2.1.1 Root numbers

Again in the context of class number one we have $\unicode[STIX]{x1D712}(p)=-1$ for all $p|N$ , so that $\unicode[STIX]{x1D712}(N)=(-1)^{\unicode[STIX]{x1D6FA}(N)}$ and since $\unicode[STIX]{x1D712}(-1)=-1$ for odd characters (applicable for imaginary quadratic fields), we get $\unicode[STIX]{x1D716}_{2}=\unicode[STIX]{x1D712}(-N)\unicode[STIX]{x1D716}_{1}=\unicode[STIX]{x1D716}_{1}(-1)^{\unicode[STIX]{x1D6FA}(N)+1}$ . We write

(2) $$\begin{eqnarray}0=I_{f}(u)+\unicode[STIX]{x1D705}(u)I_{f}(u),\quad \text{where }\unicode[STIX]{x1D705}(u)=\unicode[STIX]{x1D716}_{1}^{2}(-1)^{u+\unicode[STIX]{x1D6FA}(N)}\end{eqnarray}$$

and in our orthogonal case we have $\unicode[STIX]{x1D716}_{1}=\pm 1$ , so we should take $\unicode[STIX]{x1D6FA}(N)$ to be of the same parity as $u$ , as else the above (2) is uninteresting. In particular, for $u=2$ we cannot use $N=389$ (or $N=433$ ).

2.2.1

The first integral (call it $J_{f}(u)$ ) can be computed by moving the path of integration to the left, with the decay from the $q^{s}$ -term dominating. The main term comes from the residue at $s=1$ , while the resulting error term is easily estimated via a standard zero-free region of $\unicode[STIX]{x1D701}$ and the vertical decay of $\unicode[STIX]{x1D6E4}(s)^{2}$ . Writing $B_{f}(2s)$ for a (possible) correction from bad Euler factors, upon combining with $f$ , we have

$$\begin{eqnarray}J_{f}(u)\approx 2L_{S^{2}f}(2)B_{f}(2)(\log q)^{u-2}.\end{eqnarray}$$

Although it will prove fruitless in the end, we can note that there is a weight 2 modular eigenform $f$ of level 446 with analytic rank two corresponding to the elliptic curve 446d, and for $f$ this main term is $J_{f}(2)\approx 2L_{S^{2}f}(2)(2\cdot 223/1\cdot 222)\approx 20.2$ .

2.3.1

Although it turns out to be fruitless, with $h=1$ one can propose an attempt to go further in this vein, upon noting Oesterlé’s method is not too sharp in the error term when it bounds the contribution of $\unicode[STIX]{x1D6F9}_{f}(s)\tilde{G}_{\!f}(s)$ for the principal form (wherein the eccentricity of the ellipse plays a substantial role with respect to lattice point counting, as the asymptotic count (as $T\rightarrow \infty$ ) of $\unicode[STIX]{x1D70B}T/\sqrt{q}$ is not too accurate for $T$ of size $q$ ). Writing $\unicode[STIX]{x1D6F9}_{f}(s)=\sum _{k}\unicode[STIX]{x1D709}_{k}/k^{2s}$ an alternative method is to expand in terms of an inverse Mellin transform as

$$\begin{eqnarray}\displaystyle E_{f}(u) & = & \displaystyle \int _{(2)}\unicode[STIX]{x1D6F9}_{f}(s)\tilde{G}_{\!f}(s)\unicode[STIX]{x1D6E4}(s)^{2}\biggl(\frac{Nq}{4\unicode[STIX]{x1D70B}^{2}}\biggr)^{s-1}\frac{\unicode[STIX]{x2202}s/2\unicode[STIX]{x1D70B}\text{i}}{(s-1)^{u}}\nonumber\\ \displaystyle & = & \displaystyle \frac{4\unicode[STIX]{x1D70B}^{2}}{qN}\mathop{\sum }_{k=1}^{\infty }\unicode[STIX]{x1D709}_{k}\mathop{\sum }_{l\geqslant q/4}g_{l}W_{u}\biggl(\frac{4\unicode[STIX]{x1D70B}^{2}k^{2}l}{Nq}\biggr),\nonumber\end{eqnarray}$$

where $W_{u}(z)$ is given by

$$\begin{eqnarray}W_{u}(z)=\int _{(2)}z^{-s}\unicode[STIX]{x1D6E4}(s)^{2}\frac{\unicode[STIX]{x2202}s/2\unicode[STIX]{x1D70B}\text{i}}{(s-1)^{u}}.\end{eqnarray}$$

Specializing to $u=2$ , in terms of $K$ -Bessel functions we have $W_{2}(z)=2K_{0}(2\sqrt{z})/z$ , so in particular $W_{2}(z)=O(\text{e}^{-\sqrt{z}})$ as $z\rightarrow \infty$ . The rapid decay of $W_{u}$ combined with the assumption of class number one together imply the $l$ that have more than one prime factor $p$ with $\unicode[STIX]{x1D712}(p)\neq -1$ give a negligible contribution, as all such $l$ will exceed $(q/4)^{2}$ . Thus with negligible error we can replace the $l$ -sum by a sum over $p$ with $\unicode[STIX]{x1D712}(p)\neq -1$ . Moreover, we have $g_{p}=[1+\unicode[STIX]{x1D712}(p)](\unicode[STIX]{x1D6FC}_{p}+\unicode[STIX]{x1D6FD}_{p})$ , which for $p$ with $\unicode[STIX]{x1D712}(p)=+1$ is twice the $p$ th Dirichlet series coefficient of  $f$ .

With $f$ of level 446 and analytic rank two as above, by using the regularity of the $p$ with $\unicode[STIX]{x1D712}(p)\neq -1$ , namely coming from $p=x^{2}+xy+((q+1)/4)y^{2}$ with $\gcd (x,y)=1$ , one can obtain a bound for $E_{f}(2)$ of about 32 and my recollection is by additionally exploiting the vanishing of $J_{f}(u)$ for $u=0$ that I was able to achieve approximately 26, still less than the goal of 20.2. One could attempt to use other modular forms, such as 159k4a, or a non-orthogonal form (with unitary root number) with single central vanishing; my investigation concluded that the above choice of 446k2d is the best, yet still insufficientFootnote ⁵ .

2.4.1

As above, we have $g_{p}=[1+\unicode[STIX]{x1D712}(p)]c_{p}$ , while the contribution from non-prime $l$ to the error $E_{f}(2)$ is negligible by the decay of $W_{2}(z)=2K_{0}(2\sqrt{z})/z$ . Thus we want to estimate

$$\begin{eqnarray}E_{1}=\mathop{\sum }_{k=1}^{\infty }\unicode[STIX]{x1D709}_{k}\mathop{\sum }_{p}c_{p}[1+\unicode[STIX]{x1D712}(p)]W_{2}\biggl(\frac{4\unicode[STIX]{x1D70B}^{2}k^{2}p}{Nq}\biggr),\end{eqnarray}$$

and show this is smaller than $q$ in size. Since the coefficients $\unicode[STIX]{x1D709}_{k}$ play no arithmetic role in the end (only their size matters), we can first multiply and divide by a Dirichlet series supported on squares that replaces the above sum over $p$ by a sum over $pm^{2}$ , with $c_{p}$ replaced by $c_{pm^{2}}$ . Indeed, an accounting shows that we have

$$\begin{eqnarray}E_{1}\approx \mathop{\sum }_{k=1}^{\infty }\tilde{\unicode[STIX]{x1D709}}_{k}\mathop{\sum }_{l=1}^{\infty }c_{l}\tilde{R}_{\unicode[STIX]{x1D712}}(l)W_{2}\biggl(\frac{4\unicode[STIX]{x1D70B}^{2}k^{2}l}{Nq}\biggr),\end{eqnarray}$$

where the $\tilde{\unicode[STIX]{x1D709}}_{k}$ are adequately bounded, and $h=1$ implies that $\tilde{R}_{\unicode[STIX]{x1D712}}(l)$ is the number of representations of $l$ by the principal form $x^{2}+xy+((q+1)/4)y^{2}$ with $y\geqslant 1$ .

Expanding $c_{l}$ from the above, we thus want to bound $\sum _{k}|\tilde{\unicode[STIX]{x1D709}}_{k}||E_{1}^{k}|$ where

$$\begin{eqnarray}E_{1}^{k}=\mathop{\sum }\nolimits_{l=(4a+1)^{2}+(2b)^{2}}\tilde{R}_{\unicode[STIX]{x1D712}}(l)(4a+1)(-1)^{b}\unicode[STIX]{x1D703}(l)W_{2}\biggl(\frac{4\unicode[STIX]{x1D70B}^{2}k^{2}l}{Nq}\biggr).\end{eqnarray}$$

We can note the $W_{2}$ -decay allows us to curtail $k$ at some height, say $K=(\log q)^{1/10^{9}}$ .

Also note that it is not merely “generic” cancellation in $\sum _{l}c_{l}\tilde{R}_{\unicode[STIX]{x1D712}}(l)$ that we require, as this would already follow when weighting by $W_{0}$ (or more generally $W_{-2k}$ for positive integers $k$ ), but this does not suffice for the desired $W_{2}$ -weighting.

2.4.2

We then consider $l=x^{2}+xy+((q+1)/4)y^{2}$ with $y\geqslant 1$ , which by rearrangement says $4l=(2x+y)^{2}+qy^{2}$ . By the accounting of $\tilde{R}_{\unicode[STIX]{x1D712}}(l)$ we have

$$\begin{eqnarray}E_{1}^{k}=\mathop{\sum }_{y=1}^{\infty }\sum _{4l=(2x+y)^{2}+qy^{2}}(4a+1)(-1)^{b}\unicode[STIX]{x1D703}(l)W_{2}\biggl(\frac{4\unicode[STIX]{x1D70B}^{2}k^{2}l}{Nq}\biggr).\end{eqnarray}$$

The above conditions on $l$ imply that

$$\begin{eqnarray}4(4a+1)^{2}-(2x+y)^{2}=-4(2b)^{2}+qy^{2}.\end{eqnarray}$$

We can factor the left side with

$$\begin{eqnarray}t=2(4a+1)+(2x+y),\qquad u=2(4a+1)-(2x+y),\end{eqnarray}$$

so that $tu=-4(2b)^{2}+qy^{2}$ while

$$\begin{eqnarray}2(4a+1)=(t+u)/2,\qquad 2x+y=(t-u)/2\end{eqnarray}$$

and

$$\begin{eqnarray}4l=(t-u)^{2}/4+qy^{2},\qquad l=(t+u)^{2}/16+(2b)^{2}.\end{eqnarray}$$

2.4.3

We then switch variables from $(a,x)$ to $(t,u)$ . An analysis modulo 16 gives conditions ensuring the integrality of $(a,x)$ , and when combined with a similar analysis modulo 17 suffices to fix the effect of the character  $\unicode[STIX]{x1D703}$ .

