Proof. This is [Reference Wooley15, Lemma 3.1]. ◻

Proof. This is [Reference Ford and Wooley4, Corollary 2.2]. ◻

Proof. We are able to follow the arguments of the proofs of [Reference Wooley17, Lemma 3.5] and [Reference Ford and Wooley4, Lemma 3.3], with some modifications. For the sake of transparency of exposition, we provide an essentially complete account. We begin by considering the situation with $a\geqslant 1$ , the alterations required when $a=0$ being easily accommodated within our basic argument. Consider fixed natural numbers $a$ and $b$ with $1\leqslant a<b$ and $(k-m)b\geqslant (m+1)a$ , and fixed integers $\unicode[STIX]{x1D709}$ and $\unicode[STIX]{x1D702}$ with $1\leqslant \unicode[STIX]{x1D709}\leqslant p^{a}$ , $1\leqslant \unicode[STIX]{x1D702}\leqslant p^{b}$ and $\unicode[STIX]{x1D702}\not \equiv \unicode[STIX]{x1D709}~(\text{mod}~p)$ . Note that in view of the relation (3.4), the conclusion of the lemma follows in general if we can establish it when $h=(k-m)b-ma$ . We henceforth assume that the latter is the case. We then denote by ${\mathcal{D}}_{1}(\mathbf{n})$ the set of ${\mathcal{R}}(h)$ -equivalence classes of solutions of the system of congruences

(3.5) $$\begin{eqnarray}\mathop{\sum }_{i=1}^{m+1}(z_{i}-\unicode[STIX]{x1D702})^{j}\equiv n_{j}~(\text{mod}~p^{(k-m)b})\quad (k-m\leqslant j\leqslant k),\end{eqnarray}$$

with $1\leqslant \mathbf{z}\leqslant p^{kb}$ and $\mathbf{z}\equiv \unicode[STIX]{x1D743}~(\text{mod}~p^{a+1})$ for some $\unicode[STIX]{x1D743}\in \unicode[STIX]{x1D6EF}_{a}^{m+1}(\unicode[STIX]{x1D709})$ . To any solution $\mathbf{z}$ of (3.1) there corresponds a unique $(m+1)$ -tuple $(n_{k-m},\ldots ,n_{k})$ , satisfying $1\leqslant \mathbf{n}\leqslant p^{(k-m)b}$ , for which (3.5) holds and

$$\begin{eqnarray}n_{j}\equiv m_{j}~(\text{mod}~p^{(k-m)b})\quad (k-m\leqslant j\leqslant k).\end{eqnarray}$$

Since $h/b\leqslant k-m$ , we therefore infer that

(3.6) $$\begin{eqnarray}\text{card}({\mathcal{C}}_{a,b}^{m+1,h/b}(\mathbf{m};\unicode[STIX]{x1D709},\unicode[STIX]{x1D702}))\leqslant \max _{1\leqslant \mathbf{n}\leqslant p^{(k-m)b}}\text{card}({\mathcal{D}}_{1}(\mathbf{n})).\end{eqnarray}$$

We fix any choice of $\mathbf{n}$ for which the maximum is achieved in (3.6). There is plainly no loss of generality in supposing that ${\mathcal{D}}_{1}(\mathbf{n})$ is non-empty. Observe that for any solution $\mathbf{z}^{\prime }$ of (3.5) there is an ${\mathcal{R}}(k-m)$ -equivalent solution $\mathbf{z}$ satisfying $1\leqslant \mathbf{z}\leqslant p^{(k-m)b}$ . Rewriting each variable $z_{i}$ in (3.5) in the shape $z_{i}=p^{a}y_{i}+\unicode[STIX]{x1D709}$ , we infer from the hypothesis that $\mathbf{z}\equiv \unicode[STIX]{x1D743}~(\text{mod}~p^{a+1})$ for some $\unicode[STIX]{x1D743}\in \unicode[STIX]{x1D6EF}_{a}^{m+1}(\unicode[STIX]{x1D709})$ that the $(m+1)$ -tuple $\mathbf{y}$ necessarily satisfies

(3.7) $$\begin{eqnarray}y_{i}\not \equiv y_{l}~(\text{mod}~p)\quad (1\leqslant i<l\leqslant m+1).\end{eqnarray}$$

Write $\unicode[STIX]{x1D701}=\unicode[STIX]{x1D709}-\unicode[STIX]{x1D702}$ , and note that the constraint $\unicode[STIX]{x1D702}\not \equiv \unicode[STIX]{x1D709}~(\text{mod}~p)$ ensures that $p\nmid \unicode[STIX]{x1D701}$ . We denote the multiplicative inverse of $\unicode[STIX]{x1D701}$ modulo $p^{(k-m)b}$ by $\unicode[STIX]{x1D701}^{-1}$ . Then we deduce from (3.5) that $\text{card}({\mathcal{D}}_{1}(\mathbf{n}))$ is bounded above by the number of ${\mathcal{R}}(h-a)$ -equivalence classes of solutions of the system of congruences

(3.8) $$\begin{eqnarray}\mathop{\sum }_{i=1}^{m+1}(p^{a}y_{i}\unicode[STIX]{x1D701}^{-1}+1)^{j}\equiv n_{j}(\unicode[STIX]{x1D701}^{-1})^{j}~(\text{mod}~p^{(k-m)b})\quad (k-m\leqslant j\leqslant k),\end{eqnarray}$$

with $1\leqslant \mathbf{y}\leqslant p^{(k-m)b-a}$ satisfying (3.7). Let $\mathbf{y}=\mathbf{w}$ be any solution of the system (3.8). Then we find that all other solutions $\mathbf{y}$ satisfy the system

(3.9) $$\begin{eqnarray}\mathop{\sum }_{i=1}^{m+1}((p^{a}y_{i}\unicode[STIX]{x1D701}^{-1}+1)^{j}-(p^{a}w_{i}\unicode[STIX]{x1D701}^{-1}+1)^{j})\equiv 0~(\text{mod}~p^{(k-m)b})\quad (k-m\leqslant j\leqslant k).\end{eqnarray}$$

We next apply [Reference Wooley17, Lemma 3.2], just as in the arguments of the proofs of [Reference Wooley17, Lemmas 3.3 to 3.6]. Consider an index $j$ with $k-m\leqslant j\leqslant k$ , and apply the latter lemma with $\unicode[STIX]{x1D6FC}=k-m-1$ and $\unicode[STIX]{x1D6FD}=j-k+m+1$ . We find that there exist integers $c_{j,l}$ $(k-m-1\leqslant l\leqslant j)$ and $d_{j,u}$ $(j-k+m+1\leqslant u\leqslant j)$ , bounded in terms of $k$ , and with $d_{j,j-k+m+1}\neq 0$ , for which one has the polynomial identity

(3.10) $$\begin{eqnarray}c_{j,k-m-1}+\mathop{\sum }_{l=k-m}^{j}c_{j,l}(x+1)^{l}=\mathop{\sum }_{u=j-k+m+1}^{j}d_{j,u}x^{u}.\end{eqnarray}$$

Since we may assume $p$ to be large, we may suppose that $p\nmid d_{j,j-k+m+1}$ . Thus, multiplying (3.10) through by the multiplicative inverse of $d_{j,j-k+m+1}$ modulo $p^{(k-m)b}$ , we see that there is no loss in supposing that

$$\begin{eqnarray}d_{j,j-k+m+1}\equiv 1~(\text{mod}~p^{(k-m)b}).\end{eqnarray}$$

Consequently, by taking suitable linear combinations of the congruences comprising (3.9), we discern that any solution of this system satisfies

$$\begin{eqnarray}(\unicode[STIX]{x1D701}^{-1}p^{a})^{j-k+m+1}\mathop{\sum }_{i=1}^{m+1}(\unicode[STIX]{x1D713}_{j}(y_{i})-\unicode[STIX]{x1D713}_{j}(w_{i}))\equiv 0~(\text{mod}~p^{(k-m)b})\quad (k-m\leqslant j\leqslant k),\end{eqnarray}$$

in which

$$\begin{eqnarray}\unicode[STIX]{x1D713}_{j}(z)=z^{j-k+m+1}+\mathop{\sum }_{u=j-k+m+2}^{j}d_{j,u}(\unicode[STIX]{x1D701}^{-1}p^{a})^{u-j+k-m-1}z^{u}.\end{eqnarray}$$

Note here that

(3.11) $$\begin{eqnarray}\unicode[STIX]{x1D713}_{j}(z)\equiv z^{j-k+m+1}~(\text{mod}~p).\end{eqnarray}$$

