2.1 Proof of Proposition 2.1

Fix a prime $\ell \geqslant 3$ and a square-free integer $X\leqslant d<2X$ . Let $H=\text{Cl}(-d)$ be the class group of $\mathbb{Q}(\sqrt{-d})$ , with class number $h(-d)=\#\text{Cl}(-d)$ . Let $H_{\ell }$ denote the maximal elementary abelian $\ell$ -group in $H$ , with $h_{\ell }(-d)=\#H_{\ell }$ . Since

(2.2) $$\begin{eqnarray}\#H/H_{\ell }=\frac{h(-d)}{h_{\ell }(-d)},\end{eqnarray}$$

in order to show that $h_{\ell }(-d)$ is small it suffices to show that there are many cosets of $H_{\ell }$ in $H$ . Let $\unicode[STIX]{x1D712}_{d}(\cdot )$ denote the quadratic character associated to $\mathbb{Q}(\sqrt{-d})$ . Picking a prime $p\nmid 2d$ such that $\unicode[STIX]{x1D712}_{d}(p)=1$ , it follows that $p$ splits in $\mathbb{Q}(\sqrt{-d})$ as $\mathfrak{p}\mathfrak{p}^{\unicode[STIX]{x1D70E}}$ , say, where $\unicode[STIX]{x1D70E}$ is the nontrivial Galois automorphism of $\mathbb{Q}(\sqrt{-d})$ . Suppose that two distinct primes $p,p^{\prime }$ split in this manner as $\mathfrak{p}\mathfrak{p}^{\unicode[STIX]{x1D70E}}$ and $\mathfrak{p}^{\prime }{\mathfrak{p}^{\prime }}^{\unicode[STIX]{x1D70E}}$ respectively, and suppose that $\mathfrak{p}$ and $\mathfrak{p}^{\prime }$ represent the same class in $H/H_{\ell }$ , so that $\mathfrak{p}H_{\ell }=\mathfrak{p}^{\prime }H_{\ell }$ . It follows that $\mathfrak{p}^{-1}\mathfrak{p}^{\prime }\in H_{\ell }$ , so that $(\mathfrak{p}^{-1}\mathfrak{p}^{\prime })^{\ell }$ is a principal ideal. Thus $(\mathfrak{p}^{\unicode[STIX]{x1D70E}}\mathfrak{p}^{\prime })^{\ell }$ is also a principal ideal, say

(2.3) $$\begin{eqnarray}(\mathfrak{p}^{\unicode[STIX]{x1D70E}}\mathfrak{p}^{\prime })^{\ell }=\biggl(\frac{u+v\sqrt{-d}}{2}\biggr),\end{eqnarray}$$

for some $u,v\in \mathbb{Z}$ . Hence taking norms, it follows that

(2.4) $$\begin{eqnarray}4(pp^{\prime })^{\ell }=u^{2}+dv^{2}.\end{eqnarray}$$

Note that we may require that $\gcd (v,pp^{\prime })=1$ (and in particular that $v\neq 0$ ). For supposing that $p\mid v$ , say, then by (2.4) we see that also $p\mid u$ so that $p\mid ((u+v\sqrt{-d})/2)$ . Hence $\mathfrak{p}\mid ((u+v\sqrt{-d})/2)$ , which by (2.3) implies that $\mathfrak{p}\mid (\mathfrak{p}^{\unicode[STIX]{x1D70E}}\mathfrak{p}^{\prime })^{\ell }$ . Since $p$ is unramified this would then imply that $\mathfrak{p}\mid \mathfrak{p}^{\prime }$ , which contradicts the fact that $p\neq p^{\prime }$ . A similar argument shows that we may require that $u\neq 0$ .

We will show that for all but a small number of ‘exceptional’ $d$ there are many primes $p,p^{\prime }$ that split in this manner, while also showing there can only be few solutions $(u,v)$ to (2.4) with $\gcd (v,pp^{\prime })=1$ and $u,v$ in an appropriate range. This forces there to be many distinct cosets of $H_{\ell }$ in $H$ , and provides an upper bound for $h_{\ell }(-d)$ , as long as $d$ is not exceptional.

We first fix $X\leqslant d<2X$ and count the number of primes $p$ that split appropriately, with

$$\begin{eqnarray}Z\leqslant p<2Z\end{eqnarray}$$

for some parameter $Z$ with ${\textstyle \frac{1}{4}}X^{1/2\ell }\leqslant Z\leqslant X$ (to be chosen precisely in applications). We see that

$$\begin{eqnarray}\#\{Z\leqslant p<2Z:\unicode[STIX]{x1D712}_{d}(p)=1\}=\frac{1}{2}\mathop{\sum }_{Z\leqslant p<2Z}(1+\unicode[STIX]{x1D712}_{d}(p))+O(\unicode[STIX]{x1D714}(d)),\end{eqnarray}$$

where the last term reflects the contribution of the primes that divide $d$ , and contributes no more than $O(\log X)=O(\log Z)$ . We now separate the two terms within the sum over $p$ and apply the prime number theorem, obtaining

$$\begin{eqnarray}\#\{Z\leqslant p<2Z:\unicode[STIX]{x1D712}_{d}(p)=1\}={\textstyle \frac{1}{2}}Z(\log Z)^{-1}+{\textstyle \frac{1}{2}}M(d;Z)+O(Z(\log Z)^{-2}),\end{eqnarray}$$

say, where

$$\begin{eqnarray}M(d;Z)=\mathop{\sum }_{Z\leqslant p<2Z}\unicode[STIX]{x1D712}_{d}(p).\end{eqnarray}$$

Thus the number of split primes in this range is at least of order $Z(\log Z)^{-1}$ , unless we have $|M(d;Z)|\geqslant {\textstyle \frac{1}{4}}Z(\log Z)^{-1}$ ; we will show this exceptional scenario can occur for only a small number of $d$ .

Given a character $\unicode[STIX]{x1D712}$ , set

$$\begin{eqnarray}V(\unicode[STIX]{x1D712})=\biggl(\mathop{\sum }_{Z\leqslant p<2Z}\unicode[STIX]{x1D712}(p)\biggr)^{4\ell }.\end{eqnarray}$$

Upon unfolding the product, we see that this is a character sum of the form

$$\begin{eqnarray}\mathop{\sum }_{Z^{4\ell }\leqslant n<(2Z)^{4\ell }}a_{n}\unicode[STIX]{x1D712}(n)\end{eqnarray}$$

for some coefficients $|a_{n}|\ll d(n)^{4\ell }\ll Z^{\unicode[STIX]{x1D700}}$ . Now we note that with the particular choice $\unicode[STIX]{x1D712}=\unicode[STIX]{x1D712}_{d}$ ,

(2.5) $$\begin{eqnarray}\mathop{\sum }_{X\leqslant d<2X}|M(d;Z)|^{8\ell }=\mathop{\sum }_{X\leqslant d<2X}|V(\unicode[STIX]{x1D712}_{d})|^{2}.\end{eqnarray}$$

By positivity, we can enlarge the sum on the right-hand side of (2.5) to include all primitive characters modulo $d$ and apply the large sieve (see, for example, [Reference DavenportDav00, Theorem 4, ch. 27]), to obtain

(2.6) $$\begin{eqnarray}\displaystyle \mathop{\sum }_{X\leqslant d<2X}|M(d;Z)|^{8\ell } & {\leqslant} & \displaystyle \mathop{\sum }_{X\leqslant d<2X}\;\sum ^{\ast }_{\unicode[STIX]{x1D712}\;(\text{mod}\;d)}|V(\unicode[STIX]{x1D712})|^{2}\nonumber\\ \displaystyle & \ll & \displaystyle (X^{2}+Z^{4\ell })\biggl(\mathop{\sum }_{Z^{4\ell }\leqslant n<(2Z)^{4\ell }}|a_{n}|^{2}\biggr)\nonumber\\ \displaystyle & \ll & \displaystyle Z^{4\ell +2\unicode[STIX]{x1D700}}(X^{2}+Z^{4\ell })\ll Z^{8\ell +2\unicode[STIX]{x1D700}},\end{eqnarray}$$

since $X^{1/2\ell }\ll Z$ by assumption. Let $E(Z;X)$ denote the exceptional set,

(2.7) $$\begin{eqnarray}E(Z;X)=\{X\leqslant d<2X:|M(d;Z)|\geqslant {\textstyle \frac{1}{4}}Z(\log Z)^{-1}\}.\end{eqnarray}$$

Then we may conclude from (2.6) that the exceptional set is small:

$$\begin{eqnarray}\#E(Z;X)\ll X^{\unicode[STIX]{x1D700}},\end{eqnarray}$$

for any $\unicode[STIX]{x1D700}>0$ .