We are then left to consider sums (for $k\leqslant K$ ) such as

$$\begin{eqnarray}\mathop{\sum }_{y\leqslant K}\mathop{\sum }_{r_{b}=1}^{34}\mathop{\sum }_{r_{t}=1}^{272}z(y,r_{b},r_{t})\sum _{-4(68\tilde{b}+2r_{b})^{2}+qy^{2}\equiv 0\,(t)}\frac{t+u}{4}W_{2}\biggl(\frac{4\unicode[STIX]{x1D70B}^{2}k^{2}l}{Nq}\biggr),\end{eqnarray}$$

where $|z(y,r_{b},r_{t})|\leqslant 4$ , while $u$ and $l$ are derived from $t$ as above. Here we wish to show that the inner double sum over $t$ and $\tilde{b}$ has some cancellation. Let us note that its crude bounding would be $\ll q$ , coming from noting that $|t|$ and $|u|$ can be curtailed by the $W$ -decay at size around $\sqrt{q}$ , while the expectation is that there are a constant number of $\tilde{b}$ -roots of the congruence on average. Thus (roughly) the double sum over $(t,\tilde{b})$ has $\ll \sqrt{q}$ members, each of size $\ll |t+u|\ll \sqrt{q}$ .

2.4.4

In order to rigidify the contributions from $(t+u)/4$ and the $W$ -term, we split $t$ and $\tilde{b}$ into intervals of size $Z=\sqrt{q}/S$ , where $S=(\log q)^{1/100}$ , in practice using a smooth partition of unity on $\tilde{b}$ . The small $t$ , say those with $|t|\ll \sqrt{q}/S^{1/5}$ , contribute a negligible amount (by a crude estimate), and we are essentially left to consider for various $(T,B)$ the expressions

$$\begin{eqnarray}\frac{G(T,B)}{4}\,\times \sum _{-4(68\tilde{b}+2r_{b})^{2}+qy^{2}\equiv 0\,(t)}1,\end{eqnarray}$$

where

$$\begin{eqnarray}\displaystyle G(T,B) & = & \displaystyle \biggl(T+\frac{qy^{2}-136^{2}B^{2}}{T}\biggr)\nonumber\\ \displaystyle & & \displaystyle \times \,W_{2}\biggl(\frac{4\unicode[STIX]{x1D70B}^{2}k^{2}}{Nq}\biggl[\frac{1}{4}\biggl(T-\frac{qy^{2}-136^{2}B^{2}}{T}\biggr)^{2}+qy^{2}\biggr]\biggr)\nonumber\end{eqnarray}$$

is an odd function of  $T$ .

Our adaptation of Hooley’s methods then gives an equi-distribution of the roots of the congruence $-4(68\tilde{b}+2r_{b})^{2}+qy^{2}\equiv 0\,(t)$ as $t$ varies, namely that the above double sum over $(t,\tilde{b})$ is sufficiently well-approximated as proportional to $Z^{2}/|T|$ , and this shows the necessary cancellation, e.g. via pairing $T$ with $-T$ .

In particular, returning to

$$\begin{eqnarray}E_{1}\approx \mathop{\sum }_{k=1}^{\infty }\tilde{\unicode[STIX]{x1D709}}_{k}\mathop{\sum }_{l=1}^{\infty }c_{l}\tilde{R}_{\unicode[STIX]{x1D712}}(l)W_{2}\biggl(\frac{4\unicode[STIX]{x1D70B}^{2}k^{2}l}{Nq}\biggr),\end{eqnarray}$$

it appears that [Reference Templier and Tsimerman72, (1.5)] almost shows the type of cancellation that we desire. Replicating their result directly, for $X\gg \sqrt{d}$ they show

$$\begin{eqnarray}\mathop{\sum }_{n}a_{\unicode[STIX]{x1D70B}}(n^{2}+d)V(n/X)\ll _{\unicode[STIX]{x1D70B},\unicode[STIX]{x1D716}}X^{1/2+\unicode[STIX]{x1D716}}\,d^{\unicode[STIX]{x1D6FF}}\biggl(1+\frac{X^{2}}{d}\biggr)^{\unicode[STIX]{x1D703}/2},\end{eqnarray}$$

where $a_{\unicode[STIX]{x1D70B}}$ are unitary normalized coefficients of an automorphic cuspidal representation $\unicode[STIX]{x1D70B}$ and $V$ is a smooth function on $[1,2]$ , while $\unicode[STIX]{x1D6FF}=3/16$ comes from bounds on half-integral weight coefficients (alternatively interpreted as a $q$ -subconvexity parameter) and $\unicode[STIX]{x1D703}=7/64$ involves Selberg’s eigenvalue conjecture. Transposing to our genre with $X\approx \sqrt{d}$ , this analogously indicates $|E_{1}|\ll \sqrt{q}\cdot q^{1/4+\unicode[STIX]{x1D716}}q^{\unicode[STIX]{x1D6FF}}\ll q^{15/16+\unicode[STIX]{x1D716}}$ and moreover there is no restriction concerning complex multiplication.

Taking $d$ to be our $qy^{2}$ , it is only the fact that $n^{2}+qy^{2}$ does not quite give us our desired representations (involving an extra factor of 4 when e.g. $n$ and $y$ are odd) that prevents us from using this result in the form given. In [Reference Templier and Tsimerman72, Remark (ii)] they mention that their methods should be applicable to other quadratic polynomials, and indeed $n^{2}+n+d$ can be handled via using the $\unicode[STIX]{x1D6E9}$ -function of the odd squares, namely $\unicode[STIX]{x1D6E9}(z)-\unicode[STIX]{x1D6E9}(4z)$ . We give the details in [Reference Watkins76], though for technical reasons we use $\unicode[STIX]{x1D6FF}=3/14$ coming from the work of Iwaniec [Reference Iwaniec39] and Duke [Reference Duke21]Footnote ⁶ .

Note that their result gives a power-saving (since $\unicode[STIX]{x1D6FF}<1/4$ ) in the region of interest; however, for larger class numbers the difficulty typically is with the minima $a$ near $\sqrt{q/3}$ , where uniformity considerations will arise since the spectral analysis would then be applied on $\unicode[STIX]{x1D6E4}_{0}(aN)$ . As in our proof here, I do not see any way to generalize beyond class number one (where control of $a$ is obvious).

2.5.1

We briefly describe Hooley’s method. He writes each modulus as $m=m_{1}m_{2}$ , where $m_{1}$ consists of primes less than $x^{1/200\log \log x}$ with $m_{2}$ the complement, and notes that an elementary upper-bound sieve shows that the contribution from $m$ with $m_{1}\geqslant x^{1/3}$ (say) is negligible. For the remaining $m$ , he applies the factorization

$$\begin{eqnarray}S_{P}(h,m)=S_{P}(h\bar{m}_{2},m_{1})S_{P}(h\bar{m}_{1},m_{2}),\end{eqnarray}$$

so that (with $m_{1},m_{2}$ having the above meaning)

$$\begin{eqnarray}\displaystyle |R_{P,h}(x)| & = & \displaystyle \biggl|\sum _{m_{1}\leqslant x^{1/3}}S_{P}(h\bar{m}_{2},m_{1})S_{P}(h\bar{m}_{1},m_{2})\biggr|\nonumber\\ \displaystyle & {\leqslant} & \displaystyle \sum _{m_{1}\leqslant x^{1/3}}\unicode[STIX]{x1D70C}_{P}(m_{2})|S_{P}(h\bar{m}_{2},m_{1})|,\nonumber\end{eqnarray}$$

where $\unicode[STIX]{x1D70C}_{P}(k)$ is number of roots of $P$ modulo $k$ , and the $m_{2}$ -sum is (much) longer than the modulus $m_{1}$ . He then applies Cauchy’s inequality and the basic estimate

$$\begin{eqnarray}\mathop{\sum }_{a=1}^{k}|S_{P}(ah,k)|^{2}\leqslant k\unicode[STIX]{x1D70C}_{P}(k)\cdot \gcd (h,k),\end{eqnarray}$$

which gives a (slight) saving on average compared to the trivial bound $k\unicode[STIX]{x1D70C}_{P}(k)^{2}$ . Combined with a result on the number of primes that split, this then gives the stated savings of a small power of $\log x$ .

It could be noted that the smallness of the upper bound does not come from equi-distribution to varying moduli as such, but rather: to each fixed modulus the $S_{P}$ -sum over the roots evinces some cancellation on averageFootnote ⁷ , and this is then used with a factorization of $S_{P}(h,m)$ with respect to $m=m_{1}m_{2}$ to conclude the result.

2.5.2

To handle the $q$ -dependence (in $P$ ) we need to show that there are sufficiently many small primes that do not split in the real quadratic field $\mathbf{Q}(\sqrt{q})$ . Here we can note that class number one (together with quadratic reciprocity) implies that up to $q/4$ the splitting primes are exactly the primes that are 3 mod 4.

Similarly, in the application back to our above double $(t,\tilde{b})$ sum, we proceed by a Fourier expansion in $h$ , with the main term from $h=0$ . This main term then is simply the number of roots of the congruence when the modulus varies (over an arithmetic progression mod 272, though this aspect is essentially harmless). The effective range here is for moduli of roughly size $\sqrt{q}$ , and to achieve an asymptotic formula we need a subconvexity bound for Dirichlet $L$ -functions in the $q$ -aspect. Although we could just quote such a result from the literature ([Reference Burgess10] for instance), again in the context of class number one we argue alternatively, using that $L_{\unicode[STIX]{x1D702}}(s)L_{\unicode[STIX]{x1D702}\unicode[STIX]{x1D712}}(s)$ is well-approximated as an Euler product by $L_{\unicode[STIX]{x1D702}^{2}}(2s)$ (up to bad factors).