Denote by ${\mathcal{D}}_{2}(\mathbf{u})$ the set of ${\mathcal{R}}(h-a)$ -equivalence classes of solutions of the system of congruences

$$\begin{eqnarray}\mathop{\sum }_{i=1}^{m+1}\unicode[STIX]{x1D713}_{j}(y_{i})\equiv u_{j}~(\text{mod}~p^{(k-m)b-(j-k+m+1)a})\quad (k-m\leqslant j\leqslant k),\end{eqnarray}$$

with $1\leqslant \mathbf{y}\leqslant p^{(k-m)b-a}$ satisfying (3.7). Then we have shown thus far that

(3.12) $$\begin{eqnarray}\text{card}({\mathcal{D}}_{1}(\mathbf{n}))\leqslant \max _{1\leqslant \mathbf{u}\leqslant p^{(k-m)b}}\text{card}({\mathcal{D}}_{2}(\mathbf{u})).\end{eqnarray}$$

Let ${\mathcal{D}}_{3}(\mathbf{v})$ denote the set of solutions of the congruence

$$\begin{eqnarray}\mathop{\sum }_{i=1}^{m+1}\unicode[STIX]{x1D713}_{j}(y_{i})\equiv v_{j}~(\text{mod}~p^{h-a})\quad (k-m\leqslant j\leqslant k),\end{eqnarray}$$

with $1\leqslant \mathbf{y}\leqslant p^{h-a}$ satisfying (3.7). For $k-m\leqslant j\leqslant k$ , we have

$$\begin{eqnarray}(k-m)b-(j-k+m+1)a\geqslant (k-m)b-(m+1)a=h-a.\end{eqnarray}$$

Then we arrive at the upper bound

(3.13) $$\begin{eqnarray}\text{card}({\mathcal{D}}_{2}(\mathbf{u}))\leqslant \max _{1\leqslant \mathbf{v}\leqslant p^{h-a}}\text{card}({\mathcal{D}}_{3}(\mathbf{v})).\end{eqnarray}$$

By combining (3.6), (3.12) and (3.13), we discern at this point that

(3.14) $$\begin{eqnarray}\text{card}({\mathcal{C}}_{a,b}^{m+1,h/b}(\mathbf{m};\unicode[STIX]{x1D709},\unicode[STIX]{x1D702}))\leqslant \max _{1\leqslant \mathbf{v}\leqslant p^{h-a}}\text{card}({\mathcal{D}}_{3}(\mathbf{v})).\end{eqnarray}$$

Define the determinant

$$\begin{eqnarray}J(\unicode[STIX]{x1D74D};\mathbf{x})=\det (\unicode[STIX]{x1D713}_{k-m+l-1}^{\prime }(x_{i}))_{1\leqslant i,l\leqslant m+1}.\end{eqnarray}$$

In view of (3.11), one has $\unicode[STIX]{x1D713}_{k-m+l-1}^{\prime }(y_{i})\equiv ly_{i}^{l-1}~(\text{mod}~p)$ . It follows from (3.7) that

$$\begin{eqnarray}\det (y_{i}^{l-1})_{1\leqslant i,l\leqslant m+1}=\mathop{\prod }_{1\leqslant i<u\leqslant m+1}(y_{i}-y_{u})\not \equiv 0~(\text{mod}~p),\end{eqnarray}$$

so that, since $p>k$ , we have $(J(\unicode[STIX]{x1D74D};\mathbf{y}),p)=1$ . We therefore deduce from [Reference Wooley13, Theorem 1], just as in the corresponding argument of the proof of [Reference Wooley17, Lemma 3.3] following [Reference Wooley17, equation (3.17)], that

$$\begin{eqnarray}\text{card}({\mathcal{D}}_{3}(\mathbf{v}))\leqslant (k-m)(k-m+1)\cdots k\leqslant k!\end{eqnarray}$$

and thus the conclusion of the lemma when $a\geqslant 1$ follows at once from (3.2) and (3.14).

When $a=0$ , we must apply some minor modifications to the above argument. In this case, we denote by ${\mathcal{D}}_{1}(\mathbf{n};\unicode[STIX]{x1D702})$ the set of ${\mathcal{R}}(h)$ -equivalence classes of solutions of the system of congruences (3.5) with $1\leqslant \mathbf{z}\leqslant p^{kb}$ and $\mathbf{z}\equiv \unicode[STIX]{x1D743}~(\text{mod}~p)$ for some $\unicode[STIX]{x1D743}\in \unicode[STIX]{x1D6EF}_{0}^{m+1}(0)$ , and for which in addition $z_{i}\not \equiv \unicode[STIX]{x1D702}~(\text{mod}~p)$ for $1\leqslant i\leqslant m+1$ . Then, as in the opening paragraph of our proof, it follows from (3.1) that

(3.15) $$\begin{eqnarray}\text{card}({\mathcal{C}}_{0,b}^{m+1,h/b}(\mathbf{m};0,\unicode[STIX]{x1D702}))\leqslant \max _{1\leqslant \mathbf{n}\leqslant p^{(k-m)b}}\text{card}({\mathcal{D}}_{1}(\mathbf{n};\unicode[STIX]{x1D702})).\end{eqnarray}$$

But $\text{card}({\mathcal{D}}_{1}(\mathbf{n};\unicode[STIX]{x1D702}))=\text{card}({\mathcal{D}}_{1}(\mathbf{n};0))$ , and $\text{card}({\mathcal{D}}_{1}(\mathbf{n};0))$ counts the solutions of the system of congruences

$$\begin{eqnarray}\mathop{\sum }_{i=1}^{m+1}y_{i}^{j}\equiv n_{j}~(\text{mod}~p^{(k-m)b})\quad (k-m\leqslant j\leqslant k),\end{eqnarray}$$

with $1\leqslant \mathbf{y}\leqslant p^{(k-m)b}$ satisfying (3.7), and in addition $p\nmid y_{i}$ $(1\leqslant i\leqslant m+1)$ . Write

$$\begin{eqnarray}J(\mathbf{y})=\det ((k-m+j-1)y_{i}^{k-m+j-2})_{1\leqslant i,j\leqslant m+1}.\end{eqnarray}$$

Then, since $p>k$ , we have

$$\begin{eqnarray}J(\mathbf{y})=\frac{k!}{(k-m-1)!}(y_{1}\cdots y_{m+1})^{k-m-1}\mathop{\prod }_{1\leqslant i<j\leqslant m+1}(y_{i}-y_{j})\not \equiv 0~(\text{mod}~p).\end{eqnarray}$$

We therefore conclude from [Reference Wooley13, Theorem 1] that

$$\begin{eqnarray}\text{card}({\mathcal{D}}_{1}(\mathbf{n};0))\leqslant (k-m)(k-m+1)\cdots k\leqslant k!.\end{eqnarray}$$

In view of (3.3), the conclusion of the lemma therefore follows from (3.15) when $a=0$ . This completes our account of the proof of the lemma.◻

Proof. Our argument follows the proof of [Reference Wooley19, Lemma 4.1] with minor adjustments. Consider fixed integers $\unicode[STIX]{x1D709}$ and $\unicode[STIX]{x1D702}$ with $\unicode[STIX]{x1D702}\not \equiv \unicode[STIX]{x1D709}~(\text{mod}~p)$ . Let $T_{1}$ denote the number of integral solutions $\mathbf{x}$ , $\mathbf{y}$ , $\mathbf{v}$ , $\mathbf{w}$ of the system (2.13) counted by $I_{a,b}^{r,r}(X;\unicode[STIX]{x1D709},\unicode[STIX]{x1D702})$ in which $v_{1},\ldots ,v_{s}$ together occupy at least $r$ distinct residue classes modulo $p^{b+1}$ , and let $T_{2}$ denote the corresponding number of solutions in which these integers together occupy at most $r-1$ distinct residue classes modulo $p^{b+1}$ . Then

(4.1) $$\begin{eqnarray}I_{a,b}^{r,r}(X;\unicode[STIX]{x1D709},\unicode[STIX]{x1D702})=T_{1}+T_{2}.\end{eqnarray}$$

From (2.10) to (2.12), orthogonality and Schwarz’s inequality, one has

(4.2) $$\begin{eqnarray}\displaystyle T_{1} & {\leqslant} & \displaystyle \binom{s}{r}\oint |\mathop{\mathfrak{F}}_{a}^{r}(\unicode[STIX]{x1D736};\unicode[STIX]{x1D709})|^{2}\mathfrak{F}_{b}^{r}(\unicode[STIX]{x1D736};\unicode[STIX]{x1D702})\mathfrak{f}_{b}(\unicode[STIX]{x1D736};\unicode[STIX]{x1D702})^{s-r}\mathfrak{f}_{b}(-\unicode[STIX]{x1D736};\unicode[STIX]{x1D702})^{s}\,d\unicode[STIX]{x1D736}\nonumber\\ \displaystyle & \ll & \displaystyle (K_{a,b}^{r,r}(X;\unicode[STIX]{x1D709},\unicode[STIX]{x1D702}))^{1/2}(I_{a,b}^{r,r}(X;\unicode[STIX]{x1D709},\unicode[STIX]{x1D702}))^{1/2}.\end{eqnarray}$$