We now fix a $d$ with $X\leqslant d<2X$ such that $d\not \in E(X,Z)$ ; the above argument shows that there are at least of order $Z(\log Z)^{-1}$ split primes for this $d$ . In particular, summing over all cosets of $H_{\ell }$ in $H$ shows that, for this $d$ ,

$$\begin{eqnarray}\displaystyle \mathop{\sum }_{C\in H/H_{\ell }}\#\{Z\leqslant p<2Z:\unicode[STIX]{x1D712}_{d}(p)=1,p=\mathfrak{p}\mathfrak{p}^{\unicode[STIX]{x1D70E}},\mathfrak{p}\in C\}\,=\,\#\{Z\leqslant p<2Z:\unicode[STIX]{x1D712}_{d}(p)=1\}\,\gg \,Z(\log Z)^{-1}. & & \displaystyle \nonumber\end{eqnarray}$$

On the other hand, applying the Cauchy–Schwarz inequality to the left-hand side shows that

(2.8) $$\begin{eqnarray}(\#H/H_{\ell })^{1/2}(S_{\ell }^{(1)}(d;Z))^{1/2}\gg Z(\log Z)^{-1},\end{eqnarray}$$

where we define

$$\begin{eqnarray}S_{\ell }^{(1)}(d;Z)=\mathop{\sum }_{C\in H/H_{\ell }}\#\{Z\leqslant p<2Z:\unicode[STIX]{x1D712}_{d}(p)=1,p=\mathfrak{p}\mathfrak{p}^{\unicode[STIX]{x1D70E}},\mathfrak{p}\in C\}^{2}.\end{eqnarray}$$

By the above discussion, we know that

(2.9) $$\begin{eqnarray}S_{\ell }^{(1)}(d;Z)\ll \#\{Z\leqslant p,p^{\prime }<2Z:4(pp^{\prime })^{\ell }=u^{2}+dv^{2}\;\text{for some }u,v\in \mathbb{Z}\},\end{eqnarray}$$

where in the case that $p\neq p^{\prime }$ we may impose the additional conditions that $u,v\neq 0$ and $(v,pp^{\prime })=1$ . Combining (2.8) and (2.2), we may conclude that

$$\begin{eqnarray}h_{\ell }(-d)\ll d^{1/2+\unicode[STIX]{x1D700}}Z^{-2}(\log Z)^{2}S_{\ell }^{(1)}(d;Z),\end{eqnarray}$$

still under the assumption that $d$ is not exceptional. Finally, we write

$$\begin{eqnarray}S_{\ell }^{(1)}(d;Z)=S_{\ell }^{(0)}(d;Z)+S_{\ell }(d;Z),\end{eqnarray}$$

where $S_{\ell }^{(0)}(d;Z)$ is the contribution to the set (2.9) from pairs $p=p^{\prime }$ and $S_{\ell }(d;Z)$ is the contribution from pairs $p\neq p^{\prime }$ . Trivially, $S_{\ell }^{(0)}(d;Z)\ll Z$ , and we see that Proposition 2.1 holds.

To deduce the corollary we take $Z={\textstyle \frac{1}{4}}X^{1/2\ell }$ , and note that any pairs of primes $p,p^{\prime }$ counted by $S_{\ell }(d;Z)$ would satisfy

$$\begin{eqnarray}X\leqslant d\leqslant u^{2}+dv^{2}=4(pp^{\prime })^{\ell }\leqslant 4(4Z^{2})^{\ell }=4^{1-\ell }X<X.\end{eqnarray}$$

Thus $S_{\ell }(d;Z)$ must vanish, so that $h_{\ell }(-d)\ll X^{\unicode[STIX]{x1D700}}d^{1/2}Z^{-1}\ll d^{1/2-1/(2\ell )+\unicode[STIX]{x1D700}}$ unless $d$ lies in $E(Z;X)$ . The result then follows.

4.1 Proof of Proposition 4.1: a first bound for $T_{\ell }$

We now prove (4.3). We recall the parameters $U,V,W$ of (3.1) and note that $T_{\ell }$ is at most the number of 6-tuples $(w_{1},w_{2},u_{1},u_{2},v_{1},v_{2})$ in the ranges

$$\begin{eqnarray}W\leqslant w_{1},w_{2}<4W,\quad 1\leqslant u_{1},u_{2}\leqslant U,\quad 1\leqslant v_{1},v_{2}\leqslant V\end{eqnarray}$$

that satisfy the conditions

(4.5) $$\begin{eqnarray}\displaystyle & (u_{1},v_{1},w_{1})\neq (u_{2},v_{2},w_{2}), & \displaystyle\end{eqnarray}$$

(4.6) $$\begin{eqnarray}\displaystyle & \gcd (w_{1},v_{1})=\gcd (w_{2},v_{2})=1, & \displaystyle\end{eqnarray}$$

(4.7) $$\begin{eqnarray}\displaystyle & v_{1}^{2}\mid (4w_{1}^{\ell }-u_{1}^{2})\quad \text{and}\quad v_{2}^{2}\mid (4w_{2}^{\ell }-u_{2}^{2}), & \displaystyle\end{eqnarray}$$

(4.8) $$\begin{eqnarray}\displaystyle & v_{1}^{2}(4w_{2}^{\ell }-u_{2}^{2})=v_{2}^{2}(4w_{1}^{\ell }-u_{1}^{2})\neq 0. & \displaystyle\end{eqnarray}$$

We will obtain a first upper bound for $T_{\ell }$ by following the approach of [Reference SoundararajanSou00], ignoring the divisibility conditions (4.7); note that we are also now ignoring the fact that each of $w_{1},w_{2}$ is a product of two distinct primes. We claim that for tuples satisfying the above conditions,

(4.9) $$\begin{eqnarray}v_{1}^{2}w_{2}^{\ell }-v_{2}^{2}w_{1}^{\ell }\neq 0.\end{eqnarray}$$

To prove this we recall that $\gcd (w_{i},v_{i})=1$ for $i=1,2$ , whence $v_{1}^{2}w_{2}^{\ell }=v_{2}^{2}w_{1}^{\ell }$ would imply that $v_{1}=v_{2}$ and $w_{1}=w_{2}$ , and hence $u_{1}=u_{2}$ . This would then contradict (4.5).

We now observe that once $v_{1},v_{2},w_{1},w_{2}$ are fixed then $u_{1},u_{2}$ are fixed up to $X^{\unicode[STIX]{x1D700}}$ choices. For indeed, fixing $v_{1},v_{2},w_{1},w_{2}$ in (4.8) gives

(4.10) $$\begin{eqnarray}4(v_{2}^{2}w_{1}^{\ell }-v_{1}^{2}w_{2}^{\ell })=(v_{2}u_{1}-v_{1}u_{2})(v_{2}u_{1}+v_{1}u_{2}).\end{eqnarray}$$

The left-hand side is a nonzero integer by (4.9), so that $u_{1},u_{2}$ are fixed up to $X^{\unicode[STIX]{x1D700}}$ choices. Thus we obtain

$$\begin{eqnarray}T_{\ell }\ll W^{2}V^{2}X^{\unicode[STIX]{x1D700}}\ll Z^{2\ell +4}X^{-1+\unicode[STIX]{x1D700}},\end{eqnarray}$$

which is the bound given in (4.3).

4.2 Proof of Proposition 4.1: a second bound for $T_{\ell }$

We may obtain the alternative upper bound (4.4) for $T_{\ell }$ by following the method of [Reference Heath-BrownHB07], but with the addition of certain technical considerations because in the present case the variables $v_{i}$ are not restricted to be primes. Although it is easy enough to do this for general odd primes $\ell$ we shall confine our attention to $\ell =3$ , since this is the only case we shall use.