What is likely meant here is that Deuring’s theorem gives a regular spacing of zeros of the Dedekind $\unicode[STIX]{x1D701}$ -function of $\mathbf{Q}(\sqrt{-q})$ , which can be compared to computations of zeros of the Riemann $\unicode[STIX]{x1D701}$ -factor. Indeed, Stark in his thesisFootnote ⁸ used the computation of the first two zeros of the Riemann $\unicode[STIX]{x1D701}$ -function to 20 digits to show [Reference Stark61] that a putative 10th fundamental discriminant with class number one had to satisfy the bound $q>\exp (2.2\cdot 10^{7})$ . However, in his subsequent actual proof [Reference Stark62, §4] of class number one he only needs to assume $q>200$ , which leaves a mere hand calculation (already done by Gauss!) to give the entire list. The other proofs in the genre of “modular functions” similarly do not rely on Deuring’s theorem to handle small  $q$ .

A bound of $q\leqslant e^{160\,000}$ from an explicit Baker’s method was given by Bundschuh and Hock [Reference Bundschuh and Hock9], though in a comparable restricted problem (for class number two) Baker [Reference Baker3] obtains a rather smaller bound of $q\leqslant 10^{500}$ ; anyway, Deuring’s theorem and computations of zeros would indeed be needed to finish the proof here.

Finally, Goldfeld’s method as finished by Gross and Zagier (and computed by Oesterlé) implies directly that $h>1$ for $q>e^{55}$ , though my recollection (see the comparison with §2.3, where $L_{S^{2}f}(2)\approx 7.546$ for $f=5077a$ while $E_{f}(3)\approx 736.43$ ) is that with some effort one can reduce 55 to 50 (or 49) if desired, and a more exacting analysis for the principal form as in §2.3.1 would certainly allow the class number one proof to be finished by direct sieving. Here even $e^{55}\approx 10^{24}\approx 2^{80}$ can likely be handled by a computational sieve if necessaryFootnote ⁹ , though this is of course heavily exploiting technological advances of the past few decades.

As noted in §1.4.2, the bound given by our proof is likely quite large, yet still should succumb to the method of Stark’s thesis with (routine) computation of the first two Riemann $\unicode[STIX]{x1D701}$ -zeros to 200 or perhaps 20 000 digits.

3.1.1

The theory of genera [Reference Gauss24, §257] corresponds to divisors of $q$ . It says that the class number is divisible by $2^{\unicode[STIX]{x1D714}(q)-1}$ , and perhaps more directly in the context of class number one, one can simply write down a non-principal reduced form corresponding to a non-trivial factorization of $q$ .

Writing $(a,b,c)$ for a form, when $q=tu$ is odd with $1<t<u$ , we find that either $(t,t,(t+u)/4)$ or $((t+u)/4,(u-t)/2,(t+u)/4)$ is reduced and non-principal.

For $4\Vert q$ and $q>4$ we always have the reduced non-principal form $(2,2,(q+4)/8)$ , and factorizations $q/4=tu$ with $1<t<u$ yield $(t,0,u)$ and either $(2t,2t,(t+u)/2)$ or $((t+u)/2,(u-t),(t+u)/2)$ .

For $8\Vert q$ and $q>8$ we always have the reduced non-principal form $(2,0,q/8)$ , and factorizations $q/8=tu$ by $1<t<u$ yield $(t,0,2u)$ and either $(2t,0,u)$ or $(u,0,2t)$ .

Thus we find that if the class number is 1 and $q>8$ , then $q$ is prime. Similarly, we see 2 splits for $q$ that are 7 mod 8, namely $(2,1,(q+1)/8)$ is a non-principal reduced form for $q>7$ , so the class number one assumption implies that $q$ is 3 mod 8.

3.1.2

Writing $\unicode[STIX]{x1D712}$ for the quadratic character associated to $K=\mathbf{Q}(\sqrt{-q})$ , we then put $R_{\unicode[STIX]{x1D712}}^{\star }(l)$ for half the number of primitive representations of $l$ with $\gcd (x,y)=1$ by reduced binary quadratic forms of discriminant $-q$ , so that

$$\begin{eqnarray}R_{\unicode[STIX]{x1D712}}^{\star }(l)=\mathop{\prod }_{p|l}[1+\unicode[STIX]{x1D712}(p)]\cdot \mathop{\prod }_{p^{2}|l}\unicode[STIX]{x1D712}(p).\end{eqnarray}$$

This is 0 if $\unicode[STIX]{x1D712}(p)=-1$ for some $p|l$ . Moreover, we have $\unicode[STIX]{x1D701}_{K}(s)/\unicode[STIX]{x1D701}(2s)=\sum _{l}R_{\unicode[STIX]{x1D712}}^{\star }(l)/l^{s}$ .

The class number one assumption (for $q>8$ ) thus says $R_{\unicode[STIX]{x1D712}}^{\star }(l)=0$ for $1<l\leqslant q/4$ , as for $y\neq 0$ the values of the principal form $x^{2}+xy+((q+1)/4)y^{2}$ exceed $q/4$ . In particular, we have $\unicode[STIX]{x1D712}(p)=-1$ for all $p\leqslant q/4$ . Similarly, the $l$ with $q/4\leqslant l\leqslant (q/4)^{2}$ that are primitively represented must be prime, due to the multiplicativity of $R_{\unicode[STIX]{x1D712}}^{\star }(l)$ .

4.1.1

The Chinese remainder theorem implies $\unicode[STIX]{x1D70C}_{P}$ is multiplicative, and for prime powers $p^{e}$ with $p\nmid 34y$ we have $\unicode[STIX]{x1D70C}_{P}(p^{e})=\unicode[STIX]{x1D70C}_{P}(p)=\bar{\unicode[STIX]{x1D70C}}_{P}(p)$ .

For $p\not \in \{2,17\}$ with $p|y$ we write $v=v_{p}(y)$ and then get $\unicode[STIX]{x1D70C}_{P}(p^{e})=p^{\lceil e/2\rceil }$ for $e\leqslant 2v$ , while $\unicode[STIX]{x1D70C}_{P}(p^{e})=p^{v}\bar{\unicode[STIX]{x1D70C}}_{P}(p)$ for $e>2v$ .

One can write down similar bounds for $\unicode[STIX]{x1D70C}_{P}(p^{e})$ with $p\in \{2,17\}$ , but we can avoid needing such in the following in any event.

By multiplicativity, for $t$ coprime to 34 we thus have $\unicode[STIX]{x1D70C}_{P}(t)\leqslant \gcd (t,y)2^{\unicode[STIX]{x1D714}(t)}$ .

4.1.2

We next introduce some notation to deal with imprimitivities.

We write $E_{P}(s)$ for the distinction between $\bar{\unicode[STIX]{x1D70C}}_{P}$ and $\unicode[STIX]{x1D70C}_{P}$ , so that as Dirichlet series

$$\begin{eqnarray}\mathop{\sum }_{m}\frac{\unicode[STIX]{x1D70C}_{P}(m)}{m^{s}}=E_{P}(s)\mathop{\sum }_{m}\frac{\bar{\unicode[STIX]{x1D70C}}_{P}(m)}{m^{s}}\end{eqnarray}$$

and we have

(3) $$\begin{eqnarray}E_{P}(s)=\mathop{\prod }_{p|34y}\biggl(U_{p}(s)^{-1}\mathop{\sum }_{e=0}^{\infty }\frac{\unicode[STIX]{x1D70C}_{P}(p^{e})}{p^{es}}\biggr),\end{eqnarray}$$

where the $U_{p}(s)^{-1}$ removes the Euler factor for $\bar{\unicode[STIX]{x1D70C}}_{P}$ corresponding to $\unicode[STIX]{x1D701}_{\sqrt{q}}(s)/\unicode[STIX]{x1D701}(2s)$ , and the sum over $e$ replaces it by the enumeration for $\unicode[STIX]{x1D70C}_{P}$ . We can also take a version of this when twisted by a Dirichlet character $\unicode[STIX]{x1D702}$ , namely that

$$\begin{eqnarray}\mathop{\sum }_{m}\frac{\unicode[STIX]{x1D70C}_{P}(m)\unicode[STIX]{x1D702}(m)}{m^{s}}=E_{P}^{\unicode[STIX]{x1D702}}(s)\mathop{\sum }_{m}\frac{\bar{\unicode[STIX]{x1D70C}}_{P}(m)\unicode[STIX]{x1D702}(m)}{m^{s}},\end{eqnarray}$$

where the Euler factor of $E_{P}^{\unicode[STIX]{x1D702}}(s)$ at $p|34y$ is

$$\begin{eqnarray}U_{p}^{\unicode[STIX]{x1D702}}(s)^{-1}\mathop{\sum }_{e=0}^{\infty }\frac{\unicode[STIX]{x1D70C}_{P}(p^{e})\unicode[STIX]{x1D702}(p^{e})}{p^{es}}\quad \text{with }U_{p}^{\unicode[STIX]{x1D702}}(s)=\left\{\begin{array}{@{}ll@{}}{\displaystyle \frac{1+\unicode[STIX]{x1D702}(p)/p^{s}}{1-\unicode[STIX]{x1D702}(p)/p^{s}}}\quad & \text{for }\bar{\unicode[STIX]{x1D70C}}_{P}(p)=2,\\ 1+\unicode[STIX]{x1D702}(p)/p^{s}\quad & \text{for }\bar{\unicode[STIX]{x1D70C}}_{P}(p)=1,\\ 1\quad & \text{for }\bar{\unicode[STIX]{x1D70C}}_{P}(p)=0.\end{array}\right.\end{eqnarray}$$

We write $\tilde{E}_{P}(s)$ for $E_{P}(s)$ twisted by the principal character modulo 272, which has the effect of removing the Euler factors at 2 and 17.

We can moreover derive a simple bound for $|E_{P}^{\unicode[STIX]{x1D702}}(s)|$ for $\unicode[STIX]{x1D70E}\geqslant 4/5$ when twisting by a character $\unicode[STIX]{x1D702}$ of modulus divisible by 2 and 17 (which again removes those Euler factors), namely $\unicode[STIX]{x1D70C}_{P}(p^{e})\leqslant 2p^{\lceil e/2\rceil }$ implies the $p$ th Euler factor is bounded by $10p^{1/5}$ (the $e=1$ term dominating), and then multiplying over $p|y$ crudely gives $\ll |y|^{1/4}$ .