In order to treat $T_{2}$ , we observe first that the hypothesis $s\geqslant 2r-1$ ensures that there is an integer $\unicode[STIX]{x1D701}\equiv \unicode[STIX]{x1D702}~(\text{mod}~p^{b})$ having the property that three at least of the variables $v_{1},\ldots ,v_{s}$ are congruent to $\unicode[STIX]{x1D701}$ modulo $p^{b+1}$ . Hence, recalling the definitions (2.10) and (2.11), one finds from orthogonality and Hölder’s inequality that

(4.3) $$\begin{eqnarray}\displaystyle T_{2} & {\leqslant} & \displaystyle \binom{s}{3}\mathop{\sum }_{\substack{ 1\leqslant \unicode[STIX]{x1D701}\leqslant p^{b+1} \\ \unicode[STIX]{x1D701}\equiv \unicode[STIX]{x1D702}~(\text{mod}~p^{b})}}\oint |\mathop{\mathfrak{F}}_{a}^{r}(\unicode[STIX]{x1D736};\unicode[STIX]{x1D709})|^{2}\mathfrak{f}_{b+1}(\unicode[STIX]{x1D736};\unicode[STIX]{x1D701})^{3}\mathfrak{f}_{b}(\unicode[STIX]{x1D736};\unicode[STIX]{x1D702})^{s-3}\mathfrak{f}_{b}(-\unicode[STIX]{x1D736};\unicode[STIX]{x1D702})^{s}\,d\unicode[STIX]{x1D736}\nonumber\\ \displaystyle & \ll & \displaystyle M\max _{\substack{ 1\leqslant \unicode[STIX]{x1D701}\leqslant p^{b+1} \\ \unicode[STIX]{x1D701}\equiv \unicode[STIX]{x1D702}~(\text{mod}~p^{b})}}(I_{a,b}^{r,r}(X;\unicode[STIX]{x1D709},\unicode[STIX]{x1D702}))^{1-3/(2s)}(I_{a,b+1}^{r,r}(X;\unicode[STIX]{x1D709},\unicode[STIX]{x1D701}))^{3/(2s)}.\end{eqnarray}$$

By substituting (4.2) and (4.3) into (4.1), and recalling (2.16) and (2.17), we therefore conclude that

$$\begin{eqnarray}I_{a,b}^{r,r}(X)\ll (K_{a,b}^{r,r}(X))^{1/2}(I_{a,b}^{r,r}(X))^{1/2}+M(I_{a,b}^{r,r}(X))^{1-3/(2s)}(I_{a,b+1}^{r,r}(X))^{3/(2s)},\end{eqnarray}$$

whence

$$\begin{eqnarray}I_{a,b}^{r,r}(X)\ll K_{a,b}^{r,r}(X)+M^{2s/3}I_{a,b+1}^{r,r}(X).\end{eqnarray}$$

This completes the proof of the lemma. ◻

Proof. Repeated application of Lemma 4.1 yields the upper bound

(4.4) $$\begin{eqnarray}I_{a,b}^{r,r}(X)\ll \mathop{\sum }_{h=0}^{H-1}(M^{h})^{2s/3}K_{a,b+h}^{r,r}(X)+(M^{H})^{2s/3}I_{a,b+H}^{r,r}(X).\end{eqnarray}$$

On considering the underlying Diophantine systems, it follows from Lemma 2.3 that for each $\unicode[STIX]{x1D709}$ and $\unicode[STIX]{x1D702}$ , one has

$$\begin{eqnarray}\displaystyle I_{a,b+H}^{r,r}(X;\unicode[STIX]{x1D709},\unicode[STIX]{x1D702}) & {\leqslant} & \displaystyle \oint |\mathfrak{f}_{a}(\unicode[STIX]{x1D736};\unicode[STIX]{x1D709})^{2r}\mathfrak{f}_{b+H}(\unicode[STIX]{x1D736};\unicode[STIX]{x1D702})^{2s}|\,d\unicode[STIX]{x1D736}\nonumber\\ \displaystyle & \ll & \displaystyle (J_{s+r}(X/M^{a}))^{r/(s+r)}(J_{s+r}(X/M^{b+H}))^{s/(s+r)}.\nonumber\end{eqnarray}$$

Since $M^{b+H}=(X^{\unicode[STIX]{x1D703}})^{b+H}\leqslant X^{1/2}$ , we deduce from (2.16) and (2.24) that

$$\begin{eqnarray}\displaystyle (M^{H})^{2s/3}I_{a,b+H}^{r,r}(X) & \ll & \displaystyle X^{\unicode[STIX]{x1D6FF}}((X/M^{a})^{r/(s+r)}(X/M^{b+H})^{s/(s+r)})^{\unicode[STIX]{x1D706}}(M^{H})^{2s/3}\nonumber\\ \displaystyle & = & \displaystyle X^{\unicode[STIX]{x1D6FF}}(X/M^{b})^{s}(X/M^{a})^{\unicode[STIX]{x1D706}-s}M^{\unicode[STIX]{x1D6FA}},\nonumber\end{eqnarray}$$

where

$$\begin{eqnarray}\unicode[STIX]{x1D6FA}=\unicode[STIX]{x1D706}\biggl(a-\frac{ar}{s+r}-\frac{bs}{s+r}\biggr)+s(b-a)+Hs\biggl(\frac{2}{3}-\frac{\unicode[STIX]{x1D706}}{s+r}\biggr).\end{eqnarray}$$

Since $\unicode[STIX]{x1D706}\geqslant s+r$ , the lower bound $b>a$ leads to the estimate

$$\begin{eqnarray}\unicode[STIX]{x1D6FA}\leqslant -s(b-a)\frac{\unicode[STIX]{x1D706}}{s+r}+s(b-a)-\frac{1}{3}Hs\leqslant -\frac{1}{3}Hs.\end{eqnarray}$$

This completes the proof of the lemma, since with $H\geqslant 15$ one has

$$\begin{eqnarray}X^{\unicode[STIX]{x1D6FF}}M^{-Hs/3}\ll M^{-Hs/4}.\end{eqnarray}$$

◻

Proof. The mean value $J_{s+r}(X)$ counts the number of integral solutions of the system

(5.1) $$\begin{eqnarray}\mathop{\sum }_{i=1}^{s+r}(x_{i}^{j}-y_{i}^{j})=0\quad (1\leqslant j\leqslant k),\end{eqnarray}$$

with $1\leqslant \mathbf{x},\mathbf{y}\leqslant X$ . Let ${\mathcal{P}}$ denote a set of $\lceil (s+r)^{2}\unicode[STIX]{x1D703}^{-1}\rceil$ prime numbers lying in $(M,2M]$ . That such a set exists is a consequence of the prime number theorem. The argument of the proof of [Reference Ford and Wooley4, Lemma 6.1] leading to equation (6.2) of that paper shows that for some $p\in {\mathcal{P}}$ , one has $J_{s+r}(X)\ll T(p)$ , where $T(p)$ denotes the number of solutions of the system (5.1) counted by $J_{s+r}(X)$ in which $x_{1},\ldots ,x_{s+r}$ are distinct modulo $p$ , and likewise $y_{1},\ldots ,y_{s+r}$ are distinct modulo  $p$ .