First we consider the contribution to $T_{3}$ arising from the case in which $\gcd (w_{1},w_{2})\not =1$ . We write $T_{3}^{0}$ for the number of 6-tuples of this type. Since each of $w_{1}$ and $w_{2}$ is a product of two primes in the interval $[Z,2Z)$ this can happen only when there is at least one prime $p\in [Z,2Z)$ dividing both of $w_{1}$ and $w_{2}$ . The number of possible pairs $w_{1},w_{2}$ is thus $O(Z^{3})$ . We now follow the argument of § 4.1. There are $O(V^{2})$ pairs $v_{1},v_{2}$ , and the factorization (4.10) shows that there are $O(X^{\unicode[STIX]{x1D700}})$ possibilities for $u_{1},u_{2}$ once $w_{1},w_{2},v_{1},v_{2}$ are fixed. It follows that

$$\begin{eqnarray}T_{3}^{0}\ll Z^{3}V^{2}X^{\unicode[STIX]{x1D700}}.\end{eqnarray}$$

From now on we assume that $\gcd (w_{1},w_{2})=1$ . For each integer $1\leqslant \unicode[STIX]{x1D6FF}\leqslant V$ , we will let $T_{3}(\unicode[STIX]{x1D6FF})$ denote the contribution to $T_{3}$ from triples $(u_{1},v_{1},w_{1})$ and $(u_{2},v_{2},w_{2})$ with $w_{1},w_{2}$ coprime, such that $\gcd (v_{1},v_{2})=\unicode[STIX]{x1D6FF}$ . We will prove the following proposition.

Proposition 4.2. For each integer $1\leqslant \unicode[STIX]{x1D6FF}\leqslant V$ ,

$$\begin{eqnarray}T_{3}(\unicode[STIX]{x1D6FF})\ll X^{\unicode[STIX]{x1D700}}(W^{2}V^{2/3}\unicode[STIX]{x1D6FF}^{-2/3}+V^{3}U\unicode[STIX]{x1D6FF}^{-3}+WV\unicode[STIX]{x1D6FF}^{-1}).\end{eqnarray}$$

From this we conclude that

$$\begin{eqnarray}\displaystyle T_{3} & \ll & \displaystyle T_{3}^{0}+\mathop{\sum }_{\unicode[STIX]{x1D6FF}=1}^{V}T_{3}(\unicode[STIX]{x1D6FF})\nonumber\\ \displaystyle & \ll & \displaystyle X^{\unicode[STIX]{x1D700}}\biggl\{Z^{3}V^{2}+\mathop{\sum }_{\unicode[STIX]{x1D6FF}=1}^{V}(W^{2}V^{2/3}\unicode[STIX]{x1D6FF}^{-2/3}+V^{3}U\unicode[STIX]{x1D6FF}^{-3}+WV\unicode[STIX]{x1D6FF}^{-1})\biggr\}\nonumber\\ \displaystyle & \ll & \displaystyle X^{\unicode[STIX]{x1D700}}(Z^{3}V^{2}+VW^{2}+V^{3}U+WV)\nonumber\\ \displaystyle & \ll & \displaystyle X^{\unicode[STIX]{x1D700}}(Z^{3}V^{2}+VW^{2}+V^{3}U),\nonumber\end{eqnarray}$$

since clearly $WV\ll VW^{2}$ . Upon recalling the parameter definitions (3.1) this shows that

$$\begin{eqnarray}T_{3}\ll X^{\unicode[STIX]{x1D700}}\{Z^{9}X^{-1}+Z^{7}X^{-1/2}+Z^{12}X^{-3/2}\}.\end{eqnarray}$$

This provides the second bound for $T_{3}$ given in Proposition 4.1, since $Z\geqslant X^{1/6}$ .

Proof of Proposition 4.2 . To prove Proposition 4.2, we fix $\unicode[STIX]{x1D6FF}$ and write $v_{i}=\unicode[STIX]{x1D6FF}y_{i}$ for $i=1,2$ so that $\gcd (y_{1},y_{2})=1$ . We first isolate solutions $(u_{1},v_{1},w_{1})$ and $(u_{2},v_{2},w_{2})$ that contribute to $T_{3}(\unicode[STIX]{x1D6FF})$ such that $y_{1},y_{2}$ satisfy a relation

(4.11) $$\begin{eqnarray}y_{1}^{2}\unicode[STIX]{x1D707}_{2}^{3}=y_{2}^{2}\unicode[STIX]{x1D707}_{1}^{3}\end{eqnarray}$$

for some integers $\unicode[STIX]{x1D707}_{1},\unicode[STIX]{x1D707}_{2}$ . Given a relation of the form (4.11), we may divide both sides by $\gcd (\unicode[STIX]{x1D707}_{1},\unicode[STIX]{x1D707}_{2})^{3}$ to obtain an equivalent relation

$$\begin{eqnarray}y_{1}^{2}\unicode[STIX]{x1D706}_{2}^{3}=y_{2}^{2}\unicode[STIX]{x1D706}_{1}^{3}\end{eqnarray}$$

in which $(\unicode[STIX]{x1D706}_{1},\unicode[STIX]{x1D706}_{2})=1$ and $(y_{1},y_{2})=1$ . This implies that for each $i=1,2$ ,

(4.12) $$\begin{eqnarray}y_{i}^{2}=\unicode[STIX]{x1D706}_{i}^{3}.\end{eqnarray}$$

This implies that $y_{i}$ is itself a perfect cube, say $y_{i}=s_{i}^{3}$ . We recall from (4.10) that once $v_{1},v_{2},w_{1},w_{2}$ are fixed, $u_{1},u_{2}$ are fixed up to $X^{\unicode[STIX]{x1D700}}$ choices. Thus we count how many $v_{1},v_{2}\leqslant V$ with $\gcd (v_{1},v_{2})=\unicode[STIX]{x1D6FF}$ are of the type (4.12) by noting that there are at most $O((V\unicode[STIX]{x1D6FF}^{-1})^{1/3})$ choices for each $s_{i}$ . We bound the number of choices for $w_{1},w_{2}$ trivially by $O(W^{2})$ , and conclude that the contribution to $T_{3}(\unicode[STIX]{x1D6FF})$ of solutions for which a relation of the form (4.11) holds is at most

(4.13) $$\begin{eqnarray}\ll W^{2}V^{2/3}\unicode[STIX]{x1D6FF}^{-2/3}X^{\unicode[STIX]{x1D700}}.\end{eqnarray}$$

We now proceed to count the remaining contribution to $T_{3}(\unicode[STIX]{x1D6FF})$ ; we may assume from now on that no relation of the form (4.11) holds for $y_{1}$ and $y_{2}$ . Define

(4.14) $$\begin{eqnarray}k=y_{2}u_{1}+y_{1}u_{2}.\end{eqnarray}$$

Note that if $\unicode[STIX]{x1D6FF},w_{1},w_{2},y_{1},y_{2}$ and $k$ are fixed, then $u_{1},u_{2}$ are fixed uniquely by (4.10). Thus we will count the number of solutions $w_{1},w_{2}$ contributing to $T_{3}(\unicode[STIX]{x1D6FF})$ for each fixed $y_{1},y_{2},k$ .