For a Dirichlet character $\unicode[STIX]{x1D702}$ we denote its associated primitive character by $\unicode[STIX]{x1D702}_{\star }$ , and write

$$\begin{eqnarray}V_{\unicode[STIX]{x1D702}}(s)=\frac{L_{\unicode[STIX]{x1D702}}(s)}{L_{\unicode[STIX]{x1D702}_{\star }}(s)}=\mathop{\prod }_{p}\frac{1-\unicode[STIX]{x1D702}_{\star }(p)/p^{s}}{1-\unicode[STIX]{x1D702}(p)/p^{s}}.\end{eqnarray}$$

The Euler factor is possibly non-trivial only at $p$ dividing the modulus, and we have that $|V_{\unicode[STIX]{x1D702}}(s)|,|V_{\unicode[STIX]{x1D702}}(s)^{-1}|\ll _{\unicode[STIX]{x1D702}}1$ for $\unicode[STIX]{x1D70E}\geqslant 4/5$ .

4.2.2

For $1\leqslant x\leqslant q/4$ a truncated Perron’s formula (see [Reference Perron53, §5] or [Reference Davenport14, §17]) with $\unicode[STIX]{x1D70E}_{0}=1+1/\log 3x$ yields

$$\begin{eqnarray}\displaystyle \mathop{\sum }_{m\leqslant x}\unicode[STIX]{x1D70C}_{P}(m)\unicode[STIX]{x1D702}(m) & = & \displaystyle \int _{\unicode[STIX]{x1D70E}_{0}-\text{i}T}^{\unicode[STIX]{x1D70E}_{0}+\text{i}T}\frac{x^{s}}{s}E_{P}^{\unicode[STIX]{x1D702}}(s)\frac{L_{\unicode[STIX]{x1D702}_{\star }}(s)}{L_{(\unicode[STIX]{x1D702}\unicode[STIX]{x1D702}_{-4})_{\star }}(s)}\frac{V_{\unicode[STIX]{x1D702}}(s)}{V_{\unicode[STIX]{x1D702}\unicode[STIX]{x1D702}_{-4}}(s)}\frac{\unicode[STIX]{x2202}s}{2\unicode[STIX]{x1D70B}\text{i}}\nonumber\\ \displaystyle & & \displaystyle +\,O\biggl(\frac{x^{\unicode[STIX]{x1D70E}_{0}}Y(\log 3x)^{2}}{T}\biggr).\nonumber\end{eqnarray}$$

Writing $t^{\star }=|t|+5$ we then move the line of integration to $\unicode[STIX]{x1D70E}=1-c/\log t^{\star }$ , and by standard zero-free regions and bounds for Dirichlet $L$ -functions [Reference Davenport14] get that

$$\begin{eqnarray}\displaystyle \mathop{\sum }_{m\leqslant x}\unicode[STIX]{x1D70C}_{P}(m)\unicode[STIX]{x1D702}(m) & = & \displaystyle \unicode[STIX]{x1D6FF}_{\unicode[STIX]{x1D702}}\frac{\tilde{E}_{P}(1)}{L_{\unicode[STIX]{x1D702}_{-4}}(1)}\frac{V_{\unicode[STIX]{x1D702}_{1}}(1)}{V_{\unicode[STIX]{x1D702}_{-4}}(1)}\nonumber\\ \displaystyle & & \displaystyle +\,O(x^{1-c/\log T}Y(\log T)^{2})+O\biggl(\frac{xY(\log 3x)^{2}}{T}\biggr),\nonumber\end{eqnarray}$$

where we can take $c=1/200$ (see [Reference McCurley51]), while $\unicode[STIX]{x1D6FF}_{\unicode[STIX]{x1D702}}$ is 1 when $\unicode[STIX]{x1D702}$ is principal and else 0.

Upon choosing $T=\text{e}^{\sqrt{c\log 3x}}$ , by character orthogonality we thus get

$$\begin{eqnarray}\displaystyle \sum _{m\equiv a\,(272)}\unicode[STIX]{x1D70C}_{P}(m) & = & \displaystyle \frac{1}{\unicode[STIX]{x1D719}(272)}\mathop{\sum }_{\unicode[STIX]{x1D702}\,(272)}\bar{\unicode[STIX]{x1D702}}(a)\mathop{\sum }_{1\leqslant m\leqslant x}\unicode[STIX]{x1D70C}_{P}(m)\unicode[STIX]{x1D702}(m)\nonumber\\ \displaystyle & = & \displaystyle \frac{x}{2\unicode[STIX]{x1D719}(272)}\frac{\tilde{E}_{P}(1)}{L_{\unicode[STIX]{x1D702}_{-4}}(1)}+O(xY\text{e}^{-0.1\sqrt{\log 3x}}),\nonumber\end{eqnarray}$$

since $V_{\unicode[STIX]{x1D702}_{1}}(1)=(1-1/2)(1-1/17)$ and $V_{\unicode[STIX]{x1D702}_{-4}}(1)=(1-\unicode[STIX]{x1D702}_{-4}(17)/17)=(1-1/17)$ .◻

Corollary 4.2.3. Suppose that $\mathbf{Q}(\sqrt{-q})$ has class number one and let $P$ be as above with $0<|y|\leqslant Y=(\log q)^{1/10^{9}}$ . Then for $1\leqslant x\leqslant q/4$ we have

$$\begin{eqnarray}\sum _{\gcd (m,34)=1}\unicode[STIX]{x1D70C}_{P}(m)\ll xY.\end{eqnarray}$$

5.2.2

We then consider the $t$ with $t_{1}\leqslant x^{1/3}$ , and again can readily drop the congruential $t$ -condition, getting a contribution $\unicode[STIX]{x1D6F4}_{2}$ with

(4) $$\begin{eqnarray}|\unicode[STIX]{x1D6F4}_{2}|\leqslant \sum _{t_{1}t_{2}\leqslant x,\,t_{1}\leqslant x^{1/3}}\unicode[STIX]{x1D70C}_{P}(t_{2})|S_{P}(h\bar{t}_{2},t_{1})|=\sum _{t_{1}\leqslant x^{1/3}}T(x/t_{1},t_{1}),\end{eqnarray}$$

where (for $z\geqslant x^{2/3}$ ) we have written

$$\begin{eqnarray}T(z,t_{1})=\sum _{t_{2}\leqslant z}\unicode[STIX]{x1D70C}_{P}(t_{2})|S_{P}(h\bar{t}_{2},t_{1})|.\end{eqnarray}$$

By Cauchy’s inequality we see that

(5) $$\begin{eqnarray}T(z,t_{1})^{2}\leqslant \biggl(\,\sum _{t_{2}\leqslant z}\unicode[STIX]{x1D70C}_{P}(t_{2})^{2}\biggr)\biggl(\,\sum _{t_{2}\leqslant z}|S_{P}(h\bar{t}_{2},t_{1})|^{2}\biggr).\end{eqnarray}$$

As noted by Hooley, the first sum can be bounded by the sparsity of $t_{2}$ via an upper-bound sieve as

$$\begin{eqnarray}\sum _{t_{2}\leqslant z}\unicode[STIX]{x1D70C}_{P}(t_{2})^{2}\leqslant \sum _{t_{2}\leqslant z}4^{\unicode[STIX]{x1D714}(t_{2})}\ll \frac{z}{\log X}\cdot \biggl(\frac{\log z}{\log X}\biggr)^{4-1}\ll \frac{z(\log \log x)^{4}}{\log x}.\end{eqnarray}$$

The second sum similarly goes by an upper-bound sieve, as we have

$$\begin{eqnarray}\displaystyle \sum _{t_{2}\leqslant z}|S_{P}(h\bar{t}_{2},t_{1})|^{2} & = & \displaystyle \sum _{(a,t_{1})=1}^{t_{1}}|S_{P}(ah,t_{1})|^{2}\sum _{t_{2}\equiv a\,(t_{1})}1\nonumber\\ \displaystyle & \ll & \displaystyle \frac{z/\unicode[STIX]{x1D719}(t_{1})}{\log X}\sum _{(a,t_{1})=1}^{t_{1}}|S_{P}(ah,t_{1})|^{2}.\nonumber\end{eqnarray}$$

The remaining sum here is bounded by Lemma 5.1.1 as ${\leqslant}\unicode[STIX]{x1D70C}_{P}(t_{1})t_{1}\cdot \gcd (h,t_{1})$ which is crudely ${\leqslant}|h|t_{1}\unicode[STIX]{x1D70C}_{P}(t_{1})$ . Putting these back into (5) we find (for $h\neq 0$ ) that

$$\begin{eqnarray}|T(z,t_{1})|\ll \sqrt{|h|}\,\frac{z(\log \log x)^{5/2}}{\log x}\sum _{t_{1}\leqslant x^{1/3}}\unicode[STIX]{x1D70C}_{P}(t_{1})^{1/2}\sqrt{\frac{t_{1}}{\unicode[STIX]{x1D719}(t_{1})}}\end{eqnarray}$$

and thus an overall bound in (4) of

$$\begin{eqnarray}|\unicode[STIX]{x1D6F4}_{2}|\ll \sqrt{|h|}\,\frac{x(\log \log x)^{5/2}}{\log x}\sum _{t_{1}\leqslant x^{1/3}}\frac{\unicode[STIX]{x1D70C}_{P}(t_{1})^{1/2}}{\sqrt{t_{1}\unicode[STIX]{x1D719}(t_{1})}}.\end{eqnarray}$$

5.2.3

We can note that there is no $q$ -dependence in the above bound yet. Indeed, this only comes into play when we show there are not too many small primes $p$ with $\unicode[STIX]{x1D70C}_{P}(p)=2$ , which we shall obtain from our class number one hypothesis combined with standard results about the distribution of primes modulo 4, essentially relating the natural field $\mathbf{Q}(\sqrt{q})$ for the form $P$ to our imaginary quadratic field $\mathbf{Q}(\sqrt{-q})$ via twisting by the Dirichlet character of conductor 4.