We now examine the residue classes modulo $p^{B}$ of a given solution $\mathbf{x},\mathbf{y}$ counted by $T(p)$ . Let $\unicode[STIX]{x1D6C8}$ and $\boldsymbol{\unicode[STIX]{x1D701}}$ be $s$ -tuples with $1\leqslant \unicode[STIX]{x1D6C8},\boldsymbol{\unicode[STIX]{x1D701}}\leqslant p^{B}$ satisfying the condition that for $1\leqslant i\leqslant s$ , one has $x_{i}\equiv \unicode[STIX]{x1D702}_{i}~(\text{mod}~p^{B})$ and $y_{i}\equiv \unicode[STIX]{x1D701}_{i}~(\text{mod}~p^{B})$ . Recall the notation introduced prior to (2.18). Then, since $x_{1},\ldots ,x_{s+r}$ are distinct modulo $p$ , it follows that $(x_{s+1},\ldots ,x_{s+r})\in \unicode[STIX]{x1D6EF}^{r}(\unicode[STIX]{x1D6C8})$ , and likewise one finds that $(y_{s+1},\ldots ,y_{s+r})\in \unicode[STIX]{x1D6EF}^{r}(\boldsymbol{\unicode[STIX]{x1D701}})$ . Then, on considering the underlying Diophantine systems, one finds that

$$\begin{eqnarray}T(p)\leqslant \mathop{\sum }_{1\leqslant \unicode[STIX]{x1D6C8},\boldsymbol{\unicode[STIX]{x1D701}}\leqslant p^{B}}\oint \biggl(\mathop{\prod }_{i=1}^{s}\mathfrak{f}_{B}(\unicode[STIX]{x1D736};\unicode[STIX]{x1D702}_{i})\mathfrak{f}_{B}(-\unicode[STIX]{x1D736};\unicode[STIX]{x1D701}_{i})\biggr)\mathfrak{F}^{r}(\unicode[STIX]{x1D736};\unicode[STIX]{x1D6C8})\mathfrak{F}^{r}(-\unicode[STIX]{x1D736};\boldsymbol{\unicode[STIX]{x1D701}})\,d\unicode[STIX]{x1D736}.\end{eqnarray}$$

Write

$$\begin{eqnarray}{\mathcal{J}}(\unicode[STIX]{x1D73D},\unicode[STIX]{x1D713})=\oint |\mathfrak{F}^{r}(\unicode[STIX]{x1D736};\unicode[STIX]{x1D73D})^{2}\mathfrak{f}_{B}(\unicode[STIX]{x1D736};\unicode[STIX]{x1D713})^{2s}|\,d\unicode[STIX]{x1D736}.\end{eqnarray}$$

Then, by applying Hölder’s inequality, and again considering the underlying Diophantine systems, we discern that

$$\begin{eqnarray}\displaystyle T(p) & {\leqslant} & \displaystyle \mathop{\sum }_{1\leqslant \unicode[STIX]{x1D6C8},\boldsymbol{\unicode[STIX]{x1D701}}\leqslant p^{B}}\mathop{\prod }_{i=1}^{s}({\mathcal{J}}(\unicode[STIX]{x1D6C8},\unicode[STIX]{x1D702}_{i}){\mathcal{J}}(\boldsymbol{\unicode[STIX]{x1D701}},\unicode[STIX]{x1D701}_{i}))^{1/(2s)}\nonumber\\ \displaystyle & {\leqslant} & \displaystyle \mathop{\sum }_{1\leqslant \unicode[STIX]{x1D6C8},\boldsymbol{\unicode[STIX]{x1D701}}\leqslant p^{B}}\mathop{\prod }_{i=1}^{s}({\mathcal{J}}(\unicode[STIX]{x1D702}_{i},\unicode[STIX]{x1D702}_{i}){\mathcal{J}}(\unicode[STIX]{x1D701}_{i},\unicode[STIX]{x1D701}_{i}))^{1/(2s)}.\nonumber\end{eqnarray}$$

Hence, on recalling the definition (2.18), we obtain the upper bound

(5.2) $$\begin{eqnarray}\displaystyle T(p) & {\leqslant} & \displaystyle p^{2sB}\max _{1\leqslant \unicode[STIX]{x1D702}\leqslant p^{B}}\oint |\mathfrak{F}^{r}(\unicode[STIX]{x1D736};\unicode[STIX]{x1D702})^{2}\mathfrak{f}_{B}(\unicode[STIX]{x1D736};\unicode[STIX]{x1D702})^{2s}|\,d\unicode[STIX]{x1D736}\nonumber\\ \displaystyle & = & \displaystyle p^{2sB}\max _{1\leqslant \unicode[STIX]{x1D702}\leqslant p^{B}}\widetilde{I}_{B}^{r,r}(X;\unicode[STIX]{x1D702}).\end{eqnarray}$$

By modifying the argument of the proof of [Reference Ford and Wooley4, Lemma 6.1] leading from equation (6.3) to equation (6.6) of that paper, along the lines easily surmised from our proof of Lemma 4.1 above, one finds that

(5.3) $$\begin{eqnarray}\widetilde{I}_{c}^{r,r}(X;\unicode[STIX]{x1D702})\ll \widetilde{K}_{c}^{r,r}(X;\unicode[STIX]{x1D702})+M^{2s/3}\max _{1\leqslant \unicode[STIX]{x1D701}\leqslant p^{c+1}}\widetilde{I}_{c+1}^{r,r}(X;\unicode[STIX]{x1D701}).\end{eqnarray}$$

Iterating (5.3) in order to bound $\widetilde{I}_{B}^{r,r}(X;\unicode[STIX]{x1D702})$ , just as in the argument concluding the proof of [Reference Ford and Wooley4, Lemma 6.1] (and see also (4.4) above), we discern from (5.2) that

(5.4) $$\begin{eqnarray}\displaystyle J_{s+r}(X) & \ll & \displaystyle M^{2sB}\max _{0\leqslant h\leqslant 4B}(M^{h})^{2s/3}K_{0,B+h}^{r,r}(X)\nonumber\\ \displaystyle & & \displaystyle +\,M^{2sB+8sB/3}\max _{1\leqslant \unicode[STIX]{x1D701}\leqslant p^{5B+1}}\widetilde{I}_{5B+1}^{r,r}(X;\unicode[STIX]{x1D701}).\end{eqnarray}$$

By considering the underlying Diophantine systems, we deduce from Lemma 2.3 that

$$\begin{eqnarray}\widetilde{I}_{5B+1}^{r,r}(X;\unicode[STIX]{x1D701})\ll (J_{s+r}(X))^{r/(s+r)}(J_{s+r}(X/M^{5B+1}))^{s/(s+r)}.\end{eqnarray}$$

In this way, we find from (5.4) either that

(5.5) $$\begin{eqnarray}J_{s+r}(X)\ll M^{2sB+2sh/3}K_{0,B+h}^{r,r}(X)\end{eqnarray}$$

for some index $h$ with $0\leqslant h\leqslant 4B$ , so that the conclusion of the lemma holds, or else that

$$\begin{eqnarray}J_{s+r}(X)\ll M^{14sB/3}(J_{s+r}(X))^{r/(s+r)}(J_{s+r}(X/M^{5B+1}))^{s/(s+r)}.\end{eqnarray}$$

In the latter case, since $\unicode[STIX]{x1D706}\geqslant s+r$ , we obtain the upper bound

$$\begin{eqnarray}\displaystyle J_{s+r}(X) & \ll & \displaystyle M^{14(s+r)B/3}J_{s+r}(X/M^{5B+1})\nonumber\\ \displaystyle & \ll & \displaystyle M^{14(s+r)B/3}(X/M^{5B+1})^{\unicode[STIX]{x1D706}+\unicode[STIX]{x1D6FF}}\nonumber\\ \displaystyle & \ll & \displaystyle X^{\unicode[STIX]{x1D706}+\unicode[STIX]{x1D6FF}}M^{-(s+r)B/3}.\nonumber\end{eqnarray}$$

Invoking the definition (2.8) of $\unicode[STIX]{x1D6FF}$ , we find that $J_{s+r}(X)\ll X^{\unicode[STIX]{x1D706}-2\unicode[STIX]{x1D6FF}}$ , contradicting the lower bound (2.24) if $X=X_{l}$ is large enough. We are therefore forced to accept the former upper bound (5.5), and hence the proof of the lemma is completed by reference to (2.20).◻

Proof. We first consider the situation in which $a\geqslant 1$ . The argument associated with the case $a=0$ is very similar, and so we are able to appeal later to a highly abbreviated argument for this case in order to complete the proof of the lemma. Consider fixed integers $\unicode[STIX]{x1D709}$ and $\unicode[STIX]{x1D702}$ with

(6.3) $$\begin{eqnarray}1\leqslant \unicode[STIX]{x1D709}\leqslant p^{a},\quad 1\leqslant \unicode[STIX]{x1D702}\leqslant p^{b}\quad \text{and}\quad \unicode[STIX]{x1D702}\not \equiv \unicode[STIX]{x1D709}~(\text{mod}~p).\end{eqnarray}$$

The quantity $K_{a,b}^{m+1,r}(X;\unicode[STIX]{x1D709},\unicode[STIX]{x1D702})$ counts the number of integral solutions of the system (2.14) with $m$ replaced by $m+1$ , subject to the attendant conditions on $\mathbf{x}$ , $\mathbf{y}$ , $\mathbf{u}$ , $\mathbf{v}$ , $\mathbf{w}$ , $\mathbf{z}$ . Given such a solution of the system (2.14), the discussion leading to (2.15) shows that

(6.4) $$\begin{eqnarray}\mathop{\sum }_{i=1}^{m+1}(x_{i}-\unicode[STIX]{x1D702})^{j}\equiv \mathop{\sum }_{i=1}^{m+1}(y_{i}-\unicode[STIX]{x1D702})^{j}~(\text{mod}~p^{jb})\quad (1\leqslant j\leqslant k).\end{eqnarray}$$