Recalling the definition of $y_{1},y_{2}$ we see that the condition (4.8) now becomes

$$\begin{eqnarray}y_{1}^{2}(4w_{2}^{\ell }-u_{2}^{2})=y_{2}^{2}(4w_{1}^{\ell }-u_{1}^{2})\neq 0,\end{eqnarray}$$

and since $\gcd (y_{1},y_{2})=1$ , this implies a system of congruences

(4.15) $$\begin{eqnarray}\displaystyle 4y_{2}^{2}w_{1}^{3} & \equiv & \displaystyle k^{2}\;(\text{mod}\;y_{1}),\end{eqnarray}$$

(4.16) $$\begin{eqnarray}\displaystyle 4y_{1}^{2}w_{2}^{3} & \equiv & \displaystyle k^{2}\;(\text{mod}\;y_{2}),\end{eqnarray}$$

(4.17) $$\begin{eqnarray}\displaystyle 4y_{2}^{2}w_{1}^{3} & \equiv & \displaystyle 4y_{1}^{2}w_{2}^{3}\;(\text{mod}\;k).\end{eqnarray}$$

We first reduce this to a similar system of congruences with square-free moduli. For $i=1,2$ let $q_{i}$ denote the odd square-free kernel of $y_{i}$ , that is

$$\begin{eqnarray}q_{i}=\mathop{\prod }_{\substack{ p\mid y_{i} \\ p>2}}p.\end{eqnarray}$$

The congruence (4.15) implies that $4y_{2}^{2}w_{1}^{3}\equiv k^{2}\;(\text{mod}\;q_{1})$ . Since $(4y_{2},q_{1})=1$ this congruence may be re-written as $w_{1}^{3}\equiv a_{1}\;(\text{mod}\;q_{1})$ for some constant $a_{1}$ determined by $y_{2}$ and $k$ . A similar observation applies to (4.16). Next, we define

$$\begin{eqnarray}r=\mathop{\prod }_{\substack{ p\mid k \\ p>2}}p\end{eqnarray}$$

to be the odd square-free kernel of $k$ , and deduce from (4.17) an analogous congruence modulo $r$ . It follows that any solutions $w_{1},w_{2}$ of the system (4.15)–(4.17) must satisfy the congruences

(4.18) $$\begin{eqnarray}\displaystyle w_{1}^{3} & \equiv & \displaystyle a_{1}\;(\text{mod}\;q_{1}),\end{eqnarray}$$

(4.19) $$\begin{eqnarray}\displaystyle w_{2}^{3} & \equiv & \displaystyle a_{2}\;(\text{mod}\;q_{2}),\end{eqnarray}$$

(4.20) $$\begin{eqnarray}\displaystyle y_{2}^{2}w_{1}^{3} & \equiv & \displaystyle y_{1}^{2}w_{2}^{3}\;(\text{mod}\;r)\end{eqnarray}$$

for some constant $a_{1}$ determined by $y_{2},k\;(\text{mod}\;q_{1})$ and some constant $a_{2}$ determined by $y_{1},k\;(\text{mod}\;q_{2})$ .

Certainly $(q_{1},q_{2})=1$ . In addition, we note that $(y_{1},r)=1$ and $(y_{2},r)=1$ . For indeed, if some odd prime $p$ satisfies $p\mid k$ and $p\mid y_{1}$ , then by (4.14) it follows that $p\mid u_{1}$ , since by construction $(y_{1},y_{2})=1$ . However, by the condition $v_{1}^{2}\mid (4w_{1}^{3}-u_{1}^{2})$ , this would imply that $p\mid w_{1}$ , which contradicts the fact that $(v_{1},w_{1})=1$ . The fact that $(y_{2},r)=1$ may be shown similarly. As a consequence of these observations,

(4.21) $$\begin{eqnarray}(q_{1},q_{2})=1,\quad (q_{1},r)=1,\quad (q_{2},r)=1.\end{eqnarray}$$

The next step is to note that the conditions (4.18)–(4.20) may be interpreted as lattice conditions.

Proof. To prove this, we first consider the congruence (4.18). Fix a prime divisor $p\mid q_{1}$ ; then $w_{1}$ can only be a solution to (4.18) if

(4.22) $$\begin{eqnarray}w_{1}^{3}\equiv a_{1}\;(\text{mod}\;p).\end{eqnarray}$$

There are at most three residue classes modulo $p$ in which a solution $w_{1}$ to (4.22) may lie. We may conclude that $w_{1}$ lies in one of at most $3^{\unicode[STIX]{x1D714}(q_{1})}$ residue classes modulo $q_{1}$ . A similar argument applies to (4.19), establishing that $w_{2}$ may lie in at most $3^{\unicode[STIX]{x1D714}(q_{2})}$ residue classes modulo $q_{2}$ .

We now turn to (4.20). Since $(y_{1},r)=1$ and $(w_{1},w_{2})=1$ we must have $(w_{1},r)=1$ . (Indeed, otherwise, if we suppose $p$ is a prime factor of both $w_{1}$ and $r$ , we would see in (4.20) that $p\mid y_{1}^{2}w_{2}^{3}$ , but since $(w_{1},w_{2})=1$ we cannot have $p\mid w_{2}$ , so we would conclude $p\mid y_{1}$ . This would in turn contradict that fact we previously proved that $(y_{1},r)=1$ .) Using the fact that $(w_{1},r)=1$ we see that (4.20) implies

(4.23) $$\begin{eqnarray}w^{3}\equiv a\;(\text{mod}\;r),\end{eqnarray}$$

where $w\equiv w_{2}w_{1}^{-1}\;(\text{mod}\;r)$ and $a\equiv (y_{2}y_{1}^{-1})^{2}\;(\text{mod}\;r)$ is coprime to $r$ . Now, just as with our analysis of (4.18), we see that there is a collection of at most $3^{\unicode[STIX]{x1D714}(r)}$ residue classes $w\equiv b_{i}\;(\text{mod}\;r)$ in which $w$ must lie. This leads to a corresponding collection of lattice conditions $w_{2}\equiv b_{i}w_{1}\;(\text{mod}\;r)$ which, taken together, are equivalent to (4.23). Finally we note that the resulting lattice of pairs $(w_{1},w_{2})$ has a basis $\{(1,b_{i}),(0,r)\}$ , so that its determinant is just $r$ . This completes the proof of the lemma.◻

4.3 Counting lattice points

Since $q_{1},q_{2},r$ are coprime in pairs, we may conclude from Lemma 4.3 that $(w_{1},w_{2})$ must lie in one of at most $3^{\unicode[STIX]{x1D714}(q_{1})+\unicode[STIX]{x1D714}(q_{2})+\unicode[STIX]{x1D714}(r)}$ lattice cosets of the form $(c_{1},c_{2})+\unicode[STIX]{x1D6EC}$ , where $\unicode[STIX]{x1D6EC}$ is a lattice with $\det (\unicode[STIX]{x1D6EC})=q_{1}q_{2}r$ . We note that the total number of lattices is $\ll X^{\unicode[STIX]{x1D700}}$ , since under the assumption $Z\leqslant X$ , we have $v_{i}\leqslant V\ll X^{5/2}$ and $k\leqslant 2UV\ll X^{11/2}$ . We now fix one of these lattices, which we will denote by $\unicode[STIX]{x1D6EC}$ , and its corresponding shift $(c_{1},c_{2})$ . Note that we may choose $(c_{1},c_{2})$ such that $W\leqslant c_{i}<4W$ for $i=1,2$ , since otherwise $w_{1},w_{2}$ would lie outside the desired range $W\leqslant w_{1},w_{2}<4W$ . We now write $(z_{1},z_{2})=(w_{1},w_{2})-(c_{1},c_{2})$ , and proceed to count the number of

$$\begin{eqnarray}(z_{1},z_{2})\in \unicode[STIX]{x1D6EC},\quad |z_{i}|<3W.\end{eqnarray}$$

Let $\unicode[STIX]{x1D706}_{1}\leqslant \unicode[STIX]{x1D706}_{2}$ be the successive minima of $\unicode[STIX]{x1D6EC}$ , so that the standard Minkowski inequalities show that $\det (\unicode[STIX]{x1D6EC})\ll \unicode[STIX]{x1D706}_{1}\unicode[STIX]{x1D706}_{2}\ll \det (\unicode[STIX]{x1D6EC})$ (see, for example, Davenport [Reference DavenportDav58, Eqn (5)]). We note that in our particular case,

(4.24) $$\begin{eqnarray}\unicode[STIX]{x1D706}_{1}\ll \sqrt{\det (\unicode[STIX]{x1D6EC})}\ll \sqrt{q_{1}q_{2}r}\ll V^{3/2}U^{1/2}\unicode[STIX]{x1D6FF}^{-3/2}.\end{eqnarray}$$