We can slightly improve Hooley’s argument to save an additional log–log power, writing $\unicode[STIX]{x1D6F4}_{7}$ for the above $t_{1}$ -sum and noting that

$$\begin{eqnarray}\unicode[STIX]{x1D6F4}_{7}=\sum _{t_{1}\leqslant x^{1/3}}\frac{\unicode[STIX]{x1D70C}_{P}(t_{1})^{1/2}}{\sqrt{t_{1}\unicode[STIX]{x1D719}(t_{1})}}\leqslant \{_{p\not \in \{2,17\}}\biggl(1+\biggl(1-\frac{1}{p}\biggr)^{-1}\,\mathop{\sum }_{e=1}^{\infty }\frac{\unicode[STIX]{x1D70C}_{P}(p^{e})^{1/2}}{p^{e}}\biggr),\end{eqnarray}$$

whereas Hooley here takes $p\leqslant x$ .

We then take the logarithm of the right side, bounding it as

$$\begin{eqnarray}{\leqslant}\sum _{p\nmid 34y}\biggl(1-\frac{1}{p}\biggr)^{-2}\frac{\sqrt{\unicode[STIX]{x1D70C}_{P}(p)}}{p}+\mathop{\sum }_{p|y}\frac{O(1)}{\sqrt{p}}+O(1)\leqslant \sum _{\bar{\unicode[STIX]{x1D70C}}_{P}(p)=2}\frac{\sqrt{2}}{p}+O(\sqrt{\log Y}).\end{eqnarray}$$

Since $\mathbf{Q}(\sqrt{-q})$ has class number one and $X\leqslant q/4$ (from $x\ll q^{100}$ ), the primes $p\leqslant X$ with $\bar{\unicode[STIX]{x1D70C}}_{P}(p)=2$ are precisely those congruent to 3 mod 4. By standard results about primes in arithmetic progressions [Reference de La Vallée Poussin44] we thus have

$$\begin{eqnarray}\sum _{\bar{\unicode[STIX]{x1D70C}}(p)=2}\frac{\sqrt{2}}{p}=\frac{\sqrt{2}}{2}\sum _{p\equiv 3(4)}\frac{2}{p}=\frac{\sqrt{2}}{2}\log \log X+O(1).\end{eqnarray}$$

Re-exponentiating (crudely in $Y$ ) gives $\unicode[STIX]{x1D6F4}_{7}\ll Y(\log X)^{\sqrt{2}/2}$ , so that

$$\begin{eqnarray}|\unicode[STIX]{x1D6F4}_{2}|\ll Y\sqrt{|h|}\,\frac{x(\log \log x)^{(5-\sqrt{2})/2}}{(\log x)^{1-\sqrt{2}/2}},\end{eqnarray}$$

which is the desired result. ◻

We can also comment on spectral theory methods that obtain a power-saving, notably work of Duke et al [Reference Duke, Friedlander and Iwaniec22] in the case of negative discriminant (inspired by Bykovskii [Reference Bykovskii11], and alternatively presented in [Reference Iwaniec and Kowalski41, §21.4ff]), and Tóth [Reference Tóth73] for our case of positive discriminantFootnote ¹⁰ . It is not readily apparent to me what the uniformity in the discriminant is, but in the negative discriminant case the class number appears, while in Tóth’s work (cf. Lemma 4.2) the fundamental unit also presumably plays a role (similar to its effect on Hooley’s result).

There is also a result of Hejhal [Reference Hejhal, Hejhal, Sarnak and Terras36, (7.25)] regarding the congruence $w^{2}\equiv A$ to varying moduli, where a power-saving is similarly obtained (by spectral methods) for positive odd fundamental discriminants $A$ of the form $u^{2}-4$ with class number one (said set is finite by Siegel’s ineffective theorem [Reference Siegel58], and includes $\{5,21,77,437\}$ ).

6.2.1

Following Oesterlé [Reference Oesterlé52], we next define

(6) $$\begin{eqnarray}\unicode[STIX]{x1D6F9}_{f}(s)=L_{f}(s)L_{f\unicode[STIX]{x1D706}}(s)\quad \text{and}\quad G_{\!f}(s)=\frac{L_{f\unicode[STIX]{x1D712}}(s)}{L_{f\unicode[STIX]{x1D706}}(s)},\end{eqnarray}$$

where $\unicode[STIX]{x1D706}$ is the Liouville function, which is completely multiplicative and $-1$ on all primes. Hence

$$\begin{eqnarray}\displaystyle L_{f\unicode[STIX]{x1D706}}(s)=\mathop{\prod }_{p}\biggl(1+\frac{\unicode[STIX]{x1D6FC}_{p}}{p^{s}}\biggr)^{-1}\biggl(1+\frac{\bar{\unicode[STIX]{x1D6FC}}_{p}}{p^{s}}\biggr)^{-1} & & \displaystyle \nonumber\end{eqnarray}$$

so that

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D6F9}_{f}(s)=\mathop{\prod }_{p}\biggl(1-\frac{\unicode[STIX]{x1D6FC}_{p}^{2}}{p^{2s}}\biggr)^{-1}\biggl(1-\frac{\bar{\unicode[STIX]{x1D6FC}}_{p}^{2}}{p^{2s}}\biggr)^{-1} & & \displaystyle \nonumber\end{eqnarray}$$

is naturally a function of $2s$ , and up to bad Euler factors at $p\in \{2,17\}$ it is in fact equal to $L_{S^{2}\!f}(2s)/L_{A^{2}\!f}(2s)$ , where $S^{2}f$ and $A^{2}\,f$ are, respectively, the symmetric and alternating squares of $f$ (at the level of motivic $L$ -functions), where moreover $L_{A^{2}\,f}(2s)=\unicode[STIX]{x1D701}(2s-1)$ has a pole at $s=1$ .

More specifically, the motivic symmetric-square $L$ -function $L_{S^{2}f}(s)$ here is

$$\begin{eqnarray}L_{S^{2}f}(s)=\mathop{\prod }_{p}\biggl(1-\frac{{\tilde{\unicode[STIX]{x1D6FC}}_{p}}^{2}}{p^{s}}\biggr)^{-1}\biggl(1-\frac{\tilde{\unicode[STIX]{x1D6FC}}_{p}\bar{\tilde{\unicode[STIX]{x1D6FC}}}_{p}}{p^{s}}\biggr)^{-1}\biggl(1-\frac{\bar{\tilde{\unicode[STIX]{x1D6FC}}}_{p}^{2}}{p^{s}}\biggr)^{-1}\end{eqnarray}$$

with $\tilde{\unicode[STIX]{x1D6FC}}_{p}=\unicode[STIX]{x1D6FC}_{p}$ for all $p$ except $p=17$ , where instead we have $\tilde{\unicode[STIX]{x1D6FC}}_{17}=1+4\text{i}$ (co-opting this from the twisted modular form of level 32). Since $\tilde{\unicode[STIX]{x1D6FC}}_{p}\bar{\tilde{\unicode[STIX]{x1D6FC}}}_{p}=p$ for all $p\neq 2$ , the middle term corresponds to the Riemann $\unicode[STIX]{x1D701}$ -function, and we find that

$$\begin{eqnarray}\unicode[STIX]{x1D6F9}_{f}(s)=\frac{L_{S^{2}f}(2s)}{\unicode[STIX]{x1D701}(2s-1)}\cdot \biggl(1-\frac{2}{2^{2s}}\biggr)^{-1}\biggl(1+\frac{30}{17^{2s}}+\frac{17^{2}}{17^{4s}}\biggr),\end{eqnarray}$$

where we shall notate the contribution from the exceptional Euler factors by $B_{f}(2s)$ .

6.2.2

We define $\tilde{G}_{\!f}(s)=G_{\!f}(s)-1$ and write $g_{l}$ for the Dirichlet series coefficients of $G_{\!f}(s)$ , so that by the above Euler quotient $G_{\!f}(s)=L_{f\unicode[STIX]{x1D712}}(s)/L_{f\unicode[STIX]{x1D706}}(s)$ we have

$$\begin{eqnarray}\displaystyle G_{\!f}(s) & = & \displaystyle \mathop{\sum }_{l=1}^{\infty }\frac{g_{l}}{l^{s}}=1+\tilde{G}_{\!f}(s)\nonumber\\ \displaystyle & = & \displaystyle \biggl(1+\frac{\unicode[STIX]{x1D6FC}_{q}}{q^{s}}\biggr)\biggl(1+\frac{\bar{\unicode[STIX]{x1D6FC}}_{q}}{q^{s}}\biggr)\mathop{\prod }_{p:\unicode[STIX]{x1D712}(p)=+1}\frac{(1+\unicode[STIX]{x1D6FC}_{p}/p^{s})(1+\bar{\unicode[STIX]{x1D6FC}}_{p}/p^{s})}{(1-\unicode[STIX]{x1D6FC}_{p}/p^{s})(1-\bar{\unicode[STIX]{x1D6FC}}_{p}/p^{s})}.\nonumber\end{eqnarray}$$

We see that $g_{l}=0$ unless all $p|l$ have $\unicode[STIX]{x1D712}(p)\neq -1$ , and thus in the context of class number one we have $g_{l}=0$ for $1<l\leqslant q/4$ . For $p$ prime with $\unicode[STIX]{x1D712}(p)=+1$ we have that $g_{p}=2(\unicode[STIX]{x1D6FC}_{p}+\bar{\unicode[STIX]{x1D6FC}}_{p})=2c_{p}$ , while $c_{q}=g_{q}=0$ since $q$ is 3 mod 4. We also note that $|g_{p^{v}}|\leqslant 4vp^{v/2}$ as follows from $|\unicode[STIX]{x1D6FC}_{p}|\leqslant \sqrt{p}$ and $[(1+x)/(1-x)]^{2}=1+\sum _{n}4nx^{n}$ .

