**What new approach did Yitang Zhang try & what did the experts miss in the first place?**

Yes it is a good question as to why (say) FI did not hit upon such a result, as the two major glue components, dispersion a la BFI, and beating the square-root barrier akin as Friedlander/Iwaniec, are due to them. As Zhang puts it, last paragraph section 2 page 7, he saves a $\sqrt r$ factor in a Kloostermann type bound, and can take $r$ as a small power (well, I think he should take it larger than he does, after review).

Most everything else is the paper is rather "standard", as the idea of restricting to integers/moduli which have no small/large factors is common, the dispersion method is in BFI, maybe in S10 Zhang has to work a little to get his conditions on congruence classes to roll through. But then the Type III estimate (trilinear) in 13 and 14 is the heart, of why he wins. Indeed, he handles each $d$ separately, rather than on average (as dispersion or large sieve). Conceptually again he brings forth a Weyl shift to copy a sum many times, then turning to Fourier techniques. The extra flexibility of factoring $d=qr$ is not distressed, for he uses Chinese Remainder Theorem in repair, and in fact the Ramunajan sums pop out too. See bottom page 52, at 14.13 and below to top line page 53, where the Birch-Bombieri bound is applied. See in $J(m_1,m_2)$, the part from $r$ is a Ramanujan sum, so the double $r$-sum is bounded in essence by $r$, not $r^{3/2}$. Again the preceeding Fourier analysis has technique, but the idea is already bookish.

So I repeat, the main advance was not to apply Deligne to something like $Z(k;m_1,m_2)$ below 14.12 on page 52 directly on modulus $d$, but to first peel over factor, small perhaps but useful, as $d=qr$ leaving $q$ to geometry and $r$ to Ramanujan after spinning out the $N_3'$ sum over this modulus. Well, that is my word, I don't say I understand all, not even philosophically why this line should give the win.

The Friedlander/Iwaniec paper: http://www.jstor.org/stable/1971175

PS. It could say useful to have a rewrite of 13 and 14, independent of the rest of the work, for he chooses parameters there for his purpose, but this vitiates the generality. They are independent for the whole.

Or again, look at $P_2$ near bottom page 48. This should be of size $d_1^{3/2}K^2$ if you apply direct reasoning, but Zhang first rolls out $n$ modulo $r$. It is still mysterious for me why this avails, factoring $V$ into $W\cdot C$ in 14.7 the latter a Ramanujan.

Let me try again, and briefly sketch the whole idea of Sections 13 and 14.

Zhang wants to estimate $|\Delta(\gamma,d,c)|$ where $\gamma=\alpha\star\chi_{N_3}\star\chi_{N_2}\star\chi_{N_1}$ is a triple convolution with $N_1\ge N_2\ge N_3$ of decent size, say $N_3\ge x^{1/4-6\omega}$. The characteristic functions $\chi_N$ are for an interval say $N$ to $N+N/(\log N)^B$. Here we have the standard discrepancy $\Delta$ defined as
$$\Delta(\gamma,d,c)=\sum_{n\sim x\atop n\equiv c (d)}\gamma(n)
-{1\over\phi(d)}\sum_{n\sim x\atop (n,d)=1}\gamma(n).$$
Most importantly, $d=qr$ with $r$ convenient. The estimate $|\Delta(\gamma,d,c)|\ll x^{1-\kappa}/d$ is desired for some positive $\kappa$.

Zhang replaces $\chi_{N_1}$ by a smooth approximant. This is standard, there are various versions of this in the field, the idea being that if a function goes from 0 to 1 over an interval of length $Y$, you can control its derivatives as powers of $Y$. Then one executes the inner sum in $\Delta$ over this variable, and replaces the sum over the smoothed function by its Fourier transform. This allows the main terms in $\Delta$ to cancel from the frequency 0 contribution, leaving a deal with highers. See middle page 45 and following. Copying,
$$\sum_{n\equiv c (d)}\gamma^\star(n)=\sum_{(m,d)=1}\alpha(m)
\sum_{n_3\sim N_3\atop (n_3,d)=1}\sum_{n_2\sim N_2\atop (n_2,d)=1}\sum_{mn_3n_2n_1\equiv c (d)}f(n_1),$$
and the inner sum is
$${1\over d}\sum_{|h|\le H} \hat f(h/d)e_d(-ch\overline{mn_3n_2})+O()$$
where $H=d/N_1$ essentially. The sum over $m$ handles itself, the inner part will yield the cancel.