Here we have used the fact that $q_{i}\leqslant y_{i}\leqslant V\unicode[STIX]{x1D6FF}^{-1}$ for $i=1,2$ and hence $r\leqslant k\ll UV\unicode[STIX]{x1D6FF}^{-1}$ . Moreover, by Lemma 1 of Davenport [Reference DavenportDav58], the number of lattice points in $\unicode[STIX]{x1D6EC}$ with $|(z_{1},z_{2})|\leqslant x$ is (up to a constant) at most $(1+x/\unicode[STIX]{x1D706}_{1})(1+x/\unicode[STIX]{x1D706}_{2}).$ Thus the number of allowable $z_{1},z_{2}$ in our case is

$$\begin{eqnarray}\displaystyle & & \displaystyle \ll (1+W/\unicode[STIX]{x1D706}_{1})(1+W/\unicode[STIX]{x1D706}_{2})\nonumber\\ \displaystyle & & \displaystyle \ll 1+W^{2}/\det (\unicode[STIX]{x1D6EC})+W/\unicode[STIX]{x1D706}_{1}\nonumber\\ \displaystyle & & \displaystyle \ll 1+W^{2}/(q_{1}q_{2}r)+W/\unicode[STIX]{x1D706}_{1}.\nonumber\end{eqnarray}$$

Thus we have

(4.25) $$\begin{eqnarray}T_{3}(\unicode[STIX]{x1D6FF})\ll X^{\unicode[STIX]{x1D700}}\mathop{\sum }_{y_{1},y_{2},k}\biggl(1+\frac{W^{2}}{q_{1}q_{2}r}+\frac{W}{\unicode[STIX]{x1D706}_{1}}\biggr),\end{eqnarray}$$

where we recall that $q_{i}$ is the odd square-free kernel of $y_{i}$ and for each triple $y_{1},y_{2},k$ we take $\unicode[STIX]{x1D706}_{1}$ to be the smallest value from all the corresponding lattices $\unicode[STIX]{x1D6EC}$ . Recall that $y_{1},y_{2}\leqslant V\unicode[STIX]{x1D6FF}^{-1}$ and $k\leqslant 2UV\unicode[STIX]{x1D6FF}^{-1}$ . Then we see that the contribution of the first term in (4.25) to $T_{3}(\unicode[STIX]{x1D6FF})$ is at most

(4.26) $$\begin{eqnarray}\ll X^{\unicode[STIX]{x1D700}}V^{3}U\unicode[STIX]{x1D6FF}^{-3}.\end{eqnarray}$$

The contribution to $T_{3}(\unicode[STIX]{x1D6FF})$ from the second term in (4.25) is

(4.27) $$\begin{eqnarray}\ll X^{\unicode[STIX]{x1D700}}W^{2}\biggl(\mathop{\sum }_{y_{1}\leqslant V\unicode[STIX]{x1D6FF}^{-1}}\frac{1}{q_{1}}\biggr)\biggl(\mathop{\sum }_{y_{2}\leqslant V\unicode[STIX]{x1D6FF}^{-1}}\frac{1}{q_{2}}\biggr)\biggl(\mathop{\sum }_{k\leqslant 2UV\unicode[STIX]{x1D6FF}^{-1}}\frac{1}{r}\biggr).\end{eqnarray}$$

To bound each internal sum we apply the following minor variant of [Reference Heath-BrownHB07, Lemma 1].

Lemma 4.4. Given an integer $k$ , let $k^{\ast }$ denote its odd square-free kernel. For any fixed integer $\unicode[STIX]{x1D705}\leqslant K$ ,

$$\begin{eqnarray}\#\{k\leqslant K:k^{\ast }=\unicode[STIX]{x1D705}\}\ll K^{\unicode[STIX]{x1D700}}.\end{eqnarray}$$

We defer the proof of this lemma until § 4.4, and merely apply it now to (4.27); for example the first sum is bounded by

$$\begin{eqnarray}\mathop{\sum }_{y_{1}\leqslant V\unicode[STIX]{x1D6FF}^{-1}}\frac{1}{q_{1}}\leqslant \mathop{\sum }_{\unicode[STIX]{x1D708}\leqslant V\unicode[STIX]{x1D6FF}^{-1}}\frac{1}{\unicode[STIX]{x1D708}}\#\{v\leqslant V\unicode[STIX]{x1D6FF}^{-1}:v^{\ast }=\unicode[STIX]{x1D708}\}\ll V^{\unicode[STIX]{x1D700}}\mathop{\sum }_{\unicode[STIX]{x1D708}\leqslant V\unicode[STIX]{x1D6FF}^{-1}}\frac{1}{\unicode[STIX]{x1D708}}\ll V^{\unicode[STIX]{x1D700}}.\end{eqnarray}$$

One may handle the second and third sums in (4.27) similarly, and deduce that the second term in (4.25) is $O(W^{2}X^{\unicode[STIX]{x1D700}})$ overall. Since $W^{2}\leqslant W^{2}V^{2/3}\unicode[STIX]{x1D6FF}^{-2/3}$ for $\unicode[STIX]{x1D6FF}\leqslant V$ we see that this error is dominated by (4.13).

Finally, the contribution, say $T_{3}^{\prime }(\unicode[STIX]{x1D6FF})$ , of the third term in (4.25) may be bounded by following the same argument as in [Reference Heath-BrownHB07], which we sketch for completeness. For each triple $y_{1},y_{2},k$ , let $\unicode[STIX]{x1D6EC}$ be the lattice to which $\unicode[STIX]{x1D706}_{1}$ corresponds, and let $(\unicode[STIX]{x1D707}_{1},\unicode[STIX]{x1D707}_{2})$ be the shortest nonzero vector in $\unicode[STIX]{x1D6EC}$ , so that $\unicode[STIX]{x1D706}_{1}$ is the length of $(\unicode[STIX]{x1D707}_{1},\unicode[STIX]{x1D707}_{2})$ . Then

$$\begin{eqnarray}T_{3}^{\prime }(\unicode[STIX]{x1D6FF})\ll X^{\unicode[STIX]{x1D700}}W\mathop{\sum }_{\unicode[STIX]{x1D707}_{1},\unicode[STIX]{x1D707}_{2}}\frac{\#\{y_{1},y_{2},k\}}{\sqrt{|\unicode[STIX]{x1D707}_{1}|^{2}+|\unicode[STIX]{x1D707}_{2}|^{2}}},\end{eqnarray}$$

where we count the number of $y_{1},y_{2},k$ that generate a lattice in which $(\unicode[STIX]{x1D707}_{1},\unicode[STIX]{x1D707}_{2})$ is a vector of minimal length. We note by (4.24) that

(4.28) $$\begin{eqnarray}\unicode[STIX]{x1D707}_{1},\unicode[STIX]{x1D707}_{2}\ll V^{3/2}U^{1/2}\unicode[STIX]{x1D6FF}^{-3/2}.\end{eqnarray}$$

Since $(\unicode[STIX]{x1D707}_{1},\unicode[STIX]{x1D707}_{2})$ lies in the lattice $\unicode[STIX]{x1D6EC}$ , then by construction

(4.29) $$\begin{eqnarray}q_{1}\mid \unicode[STIX]{x1D707}_{1},\quad q_{2}\mid \unicode[STIX]{x1D707}_{2}\end{eqnarray}$$

and

(4.30) $$\begin{eqnarray}r\mid (y_{2}^{2}\unicode[STIX]{x1D707}_{1}^{3}-y_{1}^{2}\unicode[STIX]{x1D707}_{2}^{3}),\end{eqnarray}$$

as described in Lemma 4.3.