6.2.3

From $L_{f}(s)L_{f\unicode[STIX]{x1D712}}(s)=\unicode[STIX]{x1D6F9}_{f}(s)G_{\!f}(s)$ and $G_{\!f}(s)=1+\tilde{G}_{\!f}(s)$ we have

$$\begin{eqnarray}\displaystyle 0 & = & \displaystyle \int _{(2)}\unicode[STIX]{x1D6F9}_{f}(s)G_{\!f}(s)\unicode[STIX]{x1D6E4}(s)^{2}\biggl(\frac{Nq}{4\unicode[STIX]{x1D70B}^{2}}\biggr)^{s-1}\frac{\unicode[STIX]{x2202}s/2\unicode[STIX]{x1D70B}\text{i}}{(s-1)^{2}}\nonumber\\ \displaystyle & = & \displaystyle \int _{(2)}\unicode[STIX]{x1D6F9}_{f}(s)\unicode[STIX]{x1D6E4}(s)^{2}\biggl(\frac{Nq}{4\unicode[STIX]{x1D70B}^{2}}\biggr)^{s-1}\frac{\unicode[STIX]{x2202}s/2\unicode[STIX]{x1D70B}\text{i}}{(s-1)^{2}}\nonumber\\ \displaystyle & & \displaystyle +\,\int _{(2)}\unicode[STIX]{x1D6F9}_{f}(s)\tilde{G}_{\!f}(s)\unicode[STIX]{x1D6E4}(s)^{2}\biggl(\frac{Nq}{4\unicode[STIX]{x1D70B}^{2}}\biggr)^{s-1}\frac{\unicode[STIX]{x2202}s/2\unicode[STIX]{x1D70B}\text{i}}{(s-1)^{2}}.\nonumber\end{eqnarray}$$

The first integral is estimated by moving the path of integration to the left and using Cauchy’s residue theorem, with the decay from $q^{s}$ dominating the error. Via the standard zero-free region [Reference de La Vallée Poussin44] and estimates for $\unicode[STIX]{x1D701}$ (here for $|t|\ll \log q$ , as the vertical decay of $\unicode[STIX]{x1D6E4}(s)^{2}$ suffices otherwise), and analytic properties of the symmetric-square $L$ -function we get

$$\begin{eqnarray}\displaystyle & & \displaystyle \int _{(2)}\unicode[STIX]{x1D6F9}_{f}(s)\unicode[STIX]{x1D6E4}(s)^{2}\biggl(\frac{Nq}{4\unicode[STIX]{x1D70B}^{2}}\biggr)^{s-1}\frac{\unicode[STIX]{x2202}s/2\unicode[STIX]{x1D70B}\text{i}}{(s-1)^{2}}\nonumber\\ \displaystyle & & \displaystyle \quad =\int _{(2)}\frac{B_{f}(2s)L_{S^{2}f}(2s)}{\unicode[STIX]{x1D701}(2s-1)}\unicode[STIX]{x1D6E4}(s)^{2}\biggl(\frac{Nq}{4\unicode[STIX]{x1D70B}^{2}}\biggr)^{s-1}\frac{\unicode[STIX]{x2202}s/2\unicode[STIX]{x1D70B}\text{i}}{(s-1)^{2}}\nonumber\\ \displaystyle & & \displaystyle \quad =2B_{f}(2)L_{S^{2}f}(2)+O(1/q^{1/100\log \log q})\gg 1.\nonumber\end{eqnarray}$$

In fact, in this CM case we have $L_{S^{2}f}(s)=L_{\unicode[STIX]{x1D702}_{-4}}(s-1)L_{\unicode[STIX]{x1D713}_{\star }^{2}}(s)$ , where $\unicode[STIX]{x1D713}$ is the Hecke Grössencharacter corresponding to the elliptic curve isogeny class of conductor 32. Thus the analytic properties of the symmetric-square are immediate (see [Reference Hecke31], or [Reference Shimura57] in general) and in any case we can compute $L_{S^{2}f}(2)\approx 0.674969789$ .

6.3.1

The Dirichlet series $\tilde{G}_{\!f}(s)$ is not directly in a form where we can easily detect cancellation of the coefficients, and we shall make some modifications to ameliorate our efforts. What we wish to do is replace

$$\begin{eqnarray}\mathop{\sum }_{k=1}^{\infty }\unicode[STIX]{x1D709}_{k}\mathop{\sum }_{l>1}\frac{g_{l}}{(k^{2}l)^{s}}\quad \text{by}\quad \mathop{\sum }_{k=1}^{\infty }\tilde{\unicode[STIX]{x1D709}}_{k}\mathop{\sum }_{n=1}^{\infty }\frac{c_{n}\tilde{R}_{\unicode[STIX]{x1D712}}(n)}{(k^{2}n)^{s}},\end{eqnarray}$$

where $\tilde{R}_{\unicode[STIX]{x1D712}}(n)$ counts representations of $n$ by the principal form $x^{2}+xy+((q+1)/4)y^{2}$ with $y\geqslant 1$ . We similarly notate by $\tilde{R}_{\unicode[STIX]{x1D712}}^{\star }(n)$ the number of primitive such representations with $\gcd (x,y)=1$ . This resulting inner sum over $n$ is more suitable for showing cancellation by our methods.

From above we have $g_{p}=2c_{p}=\tilde{R}_{\unicode[STIX]{x1D712}}^{\star }(p)c_{p}$ for $p$ with $\unicode[STIX]{x1D712}(p)\neq -1$ . The non-prime $l$ with $g_{l}\neq 0$ or $\tilde{R}_{\unicode[STIX]{x1D712}}^{\star }(l)\neq 0$ have $l\geqslant (q/4)^{2}$ and thus contribute negligibly due to the $W$ -decay (by crude estimates $|\unicode[STIX]{x1D709}_{k}|\leqslant k\unicode[STIX]{x1D70F}_{2}(k)$ and $|g_{l}|,|c_{l}\tilde{R}_{\unicode[STIX]{x1D712}}^{\star }(l)|\leqslant \unicode[STIX]{x1D70F}_{2}(l)^{2}\sqrt{l}$ ), giving

$$\begin{eqnarray}E=\mathop{\sum }_{k=1}^{\infty }\unicode[STIX]{x1D709}_{k}\mathop{\sum }_{l>1}g_{l}W\biggl(\frac{4\unicode[STIX]{x1D70B}^{2}k^{2}l}{Nq}\biggr)=\mathop{\sum }_{k=1}^{\infty }\unicode[STIX]{x1D709}_{k}\mathop{\sum }_{l=1}^{\infty }c_{l}\tilde{R}_{\unicode[STIX]{x1D712}}^{\star }(l)W\biggl(\frac{4\unicode[STIX]{x1D70B}^{2}k^{2}l}{Nq}\biggr)+O(1/q^{99}).\end{eqnarray}$$

6.3.2

We then write

$$\begin{eqnarray}\displaystyle \mathop{\sum }_{k=1}^{\infty }\frac{\unicode[STIX]{x1D709}_{k}}{k^{2s}}\mathop{\sum }_{l=1}^{\infty }\frac{c_{l}\tilde{R}_{\unicode[STIX]{x1D712}}^{\star }(l)}{l^{s}} & = & \displaystyle \mathop{\sum }_{j=1}^{\infty }\frac{\unicode[STIX]{x1D709}_{j}}{j^{2s}}\biggl(\mathop{\sum }_{m=1}^{\infty }\frac{c_{m^{2}}}{m^{2s}}\biggr)^{-1}\cdot \mathop{\sum }_{m=1}^{\infty }\frac{c_{m^{2}}}{m^{2s}}\mathop{\sum }_{l=1}^{\infty }\frac{c_{l}\tilde{R}_{\unicode[STIX]{x1D712}}^{\star }(l)}{l^{s}}\nonumber\\ \displaystyle & = & \displaystyle \mathop{\sum }_{k=1}^{\infty }\frac{\tilde{\unicode[STIX]{x1D709}}_{k}}{k^{2s}}\cdot \mathop{\sum }_{m=1}^{\infty }\frac{c_{m^{2}}}{m^{2s}}\mathop{\sum }_{l=1}^{\infty }\frac{c_{l}\tilde{R}_{\unicode[STIX]{x1D712}}^{\star }(l)}{l^{s}},\nonumber\end{eqnarray}$$

where we have

$$\begin{eqnarray}\mathop{\sum }_{k=1}^{\infty }\frac{\tilde{\unicode[STIX]{x1D709}}_{k}}{k^{2s}}=\mathop{\prod }_{p}\biggl(\mathop{\sum }_{v=0}^{\infty }\mathop{\sum }_{e=0}^{v}\frac{\unicode[STIX]{x1D6FC}_{p}^{2e}\bar{\unicode[STIX]{x1D6FC}}_{p}^{2v-2e}}{p^{2vs}}\biggr)\biggl(\mathop{\sum }_{v=0}^{\infty }\mathop{\sum }_{e=0}^{2v}\frac{\unicode[STIX]{x1D6FC}_{p}^{a}\bar{\unicode[STIX]{x1D6FC}}_{p}^{2v-e}}{p^{2vs}}\biggr)^{-1}.\end{eqnarray}$$

We can then bound the $p^{2u}$ -coefficient of the reciprocated Dirichlet series by noting that $1/(1-3z-5z^{2}-7z^{3}+\cdots \,)=1/F(z)$ has its $u$ th coefficient bounded by $\unicode[STIX]{x1D706}^{u}$ , where $1/\unicode[STIX]{x1D706}\approx 0.2192$ satisfies $F(1/\unicode[STIX]{x1D706})=0$ , so that

$$\begin{eqnarray}|\tilde{\unicode[STIX]{x1D709}}_{p^{w}}|\leqslant \mathop{\sum }_{i=0}^{w}(\text{i}+1)p^{i}\cdot \unicode[STIX]{x1D706}^{w-i}p^{w-i}\leqslant (\unicode[STIX]{x1D706}+2)^{w}p^{w}\leqslant 7^{w}p^{w}\leqslant p^{4w},\end{eqnarray}$$

implying $|\tilde{\unicode[STIX]{x1D709}}_{k}|\leqslant k^{4}$ by multiplicativity (this can be improved, but we do not bother).