So Zhang's goal is to obtain the estimate
$$\sum_{1\le h\le H}\sum_{n_3\sim N_3\atop (n_3,d)=1}\sum_{n_2\sim N_2\atop(n_2,d)=1}\hat f(h/d)e_d(-ch\overline{mn_2n_3})\ll x^{1-\kappa}/M.$$
In fact to use a Möbius inversion device that I omit below, we need this for $d$ not just near $\sqrt x$ (beyond the Bombieri-Vinogradov range), but also divisors of $d$ so a wider range. When $d$ is small enough, or $N_2$ large enough to say another way, a one-variable estimate suffices. Actually the trickiest case is when $N_1\sim N_2\sim N_3\sim x^{5/16-9\omega/2}$ (maybe this is not the exact $\omega$ multiplier, but 5/16 is right) and $d\sim x^{5/12-9\omega}$, neither a 1-variable method or the ensuing, is going to swell. But when $d\le N_1$ the estimate for the $H$-sum is empty, and when $d^{3/2}N_3\ll x^{1-\kappa}/M$ a 1-variable bound on $n_2$ suffices. I digress. In fact, as noted per (2.4) in Friedlander/Iwaniec, a 2-variable bound can be applied from Deligne some times
(the inner double sum then bounded by $\sqrt{d^2}$). Zhang does not do this extra, I find it gives a suitable bound when $d^2\ll x^{1-\kappa}/M$.

Back to the main story. After a application of Möbius to insert a coprimality condition in the frequency variable $h$, Zhang then uses the idea of the Weyl shift. It is key that he shifts by multiples of $r$, this convenient factor of $d$. The idea of the Weyl shift is to copy a sum many times, only partially shifted by much less than its length. In the above, Zhang replaces $n_2$ by $n_2+hkr$ for $k$ up to some bound $K$, and then computes that the difference between the sum and the $hkr$-shifted sum is small (provided $K$ small enough of course). Then one wants to bound the average over the shifts
$$N(d,k)=\sum_{1\le h\le H\atop (h,d)=1}\sum_{n_3\sim N_3\atop (n_3,d)=1}\sum_{n_2\sim N_2\atop(n_2+hkr,d)=1}\hat f(h/d)e_d(-ch\overline{m(n_2+hkr)n_3}).$$
I simplified the formula on top of page 47 a bit, not including the Möbius step, the above is not truly correct, but an idea of how it goes self-contained here.
To state again, the idea is that we want to bound $N(d,0)$, we know that $N(d,k)$ is close to $N(d,0)$ for small $k$, and will establish a bound on average for ${1\over K}\sum_{k\sim K} N(d,k)$.

From Cauchy, and substituting $l\equiv \bar hn_2$ modulo $d$, one is left to estimate ($P_2$ of bottom page 48)
$$\sum_{l (d)}|\sum_{k\sim K\atop (l+kr,d)=1}\sum_{n\sim N_3\atop (n,d)=1}
e_d(b\overline{(l+kr)n})|^2.$$
Staring at this, if you expand, the $l$-sum is over $d$ and the $n$-sums are essentially incomplete sums modulo the same, so expect $d^{3/2}K^2$. But the shift by a *multiple* of $r$ will allow us to win. The idea is that $N_3$ exceeds $r$ by enough to allow sprawling $n=rn'+s$ over residue classes modulo $r$, and this to be efficient. Now normally, this should not gain, see bottom page 49 with 14.6, Zhang wants to estimate
$$\sum_{k_1\sim K}\sum_{k_2\sim K}\sum_{s_1\le r\atop (s_1,r)=1}\sum_{s_2\le r\atop (s_2,r)=1}\sum_{n_1\sim N_3/r}\sum_{n_2\sim N_3/r}\sum_{l (d)} e_d(b\overline{l(n_1r+s_1)}-b\overline{(l+kr)(n_2r+s_2)}).$$
Again one doesn't expect to win, as though the $e_d$ will factor over $d=qr$ to $e_q()e_r()$, there will still be three sums over a variable modulo $r$, and $r^{3/2}$ shall appear. But the idea is that the fact the Weyl shift was a *multiple* of $r$, so upon unwinding the CRT, the triple sum modulo $r$ is really a double sum of a Ramanujan sum. So that's why I typed out the innards of $e_d$ above.

Construing the technicalities with Fourier transforms, this results as needed. The key is that $r$ can be taken as a small power of $x$, and we win by $\sqrt r$, or really the fourth-root after Cauchy, but this is enough.

not"more than perfectly fitted here" or "perfectly legitimate" (@user32240). It is likely acceptable as an exception. But on the general principleGehard Pasemen is perfectly right.$\endgroup$ – user9072 May 20 '13 at 10:54