We first consider the case where both $\unicode[STIX]{x1D707}_{1},\unicode[STIX]{x1D707}_{2}$ are nonzero. By (4.29), once $\unicode[STIX]{x1D707}_{1},\unicode[STIX]{x1D707}_{2}$ are fixed, they determine at most $X^{\unicode[STIX]{x1D700}}$ values of $q_{1},q_{2}$ and hence at most $X^{\unicode[STIX]{x1D700}}$ values for $y_{1},y_{2}$ by Lemma 4.4. If $y_{2}^{2}\unicode[STIX]{x1D707}_{1}^{3}-y_{1}^{2}\unicode[STIX]{x1D707}_{2}^{3}$ is nonzero, then it determines at most $X^{\unicode[STIX]{x1D700}}$ values for $r$ by (4.30) and hence at most $X^{\unicode[STIX]{x1D700}}$ values for $k$ . On the other hand, if

(4.31) $$\begin{eqnarray}y_{2}^{2}\unicode[STIX]{x1D707}_{1}^{3}=y_{1}^{2}\unicode[STIX]{x1D707}_{2}^{3},\end{eqnarray}$$

then $y_{1},y_{2}$ would satisfy a relation of the form (4.11); pairs $y_{1},y_{2}$ of this type have already been treated, and are excluded from the contribution we are currently calculating. We therefore see that the contribution to $T_{3}^{\prime }(\unicode[STIX]{x1D6FF})$ from $\unicode[STIX]{x1D707}_{1},\unicode[STIX]{x1D707}_{2}$ both nonzero is

$$\begin{eqnarray}T_{3}^{\prime }(\unicode[STIX]{x1D6FF})\ll X^{4\unicode[STIX]{x1D700}}W\mathop{\sum }_{\unicode[STIX]{x1D707}_{1},\unicode[STIX]{x1D707}_{2}}\frac{1}{\sqrt{|\unicode[STIX]{x1D707}_{1}|^{2}+|\unicode[STIX]{x1D707}_{2}|^{2}}}.\end{eqnarray}$$

To bound the sum, we begin by focusing on a fixed dyadic range

$$\begin{eqnarray}{\textstyle \frac{1}{2}}B<\sqrt{|\unicode[STIX]{x1D707}_{1}|^{2}+|\unicode[STIX]{x1D707}_{2}|^{2}}\leqslant B,\end{eqnarray}$$

for any appropriate $B\geqslant 1$ ; we note that the restriction (4.28) implies that $B\ll V^{3/2}U^{1/2}\unicode[STIX]{x1D6FF}^{-3/2}.$ There are $O(B^{2})$ pairs $\unicode[STIX]{x1D707}_{1},\unicode[STIX]{x1D707}_{2}$ , each of which contribute $O(B^{-1})$ to the sum. Summing over dyadic $B\ll V^{3/2}U^{1/2}\unicode[STIX]{x1D6FF}^{-3/2}$ therefore produces a total contribution of $\ll X^{\unicode[STIX]{x1D700}}WV^{3/2}U^{1/2}\unicode[STIX]{x1D6FF}^{-3/2}$ to $T_{3}^{\prime }(\unicode[STIX]{x1D6FF})$ .

On the other hand if $\unicode[STIX]{x1D707}_{1}$ vanishes, then there are $V\unicode[STIX]{x1D6FF}^{-1}$ choices for $y_{1}$ and $O(X^{2\unicode[STIX]{x1D700}})$ choices for $q_{2},r$ , hence $O(X^{4\unicode[STIX]{x1D700}})$ choices for $y_{2},k$ . (In particular, (4.31) cannot occur, since it would force $\unicode[STIX]{x1D707}_{1}=\unicode[STIX]{x1D707}_{2}=0$ .) Thus the contribution from these terms to $T_{3}^{\prime }(\unicode[STIX]{x1D6FF})$ is

$$\begin{eqnarray}\ll X^{5\unicode[STIX]{x1D700}}VW\unicode[STIX]{x1D6FF}^{-1}\mathop{\sum }_{\unicode[STIX]{x1D707}_{2}\ll V^{3/2}U^{1/2}\unicode[STIX]{x1D6FF}^{-3/2}}\frac{1}{|\unicode[STIX]{x1D707}_{2}|}\ll X^{6\unicode[STIX]{x1D700}}VW\unicode[STIX]{x1D6FF}^{-1}.\end{eqnarray}$$

The case where $\unicode[STIX]{x1D707}_{2}$ vanishes may be treated by an analogous argument. We may conclude that

$$\begin{eqnarray}T_{3}^{\prime }(\unicode[STIX]{x1D6FF})\ll X^{\unicode[STIX]{x1D700}}(WV^{3/2}U^{1/2}\unicode[STIX]{x1D6FF}^{-3/2}+VW\unicode[STIX]{x1D6FF}^{-1}).\end{eqnarray}$$

Combining this with the contributions (4.13) and (4.26) shows that

$$\begin{eqnarray}T_{3}(\unicode[STIX]{x1D6FF})\ll X^{\unicode[STIX]{x1D700}}(W^{2}V^{2/3}\unicode[STIX]{x1D6FF}^{-2/3}+V^{3}U\unicode[STIX]{x1D6FF}^{-3}+WV^{3/2}U^{1/2}\unicode[STIX]{x1D6FF}^{-3/2}+VW\unicode[STIX]{x1D6FF}^{-1}).\end{eqnarray}$$

Since

$$\begin{eqnarray}\displaystyle WV^{3/2}U^{1/2}\unicode[STIX]{x1D6FF}^{-3/2} & = & \displaystyle \{W^{2}\}^{1/2}\{V^{3}U\unicode[STIX]{x1D6FF}^{-3}\}^{1/2}\nonumber\\ \displaystyle & {\leqslant} & \displaystyle \{W^{2}V^{2/3}\unicode[STIX]{x1D6FF}^{-2/3}\}^{1/2}\{V^{3}U\unicode[STIX]{x1D6FF}^{-3}\}^{1/2}\nonumber\end{eqnarray}$$

for $\unicode[STIX]{x1D6FF}\leqslant V$ , the third term above is dominated by the first two, so that Proposition 4.2 follows.◻

4.4 Proof of Lemma 4.4

We now prove Lemma 4.4, in the following more general form. Given any finite set ${\mathcal{P}}$ of primes (possibly empty), let

$$\begin{eqnarray}k({\mathcal{P}})=\mathop{\prod }_{\substack{ p\mid k \\ p\not \in {\mathcal{P}}}}p.\end{eqnarray}$$

Consider the set $\{k\leqslant K:k({\mathcal{P}})=\unicode[STIX]{x1D705}\}$ for a fixed positive integer $\unicode[STIX]{x1D705}$ . The set is empty unless $\unicode[STIX]{x1D705}\leqslant K$ is square-free and satisfies $(\unicode[STIX]{x1D705},\prod _{p\in {\mathcal{P}}}p)=1$ , which we now assume. Then for any $\unicode[STIX]{x1D702}>0$ ,

$$\begin{eqnarray}\displaystyle \#\{k\leqslant K:k({\mathcal{P}})=\unicode[STIX]{x1D705}\} & {\leqslant} & \displaystyle \mathop{\sum }_{\substack{ k=1 \\ k({\mathcal{P}})=\unicode[STIX]{x1D705}}}^{K}\biggl(\frac{K}{k}\biggr)^{\unicode[STIX]{x1D702}}\nonumber\\ \displaystyle & {\leqslant} & \displaystyle K^{\unicode[STIX]{x1D702}}\mathop{\sum }_{\substack{ k=1 \\ k({\mathcal{P}})=\unicode[STIX]{x1D705}}}^{\infty }k^{-\unicode[STIX]{x1D702}}\nonumber\\ \displaystyle & = & \displaystyle K^{\unicode[STIX]{x1D702}}\mathop{\prod }_{p\in {\mathcal{P}}}\biggl(\mathop{\sum }_{e=0}^{\infty }p^{-e\unicode[STIX]{x1D702}}\biggr)\mathop{\prod }_{p\mid \unicode[STIX]{x1D705}}\biggl(\mathop{\sum }_{e=1}^{\infty }p^{-e\unicode[STIX]{x1D702}}\biggr).\nonumber\end{eqnarray}$$

Setting $A(\unicode[STIX]{x1D702})=\sum _{e=0}^{\infty }2^{-e\unicode[STIX]{x1D702}}$ we then see that

$$\begin{eqnarray}\#\{k\leqslant K:k({\mathcal{P}})=\unicode[STIX]{x1D705}\}\leqslant K^{\unicode[STIX]{x1D702}}A(\unicode[STIX]{x1D702})^{\unicode[STIX]{x1D714}(\unicode[STIX]{x1D705})+\#{\mathcal{P}}}\leqslant K^{\unicode[STIX]{x1D702}}A(\unicode[STIX]{x1D702})^{(\#{\mathcal{P}}+1)\unicode[STIX]{x1D714}(\unicode[STIX]{x1D705})}.\end{eqnarray}$$

Upon recalling that $\unicode[STIX]{x1D714}(\unicode[STIX]{x1D705})\ll (\log 3\unicode[STIX]{x1D705})(\log \log 3\unicode[STIX]{x1D705})^{-1}$ and $\unicode[STIX]{x1D705}\leqslant K$ we may conclude that

$$\begin{eqnarray}\#\{k\leqslant K:k({\mathcal{P}})=\unicode[STIX]{x1D705}\}\ll _{\unicode[STIX]{x1D702}}K^{(\#{\mathcal{P}}+2)\unicode[STIX]{x1D702}}\end{eqnarray}$$

for any $\unicode[STIX]{x1D702}>0$ , which proves Lemma 4.4.◻

6.1 The case $\ell =3$

Restricting to the case $\ell =3$ , we see that (1.4) implies that

(6.2) $$\begin{eqnarray}N_{3}(H;X)\ll XH^{-1}.\end{eqnarray}$$

We also note that $A_{3}(H;X)$ is empty by (1.2) unless $H\leqslant X^{1/3+\unicode[STIX]{x1D700}}$ for some small $\unicode[STIX]{x1D700}>0$ . In general we have the following.