This Dirichlet series replacement gives us

$$\begin{eqnarray}E=\mathop{\sum }_{k=1}^{\infty }\tilde{\unicode[STIX]{x1D709}}_{k}\mathop{\sum }_{l=1}^{\infty }c_{l}\tilde{R}_{\unicode[STIX]{x1D712}}^{\star }(l)\mathop{\sum }_{m=1}^{\infty }c_{m^{2}}W\biggl(\frac{4\unicode[STIX]{x1D70B}^{2}k^{2}lm^{2}}{Nq}\biggr)+O(1/q^{99}),\end{eqnarray}$$

where the contributing $m$ with $\gcd (l,m)\neq 1$ have $lm^{2}\geqslant (q/4)^{3}$ and thus are negligible by $W$ -decay. So we can replace $c_{l}c_{m^{2}}=c_{lm^{2}}$ , and since $\tilde{R}_{\unicode[STIX]{x1D712}}^{\star }(l)=\tilde{R}_{\unicode[STIX]{x1D712}}(lm^{2})$ for $m\leqslant q/4$ , inserting this and putting $n=lm^{2}$ gives the desired rearrangement

$$\begin{eqnarray}E=\mathop{\sum }_{k=1}^{\infty }\tilde{\unicode[STIX]{x1D709}}_{k}\mathop{\sum }_{n=1}^{\infty }c_{n}\tilde{R}_{\unicode[STIX]{x1D712}}(n)W\biggl(\frac{4\unicode[STIX]{x1D70B}^{2}k^{2}n}{Nq}\biggr)+O(1/q^{99}).\end{eqnarray}$$

Moreover, using $|c_{n}|\leqslant \sqrt{n}\unicode[STIX]{x1D70F}_{2}(n)$ and the easy bound (e.g. from lattice points in an ellipse) that the summation of $\tilde{R}_{\unicode[STIX]{x1D712}}(n)$ up to $x$ is $\ll x/\sqrt{q}$ , we can curtail the $k$ -sum at $K=(\log q)^{1/10^{9}}$ by $W$ -decay, with an error that is $\ll q\text{e}^{-K/100}\ll q/(\log q)^{999}$ .

7.1.1

We then consider $l=x^{2}+xy+((q+1)/4)y^{2}$ with $y\geqslant 1$ . By rearrangement this says $4l=(2x+y)^{2}+qy^{2}$ , and so by the accounting of $\tilde{R}_{\unicode[STIX]{x1D712}}(l)$ we thus find

$$\begin{eqnarray}E_{1}^{k}=\mathop{\sum }_{y=1}^{\infty }\sum _{4l=(2x+y)^{2}+qy^{2}}(4a+1)(-1)^{b}\unicode[STIX]{x1D703}(l)\,W\biggl(\frac{4\unicode[STIX]{x1D70B}^{2}k^{2}l}{Nq}\biggr).\end{eqnarray}$$

7.1.2

Equating the above conditions on $l$ implies that

$$\begin{eqnarray}4(4a+1)^{2}-(2x+y)^{2}=-4(2b)^{2}+qy^{2}.\end{eqnarray}$$

It is a feature of working with the principal form that the left side factors in a particular nice way as $tu$ with

$$\begin{eqnarray}t=2(4a+1)+(2x+y),\qquad u=2(4a+1)-(2x+y).\end{eqnarray}$$

(In general, we can achieve a similar factorization for any form in the principal genus, though the resulting uniformity is unclear.) We thus have $tu=-4(2b)^{2}+qy^{2}$ and

(7) $$\begin{eqnarray}2(4a+1)=(t+u)/2,\qquad 2x+y=(t-u)/2.\end{eqnarray}$$

Moreover, we have

$$\begin{eqnarray}4l=\frac{(t-u)^{2}}{4}+qy^{2},\qquad l=\frac{(t+u)^{2}}{16}+(2b)^{2}.\end{eqnarray}$$

7.2.1

For odd $y$ , since $q$ is 3 mod 8 we have

$$\begin{eqnarray}tu=-4(2b)^{2}+qy^{2}\equiv qy^{2}\equiv 3,11\hspace{0.6em}({\rm mod}\hspace{0.2em}16),\end{eqnarray}$$

so in particular $t-u\equiv 2$  (4), implying that the condition $2x+y=(t-u)/2$ gives an integral $x$ since $y$ is odd.

In dealing with $a$ , when $qy^{2}\equiv 3$  (16) we see that $t\equiv 1,3,9,11$  (16) determines $u\equiv 3,1,11,9$  (16), so $(4a+1)=(t+u)/4$ gives an integral $a$ from $t+u\equiv 4$  (16). On the other hand, the other $t$ yield a non-integral $a$ by the same calculation, and so we ignore these $t$ . For $qy^{2}\equiv 11$  (16) the situation is similar, with the residue classes $t\equiv 5,15,7,13$  (16) being those that yield integral $a$ .

Thus for odd $y$ there is a set of residue classes modulo 16 (depending on $y$ ) for $t$ corresponding exactly to integral values of $a$ , while $x$ is always integral.

7.2.2

When $4|y$ , the analysis is only slightly more complicated. We have $x$ is odd, so that $(t-u)=2(2x+y)\equiv 4\,(8)$ . On the other hand, $(t+u)=4(4a+1)\equiv 4\,(16)$ and thus one of $t,u$ is 0 mod 8 and the other is 4 mod 8. By symmetry, we can assume $t$ is the latter. In particular, we find that $(t/4)(u/4)=-b^{2}+q(y/4)^{2}$ must be even, so that $b$ and $y/4$ must have the same parity. When these are both odd, we find that $(t/4)(u/4)\equiv 2\,(8)$ , implying $8\Vert u$ so that $u/4\equiv 2\,(4)$ whence $t\equiv 12\,(16)$ to ensure integrality of $a$ . Similarly, when $b$ and $y/4$ are both even we have $4|(u/4)$ and so $t\equiv 4\,(16)$ .

As a result of the above integrality considerations, with $2b=4\tilde{b}+2r_{b}$ we have

$$\begin{eqnarray}E_{1}^{k}=\mathop{\sum }_{y=1}^{\infty }\mathop{\sum }_{r_{b}=1}^{2}\mathop{\sum }_{r_{t}=1}^{16}z_{2}(y,r_{b},r_{t})\sum _{tu=-4(4\tilde{b}+2r_{b})^{2}+qy^{2}\!\!}\frac{t+u}{4}(-1)^{r_{b}}\unicode[STIX]{x1D703}(l)W\biggl(\frac{4\unicode[STIX]{x1D70B}^{2}k^{2}l}{Nq}\biggr),\end{eqnarray}$$

where

• ∙ $z_{2}(y,r_{b},r_{t})=0$ when $y$ is odd with $qy^{2}\equiv 3\,(16)$ and $r_{t}\not \in \{1,3,9,11\}$ ,

• ∙ $z_{2}(y,r_{b},r_{t})=0$ when $y$ is odd with $qy^{2}\equiv 11\,(16)$ and $r_{t}\not \in \{5,15,7,13\}$ ,

• ∙ $z_{2}(y,r_{b},r_{t})=1$ when $qy^{2}\equiv 3\,(16)$ and $r_{t}\in \{1,3,9,11\}$ ,

• ∙ $z_{2}(y,r_{b},r_{t})=1$ when $qy^{2}\equiv 11\,(16)$ and $r_{t}\in \{5,15,7,13\}$ ,

• ∙ $z_{2}(y,r_{b},r_{t})=0$ when $y\equiv 2\,(4)$ ,

• ∙ $z_{2}(y,r_{b},r_{t})=0$ when $4|y$ and $r_{b}$ and $y/4$ have differing parity,

• ∙ $z_{2}(y,r_{b},r_{t})=0$ when $4|y$ and $r_{b}$ and $y/4$ are odd and $r_{t}\neq 12$ ,

• ∙ $z_{2}(y,r_{b},r_{t})=2$ when $4|y$ and $r_{b}$ and $y/4$ are odd and $r_{t}=12$ ,

• ∙ $z_{2}(y,r_{b},r_{t})=0$ when $4|y$ and $r_{b}$ and $y/4$ are even and $r_{t}\neq 4$ ,

• ∙ $z_{2}(y,r_{b},r_{t})=2$ when $4|y$ and $r_{b}$ and $y/4$ are even and $r_{t}=4$ .

We could combine some of the congruence classes when $y$ is odd via the $t$ – $u$ symmetry (thereby doubling $z_{2}$ ), but choose not to do so.

7.3.1

We consider the effect of fixing $t$ modulo 17. When $\gcd (t,17)=1$ this determines $u$  mod 17 by the relation $tu\equiv -4(2r_{b})^{2}+qy^{2}$  (17), and thus also $a$  mod 17 from $(t+u)/4=4a+1$ . Writing $r_{a}$ for this residue class, the character contribution $\unicode[STIX]{x1D703}((4a+1)^{2}+(2r_{b})^{2})$ is thereby fixed, since $(4a+1)^{2}\equiv (4r_{a}+1)^{2}$  (136).

Similarly to the analysis with $4|y$ , when $17|t$ we can essentially switch $t$ and $u$ . Indeed, we note that $-4(2b)^{2}+qy^{2}\equiv 0$  (17) has no solutions other than the “trivial” solution $(b,y)\equiv (0,0)$  (17), due to the fact that $q$ is non-square modulo 17 (using that $-q$ is a non-square by our class number one assumption, and exploiting quadratic reciprocity since 17 is 1 mod 4).

This allows us to note that 17 divides at most one of $t$ and $u$ , or else the character evaluation is zero. Indeed, when $17|tu$ we have $17|b$ , and so if $17|t$ and $17|u$ then $17|(4a+1)$ , so that $\unicode[STIX]{x1D703}((4a+1)^{2}+(2b)^{2})=0$ as the argument is divisible by 17.

Therefore, the contribution from $17|t$ is just the contribution from $17|r_{b}$ and $17|y$ when running over $u$ -classes coprime to 17.

We then get

$$\begin{eqnarray}E_{1}^{k}=\mathop{\sum }_{y=1}^{\infty }\mathop{\sum }_{r_{b}=1}^{34}\mathop{\sum }_{r_{t}=1}^{272}z(y,r_{b},r_{t})\sum _{tu=-4(68\tilde{b}+2r_{b})^{2}+qy^{2}}\frac{t+u}{4}\,W\biggl(\frac{4\unicode[STIX]{x1D70B}^{2}k^{2}l}{Nq}\biggr),\end{eqnarray}$$

where $z(y,r_{b},r_{t})=z_{2}(y,r_{b},r_{t}\,\text{mod 16})\cdot z_{17}(y,r_{b},r_{t})\cdot \tilde{\unicode[STIX]{x1D703}}(y,r_{b},r_{t})$ and

• ∙ $z_{17}(y,r_{b},r_{t})=0$ when $17|r_{t}$ ,

• ∙ $z_{17}(y,r_{b},r_{t})=2$ when $17\nmid r_{t}$ , $17|r_{b}$ , and $17|y$ ,

• ∙ $z_{17}(y,r_{b},r_{t})=1$ otherwise,

with

$$\begin{eqnarray}\tilde{\unicode[STIX]{x1D703}}(y,r_{b},r_{t})=(-1)^{r_{b}}\unicode[STIX]{x1D703}((r_{t}+u)^{2}/16+(2r_{b})^{2}),\end{eqnarray}$$

where $u$ is determined by the congruence $r_{t}u\equiv -4(2r_{b})^{2}+qy^{2}\,(2\cdot 272)$ (here the extra 2-power is necessary when $4\Vert r_{t}$ (so $4|y$ ) to ensure $8|u$ ).