In the notation introduced in §3, it follows that for some $k$ -tuple of integers $\mathbf{m}$ , both $[\mathbf{x}~(\text{mod}~p^{kb})]$ and $[\mathbf{y}~(\text{mod}~p^{kb})]$ lie in ${\mathcal{B}}_{a,b}^{m+1}(\mathbf{m};\unicode[STIX]{x1D709},\unicode[STIX]{x1D702})$ . Write

$$\begin{eqnarray}\mathfrak{G}_{a,b}(\unicode[STIX]{x1D736};\mathbf{m})=\mathop{\sum }_{\unicode[STIX]{x1D73D}\in {\mathcal{B}}_{a,b}^{m+1}(\mathbf{m};\unicode[STIX]{x1D709},\unicode[STIX]{x1D702})}\mathop{\prod }_{i=1}^{m+1}\mathfrak{f}_{kb}(\unicode[STIX]{x1D736};\unicode[STIX]{x1D703}_{i}).\end{eqnarray}$$

Then, on considering the underlying Diophantine systems, we see from (2.12) and (6.4) that

$$\begin{eqnarray}K_{a,b}^{m+1,r}(X;\unicode[STIX]{x1D709},\unicode[STIX]{x1D702})=\mathop{\sum }_{m_{1}=1}^{p^{b}}\cdots \mathop{\sum }_{m_{k}=1}^{p^{kb}}\oint |\mathfrak{G}_{a,b}(\unicode[STIX]{x1D736};\mathbf{m})^{2}\mathfrak{F}_{m}^{\ast }(\unicode[STIX]{x1D736})^{2}|\,d\unicode[STIX]{x1D736},\end{eqnarray}$$

where we write

(6.5) $$\begin{eqnarray}\mathfrak{F}_{m}^{\ast }(\unicode[STIX]{x1D736})=\mathfrak{F}_{b}^{r}(\unicode[STIX]{x1D736};\unicode[STIX]{x1D702})\mathfrak{f}_{b}(\unicode[STIX]{x1D736};\unicode[STIX]{x1D702})^{s-m-1}.\end{eqnarray}$$

We now partition the vectors in each set ${\mathcal{B}}_{a,b}^{m+1}(\mathbf{m};\unicode[STIX]{x1D709},\unicode[STIX]{x1D702})$ into equivalence classes modulo $p^{b^{\prime }}$ as in §3. Write ${\mathcal{C}}(\mathbf{m})={\mathcal{C}}_{a,b}^{m+1,b^{\prime }/b}(\mathbf{m};\unicode[STIX]{x1D709},\unicode[STIX]{x1D702})$ . An application of Cauchy’s inequality leads via (3.2) and Lemma 3.1 to the bound

$$\begin{eqnarray}\displaystyle |\mathfrak{G}_{a,b}(\unicode[STIX]{x1D736};\mathbf{m})|^{2} & = & \displaystyle \biggl|\mathop{\sum }_{\mathfrak{C}\in {\mathcal{C}}(\mathbf{m})}\mathop{\sum }_{\unicode[STIX]{x1D73D}\in \mathfrak{C}}\mathop{\prod }_{i=1}^{m+1}\mathfrak{f}_{kb}(\unicode[STIX]{x1D736};\unicode[STIX]{x1D703}_{i})\biggr|^{2}\nonumber\\ \displaystyle & {\leqslant} & \displaystyle \text{card}({\mathcal{C}}(\mathbf{m}))\mathop{\sum }_{\mathfrak{C}\in {\mathcal{C}}(\mathbf{m})}\biggl|\mathop{\sum }_{\unicode[STIX]{x1D73D}\in \mathfrak{C}}\mathop{\prod }_{i=1}^{m+1}\mathfrak{f}_{kb}(\unicode[STIX]{x1D736};\unicode[STIX]{x1D703}_{i})\biggr|^{2}\nonumber\\ \displaystyle & {\leqslant} & \displaystyle k!\mathop{\sum }_{\mathfrak{C}\in {\mathcal{C}}(\mathbf{m})}\biggl|\mathop{\sum }_{\unicode[STIX]{x1D73D}\in \mathfrak{C}}\mathop{\prod }_{i=1}^{m+1}\mathfrak{f}_{kb}(\unicode[STIX]{x1D736};\unicode[STIX]{x1D703}_{i})\biggr|^{2}.\nonumber\end{eqnarray}$$

It may be helpful to note here that since $b^{\prime }/b\leqslant k-m-ma/b$ , then in view of the relation (3.2), it follows from Lemma 3.1 that

$$\begin{eqnarray}\text{card}({\mathcal{C}}(\mathbf{m}))=B_{a,b}^{m+1,b^{\prime }/b}(p)\leqslant k!.\end{eqnarray}$$

Hence,

$$\begin{eqnarray}K_{a,b}^{m+1,r}(X;\unicode[STIX]{x1D709},\unicode[STIX]{x1D702})\ll \mathop{\sum }_{\mathbf{m}}\mathop{\sum }_{\mathfrak{C}\in {\mathcal{C}}(\mathbf{m})}\oint \biggl|\mathop{\mathfrak{F}}_{m}^{\ast }(\unicode[STIX]{x1D6FC})\mathop{\sum }_{\unicode[STIX]{x1D73D}\in \mathfrak{C}}\mathop{\prod }_{i=1}^{m+1}\mathfrak{f}_{kb}(\unicode[STIX]{x1D736};\unicode[STIX]{x1D703}_{i})\biggr|^{2}\,d\unicode[STIX]{x1D736}.\end{eqnarray}$$

For each $k$ -tuple $\mathbf{m}$ and equivalence class $\mathfrak{C}$ , the integral above counts solutions of (2.14) with the additional constraint that both $[\mathbf{x}~(\text{mod}~p^{kb})]$ and $[\mathbf{y}~(\text{mod}~p^{kb})]$ lie in $\mathfrak{C}$ . In particular, one has $\mathbf{x}\equiv \mathbf{y}~(\text{mod}~p^{b^{\prime }})$ . Moreover, the sets ${\mathcal{B}}_{a,b}^{m+1}(\mathbf{m};\unicode[STIX]{x1D709},\unicode[STIX]{x1D702})$ are disjoint for distinct $k$ -tuples $\mathbf{m}$ with $1\leqslant m_{j}\leqslant p^{jb}$ $(1\leqslant j\leqslant k)$ , so to each pair $(\mathbf{x},\mathbf{y})$ there corresponds at most one pair $(\mathbf{m},\mathfrak{C})$ . Thus, we deduce that

$$\begin{eqnarray}K_{a,b}^{m+1,r}(X;\unicode[STIX]{x1D709},\unicode[STIX]{x1D702})\ll H^{\dagger }(X;\unicode[STIX]{x1D709},\unicode[STIX]{x1D702}),\end{eqnarray}$$

where $H^{\dagger }(X;\unicode[STIX]{x1D709},\unicode[STIX]{x1D702})$ denotes the number of solutions of (2.14) subject to the additional condition $\mathbf{x}\equiv \mathbf{y}~(\text{mod}~p^{b^{\prime }})$ . Hence, on considering the underlying Diophantine systems and recalling (6.1), we discern that

(6.6) $$\begin{eqnarray}K_{a,b}^{m+1,r}(X;\unicode[STIX]{x1D709},\unicode[STIX]{x1D702})\ll \oint \mathop{\mathfrak{H}}_{a,b^{\prime }}^{m+1}(\unicode[STIX]{x1D736};\unicode[STIX]{x1D709})|\mathfrak{F}_{m}^{\ast }(\unicode[STIX]{x1D736})|^{2}\,d\unicode[STIX]{x1D736}.\end{eqnarray}$$

An inspection of the definition of $\unicode[STIX]{x1D6EF}_{a}^{m}(\unicode[STIX]{x1D709})$ in the preamble to (2.10) reveals on this occasion that when $\unicode[STIX]{x1D743}\in \unicode[STIX]{x1D6EF}_{a}^{m+1}(\unicode[STIX]{x1D709})$ , then

$$\begin{eqnarray}(\unicode[STIX]{x1D709}_{1},\ldots ,\unicode[STIX]{x1D709}_{m})\in \unicode[STIX]{x1D6EF}_{a}^{m}(\unicode[STIX]{x1D709})\quad \text{and}\quad (\unicode[STIX]{x1D709}_{m+1})\in \unicode[STIX]{x1D6EF}_{a}^{1}(\unicode[STIX]{x1D709}).\end{eqnarray}$$

