alpha_m\sum_{1\leq |h_i|\\ H_i} K_3(a\bar{m}h_1h_2h_3;q),$
where the dual lengths are $latex H_i=Q/N_i$, so that the total number of frequencies is $latex H_1H_2H_3=Q^3/N$, and where $latex K_3$ is a normalized hyper-Kloosterman sum modulo $latex rs$:
$latex K_3(u;q)=\frac{1}{q}\sum_{xyz=u}\psi_q(x+y+z),\quad\quad \psi_q(x)=e(x/q).$
(Below I will usually not repeat the range of summations when they are unchanged from one line to the next.)
Now we sum over $latex r$ and $latex s$, move the sum over $latex m$ outside, and apply the Cauchy-Schwarz inequality to the $latex r$ sum, for a fixed $latex (s,a_s,m)$, where $latex a_s$ is $latex a$ modulo $latex s$. To prepare for this step, we use the Chinese Remainder Theorem to split the condition $latex P(a)=0\bmod rs$, and to factor the hyper-Kloosterman sum as a sum modulo $latex r$ times one modulo $latex s$.
The contribution of a fixed $latex (s,a_s,m)$ is
$latex T=\frac{1}{R}\sum_{r}\sum_{P(a_r)=0\bmod{r}}\Bigl|\sum_{h_i}K_3(a\bar{m}h_1h_2h_3;rs)\Bigr|$
and we can bound
$latex T\ll H^{1/2+\epsilon} U^{1/2},$
where
$latex U=\frac{1}{R^2}\sum_{r_1,r_2} \lambda(r_1,a_1)\lambda(r_2,a_2) \sum_{1\leq |h|\ll H} \overline{K_3(a_1\bar{s}^3\bar{m}h;r_1)}K_3(a_2\bar{s}^3\bar{m}h;r_2)\overline{K_3(a_s\bar{r}_1^3\bar{m}h;s)}K_3(a_s\bar{r}_2^3\bar{m}h;s)$
for some coefficients $latex \lambda(\cdot,\cdot)$ which are bounded.
The point of this is that we have smoothed the variable $latex h=h_1h_2h_3$ by eliminating its multiplicity, and that the range of this variable can be quite long; as long as $latex H>(R^2S)^{1/2}$, completing the sum in the Polya-Vinogradov (or Poisson) style will be useful.
Now we continue with $latex U$. It is here that it simplifies matters to have $latex R<S$ and to do as if $latex r$ and $latex s$ were primes (this is a technicality which experience shows should give no loss in the final, complete, analysis.)
So we consider the sum over $latex r_1, r_2$. The diagonal contribution where $latex r_1=r_2$ is $latex \ll H^{1+\epsilon}R^{-1+\epsilon}$.
In the non-diagonal terms, we distinguish whether $latex r_1\equiv r_2\bmod s$ or not. If not, we complete the two hyper-Kloosterman sums. This gives two complete exponential sums modulo $latex r_1, r_2$ and modulo $latex s$. The latter is the Friedlander-Iwaniec sum, in its incarnation as "Borel" correlations of hyper-Kloosterman sums (see the remark at the end of our note on these sums; this identification is already in Heath-Brown's paper, and Philippe realized recently that he had also encountered them in a paper on lower bounds for exponential sums.)
Both sums give square root cancellation (using Deligne’s work, of course) except for the $latex s$ sum if $latex r_1^3\equiv \pm r_2^3\bmod s$. But we may just push these to the second case. Thus, the contribution $latex U_0$ of these non-exceptional terms gives
$latex U_0\ll (R^2S)^{1/2+\epsilon}\Bigl(\frac{H}{R^2S}+1\Bigr)$
(counting the number of complete sums in the $latex h$-interval).
On the other hand, the exceptional $latex (r_1,r_2)$ are still controlled by the diagonal terms (because of the condition $latex R<S$; there is a minor trick involved here if the cubic roots of unity exist modulo $latex s$, but I'll gloss over that.)
Now we can gather everything, and one checks that we end up with a bound
$latex \sum_{r,s}\sum_a S_2 \ll X^{\epsilon} QM\Bigl(\frac{1}{R^{1/2}}+\frac{R^{1/4}N^{1/2}}{Q^{5/4}}\Bigr).$
We need this to be $latex \ll MNQ^{-1} (\log MN)^{-A}$ for any $latex A$, and we see that we succeed as long as
$latex Q^2R^{-1/2}<N^{1-\epsilon},\quad\quad Q^{3/2}R^{1/2}<N^{1-\epsilon},$
as stated in the theorem (the second condition is implied by the first if $latex R<S$). And as mentioned just afterwards, if we have $latex R=Q^{1/2-\epsilon}$, this gives a good distribution up to $latex X^{-\epsilon}N^{4/7}$ (note that epsilons may change from one inequality to the next.)
Remark. As the reader can see, we do not use either Weyl shifts, or cancellation in Ramanujan sums. The latter might appear in a more precise analysis, however, and give some extra gain.