Proposition 6.1. For $1\leqslant H\leqslant X^{1/3+\unicode[STIX]{x1D700}}$ ,

$$\begin{eqnarray}N_{3}(H;X)\ll X^{\unicode[STIX]{x1D700}}(X^{1/2}+X^{7/2}H^{-10}).\end{eqnarray}$$

Proof. To prove this we consider $A_{3}^{0}(H,Z;X)$ with the choice $Z=X^{1/2+2\unicode[STIX]{x1D700}}H^{-1}$ ; note in particular $Z\geqslant X^{1/6}$ when $H\leqslant X^{1/3+\unicode[STIX]{x1D700}}$ . Moreover we will have

$$\begin{eqnarray}h_{3}(-d)>H\gg d^{1/2+\unicode[STIX]{x1D700}}Z^{-1}\end{eqnarray}$$

for all $d$ in $A_{3}^{0}(H,Z;X),$ whence Proposition 2.1 shows that

$$\begin{eqnarray}h_{3}(-d)\ll d^{1/2+\unicode[STIX]{x1D700}}Z^{-2}S_{3}(d;Z).\end{eqnarray}$$

We therefore have

$$\begin{eqnarray}S_{3}(d;Z)\gg d^{-1/2-\unicode[STIX]{x1D700}}Z^{2}h_{3}(-d)\gg X^{-1/2-\unicode[STIX]{x1D700}}Z^{2}h_{3}(-d)>X^{-1/2-\unicode[STIX]{x1D700}}Z^{2}H,\end{eqnarray}$$

for all $d\in A_{3}^{0}(H,Z;X)$ . This leads to the bound

$$\begin{eqnarray}N_{3}^{0}(H;X)(X^{-1/2-\unicode[STIX]{x1D700}}Z^{2}H)^{2}\ll \mathop{\sum }_{d\in A_{3}^{0}(H,Z;X)}S_{3}(d;Z)^{2}\ll \mathop{\sum }_{X\leqslant d<2X}S_{3}(d;Z)^{2}.\end{eqnarray}$$

We can now apply the case $\ell =3$ of Proposition 2.4 to obtain

$$\begin{eqnarray}N_{3}^{0}(H,Z;X)(X^{-1/2-\unicode[STIX]{x1D700}}Z^{2}H)^{2}\ll X^{\unicode[STIX]{x1D700}}\{Z^{2}X^{1/2}+Z^{12}X^{-3/2}\},\end{eqnarray}$$

so that

$$\begin{eqnarray}N_{3}^{0}(H,Z;X)\ll X^{3\unicode[STIX]{x1D700}}H^{-2}\{Z^{-2}X^{3/2}+Z^{8}X^{-1/2}\}\ll X^{19\unicode[STIX]{x1D700}}\{X^{1/2}+X^{7/2}H^{-10}\}\end{eqnarray}$$

in view of our choice of $Z$ . This is sufficient to prove Proposition 6.1, by (6.1).◻

Proof of Theorem 1.2.

We may now derive Theorem 1.2 from Proposition 6.1. It will suffice to consider a dyadic range $X\leqslant d<2X$ . Then

$$\begin{eqnarray}\displaystyle \mathop{\sum }_{X\leqslant d<2X}h_{3}(-d)^{k} & \ll & \displaystyle \mathop{\sum }_{\substack{ H\leqslant X^{1/3+\unicode[STIX]{x1D700}} \\ \text{dyadic}}}\mathop{\sum }_{\substack{ X\leqslant d<2X \\ H<h_{3}(-d)\leqslant 2H}}h_{3}(-d)^{k}\nonumber\\ \displaystyle & {\leqslant} & \displaystyle \mathop{\sum }_{\substack{ H\leqslant X^{1/3+\unicode[STIX]{x1D700}} \\ \text{dyadic}}}N_{3}(H;X)(2H)^{k}.\nonumber\end{eqnarray}$$

In view of (6.2) we have

$$\begin{eqnarray}N_{3}(H;X)(2H)^{k}\ll XH^{k-1}.\end{eqnarray}$$

On the other hand, Proposition 6.1 yields

$$\begin{eqnarray}N_{3}(H;X)(2H)^{k}\ll X^{\unicode[STIX]{x1D700}}(X^{1/2}H^{k}+X^{7/2}H^{k-10}).\end{eqnarray}$$

In particular for $k=4$ we deduce that

$$\begin{eqnarray}\displaystyle N_{3}(H;X)(2H)^{4} & \ll & \displaystyle X^{\unicode[STIX]{x1D700}}\min \{XH^{3},X^{1/2}H^{4}+X^{7/2}H^{-6}\}\nonumber\\ \displaystyle & \ll & \displaystyle X^{\unicode[STIX]{x1D700}}\min \{XH^{3},X^{1/2}H^{4}\}+\min \{XH^{3},X^{7/2}H^{-6}\}.\nonumber\end{eqnarray}$$

For $H\leqslant X^{1/3+\unicode[STIX]{x1D700}}$ the first term is at most

$$\begin{eqnarray}X^{1/2}H^{4}\leqslant X^{11/6+4\unicode[STIX]{x1D700}}\end{eqnarray}$$

while the second term is at most

$$\begin{eqnarray}\{XH^{3}\}^{2/3}\{X^{7/2}H^{-6}\}^{1/3}=X^{11/6}.\end{eqnarray}$$

It follows that $N_{3}(H;X)(2H)^{4}\ll X^{11/6+4\unicode[STIX]{x1D700}}$ , whence

$$\begin{eqnarray}\mathop{\sum }_{X\leqslant d<2X}h_{3}(-d)^{4}\ll X^{11/6+5\unicode[STIX]{x1D700}}.\end{eqnarray}$$

This suffices to prove Theorem 1.2. ◻

As noted in the introduction, one can deduce estimates for other moments from the fourth moment. The reader may check that a direct application of the methods of this section to the general moment only reproduces these consequences of the special case $k=4$ .

6.2 The case $\ell \geqslant 5$

We now consider the $k$ th moment of $h_{\ell }(-d)$ for primes $\ell \geqslant 5$ and any real $k\geqslant 1$ . By Corollary 2.2 we see that

$$\begin{eqnarray}N_{\ell }(H;X)\ll X^{\unicode[STIX]{x1D700}}\quad \text{if }H\geqslant X^{1/2-1/2\ell +\unicode[STIX]{x1D700}}.\end{eqnarray}$$

We also record the trivial bound

(6.3) $$\begin{eqnarray}N_{\ell }(H;X)\ll X,\end{eqnarray}$$

valid for all $H$ . In addition, we claim the following.

Proposition 6.2. For any prime $\ell \geqslant 3$ and $1\leqslant H\leqslant X^{1/2-1/(2\ell )+\unicode[STIX]{x1D700}}$ ,

$$\begin{eqnarray}N_{\ell }(H;X)\ll X^{\unicode[STIX]{x1D700}}(XH^{-1}+X^{\ell /2}H^{-(\ell +1)}).\end{eqnarray}$$

With Proposition 6.2 in hand, we will prove the following.