Then a further consideration of the underlying Diophantine systems leads from (6.6) via (6.1) to the upper bound

$$\begin{eqnarray}K_{a,b}^{m+1,r}(X;\unicode[STIX]{x1D709},\unicode[STIX]{x1D702})\ll \oint \mathop{\mathfrak{H}}_{a,b^{\prime }}^{m}(\unicode[STIX]{x1D736};\unicode[STIX]{x1D709})\mathfrak{H}_{a,b^{\prime }}^{1}(\unicode[STIX]{x1D736};\unicode[STIX]{x1D709})|\mathfrak{F}_{m}^{\ast }(\unicode[STIX]{x1D736})|^{2}\,d\unicode[STIX]{x1D736}.\end{eqnarray}$$

By applying Hölder’s inequality to the integral on the right-hand side of this relation, bearing in mind the definition (6.5), we obtain the bound

(6.7) $$\begin{eqnarray}K_{a,b}^{m+1,r}(X;\unicode[STIX]{x1D709},\unicode[STIX]{x1D702})\ll U_{1}^{\unicode[STIX]{x1D714}_{1}}U_{2}^{\unicode[STIX]{x1D714}_{2}}U_{3}^{\unicode[STIX]{x1D714}_{3}},\end{eqnarray}$$

where

(6.8) $$\begin{eqnarray}\unicode[STIX]{x1D714}_{1}=\frac{s-m-1}{s-m},\qquad \unicode[STIX]{x1D714}_{2}=\frac{1}{s},\qquad \unicode[STIX]{x1D714}_{3}=\frac{m}{s(s-m)}\end{eqnarray}$$

and

(6.9) $$\begin{eqnarray}\displaystyle & \displaystyle U_{1}=\oint \mathop{\mathfrak{H}}_{a,b^{\prime }}^{m}(\unicode[STIX]{x1D736};\unicode[STIX]{x1D709})|\mathfrak{F}_{b}^{r}(\unicode[STIX]{x1D736};\unicode[STIX]{x1D702})^{2}\mathfrak{f}_{b}(\unicode[STIX]{x1D736};\unicode[STIX]{x1D702})^{2s-2m}|\,d\unicode[STIX]{x1D736}, & \displaystyle\end{eqnarray}$$

(6.10) $$\begin{eqnarray}\displaystyle & \displaystyle U_{2}=\oint |\mathop{\mathfrak{F}}_{b}^{r}(\unicode[STIX]{x1D736};\unicode[STIX]{x1D702})|^{2}\mathfrak{H}_{a,b^{\prime }}^{1}(\unicode[STIX]{x1D736};\unicode[STIX]{x1D709})^{s}\,d\unicode[STIX]{x1D736}, & \displaystyle\end{eqnarray}$$

(6.11) $$\begin{eqnarray}\displaystyle & \displaystyle U_{3}=\oint |\mathop{\mathfrak{F}}_{b}^{r}(\unicode[STIX]{x1D736};\unicode[STIX]{x1D702})|^{2}\mathfrak{H}_{a,b^{\prime }}^{m}(\unicode[STIX]{x1D736};\unicode[STIX]{x1D709})^{s/m}\,d\unicode[STIX]{x1D736}. & \displaystyle\end{eqnarray}$$

We now relate the mean values $U_{i}$ to those introduced in §2. Observe first that a consideration of the underlying Diophantine systems leads from (6.9) via (6.1) and (2.10) to the upper bound

$$\begin{eqnarray}U_{1}\leqslant \oint |\mathop{\mathfrak{F}}_{a}^{m}(\unicode[STIX]{x1D736};\unicode[STIX]{x1D709})^{2}\mathfrak{F}_{b}^{r}(\unicode[STIX]{x1D736};\unicode[STIX]{x1D702})^{2}\mathfrak{f}_{b}(\unicode[STIX]{x1D736};\unicode[STIX]{x1D702})^{2s-2m}|\,d\unicode[STIX]{x1D736}.\end{eqnarray}$$

On recalling (2.12) and (2.17), we thus deduce that

(6.12) $$\begin{eqnarray}U_{1}\leqslant K_{a,b}^{m,r}(X).\end{eqnarray}$$

Next, by employing (6.2) within (6.10) and (6.11), we find that

$$\begin{eqnarray}U_{2}+U_{3}\ll (M^{b^{\prime }-a})^{s}\max _{\substack{ 1\leqslant \unicode[STIX]{x1D701}\leqslant p^{b^{\prime }} \\ \unicode[STIX]{x1D701}\equiv \unicode[STIX]{x1D709}~(\text{mod}~p^{a})}}\oint |\mathop{\mathfrak{F}}_{b}^{r}(\unicode[STIX]{x1D736};\unicode[STIX]{x1D702})^{2}\mathfrak{f}_{b^{\prime }}(\unicode[STIX]{x1D736};\unicode[STIX]{x1D701})^{2s}|\,d\unicode[STIX]{x1D736}.\end{eqnarray}$$

Notice here that since the condition (6.3) implies that $\unicode[STIX]{x1D702}\not \equiv \unicode[STIX]{x1D709}~(\text{mod}~p)$ , and we have $\unicode[STIX]{x1D701}\equiv \unicode[STIX]{x1D709}~(\text{mod}~p^{a})$ with $a\geqslant 1$ , then once more one has $\unicode[STIX]{x1D701}\not \equiv \unicode[STIX]{x1D702}~(\text{mod}~p)$ . In this way we deduce from (2.11) and (2.16) that

(6.13) $$\begin{eqnarray}U_{2}+U_{3}\ll (M^{b^{\prime }-a})^{s}I_{b,b^{\prime }}^{r,r}(X).\end{eqnarray}$$

By substituting (6.12) and (6.13) into the relation

$$\begin{eqnarray}K_{a,b}^{m+1,r}(X;\unicode[STIX]{x1D709},\unicode[STIX]{x1D702})\ll U_{1}^{\unicode[STIX]{x1D714}_{1}}(U_{2}+U_{3})^{1-\unicode[STIX]{x1D714}_{1}},\end{eqnarray}$$

that is immediate from (6.7), and then recalling (6.8) and (2.17), the conclusion of the lemma follows when $a\geqslant 1$ . When $a=0$ , we must modify this argument slightly. In this case, from (2.19) and (2.20), we find that

$$\begin{eqnarray}K_{0,b}^{m+1,r}(X)=\max _{1\leqslant \unicode[STIX]{x1D702}\leqslant p^{b}}\oint |\mathfrak{F}^{m+1}(\unicode[STIX]{x1D736};\unicode[STIX]{x1D702})^{2}\mathfrak{F}_{b}^{r}(\unicode[STIX]{x1D736};\unicode[STIX]{x1D702})^{2}\mathfrak{f}_{b}(\unicode[STIX]{x1D736};\unicode[STIX]{x1D702})^{2s-2m-2}|\,d\unicode[STIX]{x1D736}.\end{eqnarray}$$

The desired conclusion follows in this instance by pursuing the proof given above in the case $a\geqslant 1$ , noting that the definition of $\mathfrak{F}^{m+1}(\unicode[STIX]{x1D736};\unicode[STIX]{x1D702})$ ensures that the variables resulting from the congruencing argument avoid the congruence class $\unicode[STIX]{x1D702}$ modulo $p$ . This completes our account of the proof of the lemma.◻

Proof. We prove by backwards induction that for $0\leqslant l\leqslant r-1$ , one has

(7.4) $$\begin{eqnarray}K_{a,b}^{r,r}(X)\ll (K_{a,b}^{l,r}(X))^{\unicode[STIX]{x1D719}_{l}^{\ast }}\mathop{\prod }_{m=l}^{r-1}((M^{b_{m}-a})^{s}I_{b,b_{m}}^{r,r}(X))^{\unicode[STIX]{x1D719}_{m}},\end{eqnarray}$$

where

$$\begin{eqnarray}\unicode[STIX]{x1D719}_{l}^{\ast }=(s-r)/(s-l).\end{eqnarray}$$

The conclusion of the lemma follows from the case $l=0$ of (7.4), on noting that Lemma 2.2 delivers the estimate $K_{a,b}^{0,r}(X)\ll J_{s+r}(X/M^{b})$ .