7.4.1

We proceed to eliminate the $y>K=(\log q)^{1/10^{9}}$ by $W$ -decay via the first expression in (9). For this, we first consider $y\gg (\log q)^{2}$ , where the $W$ -decay of $W(z)\ll \text{e}^{-\sqrt{z}}$ is so strong that we can simply ignore the $P(\tilde{b})\equiv 0\,(t)$ condition and sum over all $(y,t,\tilde{b})$ , getting a sufficient bound. Indeed, the same argument works when any of $|\tilde{b}|,|t|,|u|\gg \sqrt{q}(\log q)^{2}$ , as each implies $l\gg q(\log q)^{4}$ . To handle the remaining terms, we use the bounds $W(\cdot )\ll \text{e}^{-K/100}$ and $\unicode[STIX]{x1D70C}_{P}(t)\leqslant y2^{\unicode[STIX]{x1D714}(t)}$ for an overall error in $E_{1}$ of

$$\begin{eqnarray}\ll K^{5}\cdot (\log q)^{2}\cdot \frac{\sqrt{q}(\log q)^{2}}{\text{e}^{K/100}}\sum _{\gcd (t,34)=1}\unicode[STIX]{x1D70C}_{P}(t)\,\frac{\sqrt{q}(\log q)^{2}}{|t|}\ll \frac{q}{(\log q)^{999}}.\end{eqnarray}$$

7.4.2

For future reference, we can similarly note there is negligible contribution when any of $|t|,|u|,|\tilde{b}|\geqslant \sqrt{q}K$ . Indeed, by the above $l$ -expressions (9) each implies that $l\geqslant qK^{2}/16$ , while we already saw that when one of $|\tilde{b}|,|t|,|u|\gg \sqrt{q}(\log q)^{2}$ there is negligible contribution, and so as before we get a bound in $E_{1}$ of

$$\begin{eqnarray}\ll K^{5}\cdot K\cdot \frac{\sqrt{q}(\log q)^{2}}{\text{e}^{K/400}}\cdot \sum _{\gcd (t,34)=1}\unicode[STIX]{x1D70C}_{P}(t)\,\frac{\sqrt{q}(\log q)^{2}}{|t|}\ll \frac{q}{(\log q)^{999}}.\end{eqnarray}$$

7.6.1

For $s\neq 0$ the values of $w/|t|$ are restricted to a region of size $2/S$ about $s/S$ , and thus we can aim to approximate

$$\begin{eqnarray}\displaystyle u=\frac{-4(68\tilde{b}+2r_{b})^{2}+qy^{2}}{m_{i}t} & & \displaystyle \nonumber\end{eqnarray}$$

by

$$\begin{eqnarray}\displaystyle U_{j,c,s}=\frac{-4[68(c|T_{j}|+s|T_{j}|/S)]^{2}+qy^{2}}{m_{i}T_{j}}, & & \displaystyle \nonumber\end{eqnarray}$$

while for $s=0$ we instead wish to approximate by $U_{j,\tilde{c},s}$ .

This leads us to approximate $B_{3}^{j,s}$ by

(11) $$\begin{eqnarray}B_{4}^{j,s}=G_{s}^{j}\sum _{t\equiv r_{t}\,(M_{i})}\sum _{P(w)\equiv 0\,(t)}F_{s}(w/|t|),\end{eqnarray}$$

where the $w$ -sum can be over any period of length $|t|$ , and we have

$$\begin{eqnarray}G_{s}^{j}=\mathop{\sum }_{|c|\ll C_{j}}\frac{m_{i}T_{j}+U_{j,c,s}}{4}\,W\biggl(\frac{4\unicode[STIX]{x1D70B}^{2}k^{2}}{Nq}[(m_{i}T_{j}-U_{j,c,s})^{2}/4+qy^{2}]\biggr),\end{eqnarray}$$

with $G_{s}^{j}$ thus also depending on $(k,i,y)$ , which we suppress from the notation.

7.6.2

We now analyze the approximation error in $B_{4}^{j,s}$ to $B_{3}^{j,s}$ . When $s\neq 0$ we avoid endpoints and have $|w/|t|-s/S|\leqslant 1/S$ , so that $|w-s|t|/S|\ll |T_{j}|/S$ and thus $|w-s|T_{j}|/S|\ll |T_{j}|/S+\sqrt{q}/S\ll \sqrt{q}K/S$ . Since $|t-T_{j}|\ll \sqrt{q}/S\ll |T_{j}|/S^{4/5}$ we get

$$\begin{eqnarray}u=\frac{-4[68(c|T_{j}|+s|T_{j}|/S)+O(C_{j}\cdot \sqrt{q}/S)+O(\sqrt{q}K/S)+O(1)]^{2}+qy^{2}}{m_{i}T_{j}(1+O(1/S^{4/5}))}\!,\end{eqnarray}$$

so from $C_{j}=SK/|j|\ll KS^{1/5}=C$ and using $\sqrt{q}/S^{1/5}\ll |T_{j}|\ll \sqrt{q}K$ we have

$$\begin{eqnarray}\displaystyle |u-U_{j,c,s}| & \ll & \displaystyle \frac{C_{j}|T_{j}|\cdot C_{j}(\sqrt{q}/S)}{|T_{j}|}+\frac{(C_{j}T_{j})^{2}}{|T_{j}|S^{4/5}}+\frac{qK^{2}}{|T_{j}|S^{4/5}}\nonumber\\ \displaystyle & \ll & \displaystyle \frac{\sqrt{q}K^{2}}{S^{3/5}}+\frac{\sqrt{q}K^{3}}{S^{2/5}}+\frac{\sqrt{q}K^{2}}{S^{3/5}}.\nonumber\end{eqnarray}$$

Replacing the $(m_{i}t+u)$ term in $B_{3}^{j,s}$ of (10) by $(m_{i}T_{j}+U_{j,c,s})$ , for the error therein we see that summing over $s\neq 0$ still leaves $\sum _{s}F_{s}(\cdot )\leqslant 1$ everywhere, while summing over $j$ gives a bound via $|W(\cdot )|\ll 1$ and Corollary 4.2.3 of

$$\begin{eqnarray}\displaystyle & & \displaystyle \mathop{\sum }_{S^{4/5}\ll |j|\ll SK}\,\sum _{t\equiv r_{t}\,(M_{i})}\sum _{P(w)\equiv 0\,(t)}\mathop{\sum }_{|c|\ll C_{j}}\frac{\sqrt{q}K^{3}}{S^{2/5}}\nonumber\\ \displaystyle & & \displaystyle \quad \ll \frac{C\sqrt{q}K^{3}}{S^{2/5}}\sum _{(t,34)=1}\unicode[STIX]{x1D70C}_{P}(t)\ll \frac{qK^{6}}{S^{1/5}},\nonumber\end{eqnarray}$$

which is acceptable, as summing over $(k,i,y,r_{b},r_{t})$ multiplies by the bound by  $K^{6}$ .

Next we have to replace the argument of $W$ involving $t$ and $u$ by the corresponding approximations. We first note that $|t|,|u|\ll \sqrt{q}K$ by $W$ -decay, and so $(m_{i}t-u)^{2}$ differs from $(m_{i}T_{j}-U_{j,c,s})^{2}$ by $\ll \sqrt{q}K\cdot (\sqrt{q}K^{3}/S^{2/5})=qK^{4}/S^{2/5}$ . The boundedness of the derivative of $W$ along with $|m_{i}T_{j}|,|U_{j,c,s}|\ll \sqrt{q}K$ imply (again with the $s$ -sum being harmless) that the error herein when summing over all $j$ is bounded as

$$\begin{eqnarray}K^{2}\frac{K^{4}}{S^{2/5}}\cdot C\sqrt{q}K\sum _{(t,34)=1}\unicode[STIX]{x1D70C}_{P}(t)\ll \frac{qK^{10}}{S^{1/5}},\end{eqnarray}$$

which again is acceptable. Thus we find that $B_{4}^{j,s}$ is a good approximant to $B_{3}^{j,s}$ .

The same argument holds for $\tilde{B}_{3}^{j,s}$ and $s=0$ , as we start with $|\tilde{w}/|t|-s/S|\leqslant 1/S$ , and the rest follows mutatis mutandis, the re-naming of $c$ to $\tilde{c}$ in $G_{s}^{j}$ being irrelevant.

7.7.1

For each $j$ with $S^{4/5}\ll |j|\ll SK$ this leaves us with

$$\begin{eqnarray}\mathop{\sum }_{s=0}^{S-1}G_{s}^{j}\mathop{\sum }_{|h|\ll H}c_{s}(h)\sum _{t\equiv r_{t}\,(M_{i})}\sum _{P(w)\equiv 0\,(t)}\exp (2\unicode[STIX]{x1D70B}\text{i}hw/|t|),\end{eqnarray}$$

where the inner double sum for $h\neq 0$ can be estimated by Hooley’s methods. Noting that $|c_{s}(h)|\ll 1/S$ , by Lemma 5.2.1 the contribution from $h\neq 0$ is bounded as

$$\begin{eqnarray}\ll C_{j}\sqrt{q}K\cdot KH^{3/2}\biggl(\frac{|T_{j}|(\log \log |T_{j}|)^{(5-\sqrt{2})/2}}{(\log |T_{j}|)^{(2-\sqrt{2})/2}}\biggr),\end{eqnarray}$$

and crudely using $|T_{j}|\ll \sqrt{q}K$ while $|j|\ll SK$ the overall bound when summing over $j$ is

$$\begin{eqnarray}\ll SK\cdot S^{1/5}\sqrt{q}K^{2}\cdot KS^{9/2}\frac{\sqrt{q}K}{(\log q)^{29/100}}=\frac{qS^{57/10}K^{5}}{(\log q)^{29/100}},\end{eqnarray}$$

which by our choice of $S=(\log q)^{1/100}$ and $K=(\log q)^{1/10^{9}}$ is sufficient.