Proposition 6.3. For any prime $\ell \geqslant 5$ and any real number $k\geqslant 1$ ,

$$\begin{eqnarray}\mathop{\sum }_{X\leqslant d<2X}h_{\ell }(-d)^{k}\ll X^{\unicode[STIX]{x1D70E}+\unicode[STIX]{x1D700}},\end{eqnarray}$$

where

$$\begin{eqnarray}\unicode[STIX]{x1D70E}=\max \{\unicode[STIX]{x1D70E}_{1},\unicode[STIX]{x1D70E}_{2},\unicode[STIX]{x1D70E}_{3}\}\end{eqnarray}$$

and

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D70E}_{1} & = & \displaystyle 1+k\biggl(\frac{\ell -2}{2\ell +2}\biggr),\nonumber\\ \displaystyle \unicode[STIX]{x1D70E}_{2} & = & \displaystyle 1+k\biggl(\frac{\ell -1}{2\ell }\biggr)-\biggl(\frac{\ell -1}{2\ell }\biggr),\nonumber\\ \displaystyle \unicode[STIX]{x1D70E}_{3} & = & \displaystyle \frac{k}{2}.\nonumber\end{eqnarray}$$

We note that the maximum is $\unicode[STIX]{x1D70E}_{1}$ in the range $1\leqslant k\leqslant (\ell ^{2}-1)/(2\ell -1)$ ; it is $\unicode[STIX]{x1D70E}_{2}$ in the range $(\ell ^{2}-1)/(2\ell -1)\leqslant k\leqslant \ell +1$ ; and it is $\unicode[STIX]{x1D70E}_{3}$ for $k\geqslant \ell +1$ . This leads immediately to the statement of Theorem 1.5. We note that Proposition 6.2 does not imply any new results in the case of $h_{3}(-d)$ .

Proof of Proposition 6.2.

The proof of Proposition 6.2 follows similar lines to that of Proposition 6.1. As before we set $Z=X^{1/2+2\unicode[STIX]{x1D700}}H^{-1}$ , so that $Z\geqslant X^{1/(2\ell )}$ for $H\leqslant X^{1/2-1/(2\ell )+\unicode[STIX]{x1D700}}$ . We deduce that

$$\begin{eqnarray}S_{\ell }(d;Z)\gg d^{-1/2-\unicode[STIX]{x1D700}}Z^{2}h_{\ell }(-d)\gg X^{-1/2-\unicode[STIX]{x1D700}}Z^{2}h_{\ell }(-d)>X^{-1/2-\unicode[STIX]{x1D700}}Z^{2}H,\end{eqnarray}$$

again under the assumption that $d\in A_{\ell }^{0}(H,Z;X)$ . As a result,

$$\begin{eqnarray}N_{\ell }^{0}(H,Z;X)X^{-1/2-\unicode[STIX]{x1D700}}Z^{2}H\ll \mathop{\sum }_{d\in A_{\ell }^{0}(H,Z;Z)}S_{\ell }(d;Z)\ll \mathop{\sum }_{X\leqslant d<2X}S_{\ell }(d;Z).\end{eqnarray}$$

Upon applying Proposition 2.3 we obtain

$$\begin{eqnarray}N_{\ell }^{0}(H,Z;X)X^{-1/2-\unicode[STIX]{x1D700}}Z^{2}H\ll X^{\unicode[STIX]{x1D700}}(Z^{2}X^{1/2}+Z^{\ell +2}X^{-1/2}),\end{eqnarray}$$

so that

$$\begin{eqnarray}N_{\ell }^{0}(H,Z;X)\ll X^{2\unicode[STIX]{x1D700}}(XH^{-1}+Z^{\ell }H^{-1})\ll X^{(2+2\ell )\unicode[STIX]{x1D700}}(XH^{-1}+X^{\ell /2}H^{-(\ell +1)}),\end{eqnarray}$$

upon recalling the choice of $Z$ . This is sufficient to prove Proposition 6.2, by (6.1).◻

Proof of Proposition 6.3.

We turn finally to Proposition 6.3, for which we initially fix any real number $k\geqslant 1$ . We have already observed that $N_{\ell }(H;X)\ll X^{\unicode[STIX]{x1D700}}$ if

$$\begin{eqnarray}X^{1/2-1/(2\ell )+\unicode[STIX]{x1D700}}\leqslant H\leqslant X^{1/2+\unicode[STIX]{x1D700}},\end{eqnarray}$$

which shows that for such $H$ ,

(6.4) $$\begin{eqnarray}N_{\ell }(H;X)H^{k}\ll X^{k/2+\unicode[STIX]{x1D700}}.\end{eqnarray}$$

Thus we now instead assume that

(6.5) $$\begin{eqnarray}H\leqslant X^{1/2-1/(2\ell )+\unicode[STIX]{x1D700}}.\end{eqnarray}$$

Then by the trivial bound (6.3) and Proposition 6.2 we have

$$\begin{eqnarray}\displaystyle N_{\ell }(H;X)H^{k} & \ll & \displaystyle X^{\unicode[STIX]{x1D700}}\min \{XH^{k},XH^{k-1}+X^{\ell /2}H^{k-\ell -1}\}\nonumber\\ \displaystyle & \ll & \displaystyle X^{\unicode[STIX]{x1D700}}(XH^{k-1}+\min \{XH^{k},X^{\ell /2}H^{k-\ell -1}\}).\nonumber\end{eqnarray}$$

Under (6.5), the first term is $\ll X^{\unicode[STIX]{x1D70E}_{2}}$ . As long as $k\leqslant \ell +1$ , the second term is largest when $XH^{k}=X^{\ell /2}H^{k-\ell -1}$ , namely when

$$\begin{eqnarray}H=X^{(\ell -2)/(2\ell +2)}=X^{1/2-3/(2\ell +2)}.\end{eqnarray}$$

We may conclude that if $k\leqslant \ell +1$ and $H\leqslant X^{1/2-1/(2\ell )+\unicode[STIX]{x1D700}}$ then

$$\begin{eqnarray}N_{\ell }(H;X)H^{k}\ll X^{\unicode[STIX]{x1D700}}(X^{\unicode[STIX]{x1D70E}_{1}}+X^{\unicode[STIX]{x1D70E}_{2}}),\end{eqnarray}$$

with the notation of Proposition 6.3. On the other hand, if $k\geqslant \ell +1$ then

$$\begin{eqnarray}X^{\ell /2}H^{k-\ell -1}\leqslant X^{\ell /2}H^{k-\ell }\leqslant X^{\ell /2}(X^{1/2})^{k-\ell }=X^{k/2}.\end{eqnarray}$$

Thus $N_{\ell }(H;X)H^{k}\ll X^{\unicode[STIX]{x1D700}}(X^{\unicode[STIX]{x1D70E}_{2}}+X^{k/2})$ in this case; note that the second term dominates in the range $k\geqslant \ell +1$ . To conclude, we have

(6.6) $$\begin{eqnarray}N_{\ell }(H;X)H^{k}\ll X^{\unicode[STIX]{x1D700}}(X^{\unicode[STIX]{x1D70E}_{1}}+X^{\unicode[STIX]{x1D70E}_{2}}+X^{\unicode[STIX]{x1D70E}_{3}})\end{eqnarray}$$

for all $k\geqslant 1$ .

Combining (6.4) and (6.6) shows that

$$\begin{eqnarray}\displaystyle \mathop{\sum }_{X\leqslant d<2X}h_{\ell }(-d)^{k} & \ll & \displaystyle \mathop{\sum }_{\substack{ H\ll X^{1/2+\unicode[STIX]{x1D700}} \\ \text{dyadic}}}\mathop{\sum }_{\substack{ X\leqslant d<2X \\ H<h_{\ell }(-d)\leqslant 2H}}h_{\ell }(-d)^{k}\nonumber\\ \displaystyle & {\leqslant} & \displaystyle \mathop{\sum }_{\substack{ H\ll X^{1/2+\unicode[STIX]{x1D700}} \\ \text{dyadic}}}N_{\ell }(H;X)(2H)^{k}\nonumber\\ \displaystyle & \ll & \displaystyle X^{\unicode[STIX]{x1D700}}(X^{\unicode[STIX]{x1D70E}_{1}}+X^{\unicode[STIX]{x1D70E}_{2}}+X^{\unicode[STIX]{x1D70E}_{3}}).\nonumber\end{eqnarray}$$

We note that $k/2\leqslant \max \{\unicode[STIX]{x1D70E}_{1},\unicode[STIX]{x1D70E}_{2}\}$ in the range $k\leqslant \ell +1$ . This proves Proposition 6.3, and hence Theorem 1.5.◻

The reader may verify that a similar computation based on Proposition 2.4 yields no improvements.