We observe first that the inductive hypothesis (7.4) holds when $l=r-1$ , as a consequence of the case $m=r-1$ of Lemma 6.1 and the definitions (7.1) and (7.2). Suppose then that $J$ is a non-negative integer not exceeding $r-2$ , and that the inductive hypothesis (7.4) holds for $J<l\leqslant r-1$ . An application of Lemma 6.1 yields the estimate

$$\begin{eqnarray}K_{a,b}^{J+1,r}(X)\ll ((M^{b_{J}-a})^{s}I_{b,b_{J}}^{r,r}(X))^{1/(s-J)}(K_{a,b}^{J,r}(X))^{(s-J-1)/(s-J)}.\end{eqnarray}$$

On substituting this bound into the estimate (7.4) with $l=J+1$ , one obtains the new upper bound

$$\begin{eqnarray}K_{a,b}^{r,r}(X)\ll (K_{a,b}^{J,r}(X))^{\unicode[STIX]{x1D719}_{J}^{\ast }}\mathop{\prod }_{m=J}^{r-1}((M^{b_{m}-a})^{s}I_{b,b_{m}}^{r,r}(X))^{\unicode[STIX]{x1D719}_{m}}\end{eqnarray}$$

and thus the inductive hypothesis holds with $l=J$ . This completes the inductive step, so in view of our earlier remarks, the conclusion of the lemma now follows.◻

Proof. We find from Lemma 7.1 in combination with (2.23) that

(7.5) $$\begin{eqnarray}\unicode[STIX]{x27E6}K_{a,b}^{r,r}(X)\unicode[STIX]{x27E7}\ll M^{\unicode[STIX]{x1D707}^{\ast }b+\unicode[STIX]{x1D708}^{\ast }a}((X/M^{b})^{\unicode[STIX]{x1D6EC}+\unicode[STIX]{x1D6FF}})^{\unicode[STIX]{x1D719}^{\ast }}\mathop{\prod }_{m=0}^{r}\unicode[STIX]{x27E6}I_{b,b_{m}}^{r,r}(X)\unicode[STIX]{x27E7}^{\unicode[STIX]{x1D719}_{m}},\end{eqnarray}$$

where

$$\begin{eqnarray}\unicode[STIX]{x1D707}^{\ast }=s-\unicode[STIX]{x1D719}^{\ast }(s+r)-r\mathop{\sum }_{m=0}^{r-1}\unicode[STIX]{x1D719}_{m}\quad \text{and}\quad \unicode[STIX]{x1D708}^{\ast }=r-s\mathop{\sum }_{m=0}^{r-1}\unicode[STIX]{x1D719}_{m}.\end{eqnarray}$$

On recalling (7.2) and (7.3), a modicum of computation confirms that

$$\begin{eqnarray}\unicode[STIX]{x1D707}^{\ast }=s-\unicode[STIX]{x1D719}^{\ast }(s+r)-r(1-\unicode[STIX]{x1D719}^{\ast })=0\quad \text{and}\quad \unicode[STIX]{x1D708}^{\ast }=r-s(1-\unicode[STIX]{x1D719}^{\ast })=0.\end{eqnarray}$$

The conclusion of the lemma is therefore immediate from (7.5). ◻

Proof. We prove by induction on $l$ that when $1\leqslant l\leqslant R$ , one has the upper bound

(7.9) $$\begin{eqnarray}(X/M^{b})^{-\unicode[STIX]{x1D6EC}-\unicode[STIX]{x1D6FF}}\unicode[STIX]{x27E6}K_{a,b}^{r,r}(X)\unicode[STIX]{x27E7}\ll \mathop{\prod }_{\mathbf{m}\in [0,r-1]^{l}}\unicode[STIX]{x1D6E9}_{l}(\mathbf{m};\mathbf{h})^{\unicode[STIX]{x1D719}_{m_{1}}\cdots \unicode[STIX]{x1D719}_{m_{l}}}.\end{eqnarray}$$

We observe first that, as a consequence of (7.3) and Lemma 7.2, one has

(7.10) $$\begin{eqnarray}(X/M^{b})^{-\unicode[STIX]{x1D6EC}-\unicode[STIX]{x1D6FF}}\unicode[STIX]{x27E6}K_{a,b}^{r,r}(X)\unicode[STIX]{x27E7}\ll \mathop{\prod }_{m=0}^{r-1}((X/M^{b})^{-\unicode[STIX]{x1D6EC}-\unicode[STIX]{x1D6FF}}\unicode[STIX]{x27E6}I_{b,b_{m}^{\ast }}^{r,r}(X)\unicode[STIX]{x27E7})^{\unicode[STIX]{x1D719}_{m}},\end{eqnarray}$$

where

$$\begin{eqnarray}b_{m}^{\ast }=(k-m)b-m\lfloor b/\sqrt{k}\rfloor .\end{eqnarray}$$

Moreover, it follows from Lemma 4.2 that for each $m$ with $0\leqslant m\leqslant r-1$ , there exists an integer $h=h(m)$ , with $0\leqslant h\leqslant 15k^{R}b$ , having the property that

$$\begin{eqnarray}I_{b,b_{m}^{\ast }}^{r,r}(X)\ll (M^{h})^{2s/3}K_{b,b_{m}^{\ast }+h}^{r,r}(X)+(M^{15k^{R}b})^{-s/4}(X/M^{b_{m}^{\ast }})^{s}(X/M^{b})^{\unicode[STIX]{x1D706}-s},\end{eqnarray}$$

whence from (2.22) and (2.23) we discern that

(7.11) $$\begin{eqnarray}\unicode[STIX]{x27E6}I_{b,b_{m}^{\ast }}^{r,r}(X)\unicode[STIX]{x27E7}\ll M^{-hs/3}\unicode[STIX]{x27E6}K_{b,b_{1}}^{r,r}(X)\unicode[STIX]{x27E7}+M^{-3sk^{R}b}(X/M^{b})^{\unicode[STIX]{x1D6EC}+\unicode[STIX]{x1D6FF}},\end{eqnarray}$$

where

$$\begin{eqnarray}b_{1}=(k-m)b_{0}-ma_{0}+h(m).\end{eqnarray}$$

The inductive hypothesis (7.9) therefore follows in the case $R=1$ by substituting (7.11) into (7.10). Notice here that we have discarded the factor $M^{-hs/3}$ from (7.11), since at this stage of our argument it serves only as an unwelcome complication.

Suppose next that the inductive hypothesis (7.9) holds for $1\leqslant l<L$ , with some integer $L$ satisfying $L\leqslant R$ . Consider the quantity $\unicode[STIX]{x1D6E9}_{L-1}(\mathbf{m};\mathbf{h})$ for a given tuple $\mathbf{m}\in [0,r-1]^{L-1}$ and a fixed tuple $\mathbf{h}=\mathbf{h}(\mathbf{m})$ . We distinguish two possibilities. If it is the case that

$$\begin{eqnarray}\unicode[STIX]{x1D6E9}_{L-1}(\mathbf{m};\mathbf{h})\ll M^{-3sk^{R}b},\end{eqnarray}$$

then from (7.3) one finds that

$$\begin{eqnarray}\unicode[STIX]{x1D6E9}_{L-1}(\mathbf{m};\mathbf{h})\ll \biggl(\mathop{\prod }_{m_{L}=0}^{r-1}(M^{-3sk^{R}b})^{\unicode[STIX]{x1D719}_{m_{L}}}\biggr)^{s/r},\end{eqnarray}$$

so that

(7.12) $$\begin{eqnarray}\unicode[STIX]{x1D6E9}_{L-1}(\mathbf{m};\mathbf{h})\ll \mathop{\prod }_{m_{L}=0}^{r-1}(M^{-3sk^{R}b})^{\unicode[STIX]{x1D719}_{m_{L}}}\ll \mathop{\prod }_{m_{L}=0}^{r-1}\unicode[STIX]{x1D6E9}_{L}(\mathbf{m},m_{L};\mathbf{h},0)^{\unicode[STIX]{x1D719}_{m_{L}}}.\end{eqnarray}$$

When

$$\begin{eqnarray}\unicode[STIX]{x1D6E9}_{L-1}(\mathbf{m};\mathbf{h})\ll (X/M^{b})^{-\unicode[STIX]{x1D6EC}-\unicode[STIX]{x1D6FF}}\unicode[STIX]{x27E6}K_{a_{L-1},b_{L-1}}^{r,r}(X)\unicode[STIX]{x27E7},\end{eqnarray}$$

meanwhile, we apply Lemma 7.2 to obtain the bound

(7.13) $$\begin{eqnarray}\unicode[STIX]{x1D6E9}_{L-1}(\mathbf{m};\mathbf{h})\ll \biggl(\frac{X}{M^{b}}\biggr)^{-\unicode[STIX]{x1D6EC}-\unicode[STIX]{x1D6FF}}\biggl(\frac{X}{M^{b_{L-1}}}\biggr)^{(\unicode[STIX]{x1D6EC}+\unicode[STIX]{x1D6FF})(1-r/s)}\mathop{\prod }_{m_{L}=0}^{r-1}\unicode[STIX]{x27E6}I_{b_{L-1},b_{m_{L}}^{\ast }}^{r,r}(X)\unicode[STIX]{x27E7}^{\unicode[STIX]{x1D719}_{m_{L}}},\end{eqnarray}$$

where

$$\begin{eqnarray}b_{m_{L}}^{\ast }=(k-m_{L})b_{L-1}-ma_{L-1}.\end{eqnarray}$$

Two observations here serve to confirm that Lemma 7.2 is applicable. First, the hypothesis $b\leqslant (16Rk^{2R}\unicode[STIX]{x1D703})^{-1}$ ensures that $b_{L-1}\leqslant \unicode[STIX]{x1D703}^{-1}$ . Second, just as in the discussion concluding §6, an inductive argument shows that $a_{n}\leqslant b_{n}/\sqrt{k}$ for each $n$ , for the hypothesis $1\leqslant r\leqslant r_{0}=k-\lceil 2\sqrt{k}\rceil +2$ ensures that whenever $a_{n-1}\leqslant b_{n-1}/\sqrt{k}$ , then

$$\begin{eqnarray}k-m_{n}-m_{n}a_{n-1}/b_{n-1}\geqslant k-(k-2\sqrt{k}+1)-(k-2\sqrt{k}+1)/\sqrt{k}>\sqrt{k},\end{eqnarray}$$

whence $b_{n}>\sqrt{k}b_{n-1}=\sqrt{k}a_{n}$ . The claimed relation therefore follows by induction from (7.6), and in particular one obtains

$$\begin{eqnarray}(k-r+1)b_{L-1}-ra_{L-1}\geqslant b_{L-1}(k-(k-2\sqrt{k}+1)-(k-2\sqrt{k}+1)/\sqrt{k})>0.\end{eqnarray}$$

The applicability of Lemma 7.2 now confirmed, we again invoke Lemma 4.2, finding that for each integer $m_{L}$ with $0\leqslant m_{L}\leqslant r-1$ , there exists an integer $h_{L}=h_{L}(\mathbf{m},m_{L})$ , with $0\leqslant h_{L}\leqslant 15k^{R}b$ , having the property that

$$\begin{eqnarray}\unicode[STIX]{x27E6}I_{b_{L-1},b_{m_{L}}^{\ast }}^{r,r}(X)\unicode[STIX]{x27E7}\ll M^{-h_{L}s/3}\unicode[STIX]{x27E6}K_{b_{L-1},b_{L}}^{r,r}(X)\unicode[STIX]{x27E7}+M^{-3sk^{R}b}(X/M^{b_{L-1}})^{\unicode[STIX]{x1D6EC}+\unicode[STIX]{x1D6FF}}.\end{eqnarray}$$

Thus, we deduce that

$$\begin{eqnarray}\displaystyle (X/M^{b})^{-\unicode[STIX]{x1D6EC}-\unicode[STIX]{x1D6FF}}\unicode[STIX]{x27E6}I_{b_{L-1},b_{m_{L}}^{\ast }}^{r,r}(X)\unicode[STIX]{x27E7} & \ll & \displaystyle (X/M^{b})^{-\unicode[STIX]{x1D6EC}-\unicode[STIX]{x1D6FF}}\unicode[STIX]{x27E6}K_{a_{L},b_{L}}^{r,r}(X)\unicode[STIX]{x27E7}+M^{-3sk^{R}b}\nonumber\\ \displaystyle & \ll & \displaystyle \unicode[STIX]{x1D6E9}_{L}(\mathbf{m},m_{L};\mathbf{h},h_{L}).\nonumber\end{eqnarray}$$

On substituting this estimate into (7.13), we deduce from (7.3) that

(7.14) $$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D6E9}_{L-1}(\mathbf{m};\mathbf{h}) & \ll & \displaystyle (M^{b-b_{L-1}})^{(1-r/s)(\unicode[STIX]{x1D6EC}+\unicode[STIX]{x1D6FF})}\mathop{\prod }_{m_{L}=0}^{r-1}\unicode[STIX]{x1D6E9}_{L}(\mathbf{m},m_{L};\mathbf{h},h_{L})^{\unicode[STIX]{x1D719}_{m_{L}}}\nonumber\\ \displaystyle & \ll & \displaystyle \mathop{\prod }_{m_{L}=0}^{r-1}\unicode[STIX]{x1D6E9}_{L}(\mathbf{m},m_{L};\mathbf{h},h_{L})^{\unicode[STIX]{x1D719}_{m_{L}}}.\end{eqnarray}$$

Applying this estimate in combination with (7.12) within the case $l=L-1$ of the inductive hypothesis (7.9), we conclude that for some choice of the integer $h_{L}=h_{L}(\mathbf{m})$ , one has the upper bound

$$\begin{eqnarray}(X/M^{b})^{-\unicode[STIX]{x1D6EC}-\unicode[STIX]{x1D6FF}}\unicode[STIX]{x27E6}K_{a,b}(X)\unicode[STIX]{x27E7}\ll \mathop{\prod }_{\mathbf{m}\in [0,r-1]^{L-1}}\mathop{\prod }_{m_{L}=0}^{r-1}\unicode[STIX]{x1D6E9}_{L}(\mathbf{m},m_{L};\mathbf{h},h_{L})^{\unicode[STIX]{x1D719}_{m_{1}}\cdots \unicode[STIX]{x1D719}_{m_{L}}}.\end{eqnarray}$$

This confirms the inductive hypothesis (7.9) for $l=L$ , and thus the conclusion of the lemma follows by induction.◻

Proof. This is [Reference Wooley19, Lemma 8.1]. ◻

Proof. We deduce from the postulated bound (8.5) and Lemma 7.3 that there exists a choice of the tuple $\mathbf{h}=\mathbf{h}(\mathbf{m})$ , with $0\leqslant h_{n}(\mathbf{m})\leqslant 15k^{R}b$ $(1\leqslant n\leqslant R)$ , such that

$$\begin{eqnarray}X^{\unicode[STIX]{x1D6EC}}M^{\unicode[STIX]{x1D6EC}\unicode[STIX]{x1D713}}\ll X^{(c+1)\unicode[STIX]{x1D6FF}}M^{-\unicode[STIX]{x1D6FE}}(X/M^{b})^{\unicode[STIX]{x1D6EC}}\mathop{\prod }_{\mathbf{m}\in [0,r-1]^{R}}\unicode[STIX]{x1D6E9}_{R}(\mathbf{m};\mathbf{h})^{\unicode[STIX]{x1D719}_{m_{1}}\cdots \unicode[STIX]{x1D719}_{m_{R}}}.\end{eqnarray}$$

Consequently, one has

$$\begin{eqnarray}\mathop{\prod }_{\mathbf{m}\in [0,r-1]^{R}}\unicode[STIX]{x1D6E9}_{R}(\mathbf{m};\mathbf{h})^{\unicode[STIX]{x1D719}_{m_{1}}\cdots \unicode[STIX]{x1D719}_{m_{R}}}\gg X^{-(c+1)\unicode[STIX]{x1D6FF}}M^{\unicode[STIX]{x1D6EC}(\unicode[STIX]{x1D713}+b)+\unicode[STIX]{x1D6FE}}.\end{eqnarray}$$

Note that, in view of (7.3), one has

(8.7) $$\begin{eqnarray}\mathop{\sum }_{m=0}^{r-1}\unicode[STIX]{x1D719}_{m}=r/s,\end{eqnarray}$$

so that

$$\begin{eqnarray}\mathop{\sum }_{\mathbf{m}\in [0,r-1]^{R}}\unicode[STIX]{x1D719}_{m_{1}}\cdots \unicode[STIX]{x1D719}_{m_{R}}=(r/s)^{R}\leqslant r/s.\end{eqnarray}$$

Then we deduce from the definition (7.8) of $\unicode[STIX]{x1D6E9}_{n}(\mathbf{m};\mathbf{h})$ that

(8.8) $$\begin{eqnarray}\displaystyle & & \displaystyle \mathop{\prod }_{\mathbf{m}\in [0,r-1]^{R}}(X^{-\unicode[STIX]{x1D6EC}}\unicode[STIX]{x27E6}K_{a_{R},b_{R}}^{r,r}(X)\unicode[STIX]{x27E7}+M^{-3sk^{R}b})^{\unicode[STIX]{x1D719}_{m_{1}}\cdots \unicode[STIX]{x1D719}_{m_{R}}}\nonumber\\ \displaystyle & & \displaystyle \quad \gg X^{-(c+1)\unicode[STIX]{x1D6FF}}M^{\unicode[STIX]{x1D6EC}(\unicode[STIX]{x1D713}+(1-r/s)b)+\unicode[STIX]{x1D6FE}}.\end{eqnarray}$$

In preparation for our application of Lemma 8.1, we examine the exponents $\unicode[STIX]{x1D719}_{m_{1}}\cdots \unicode[STIX]{x1D719}_{m_{R}}$ occurring in the lower bound (8.8). Our plan is to apply Lemma 8.1 with exponents given by

$$\begin{eqnarray}\unicode[STIX]{x1D6FD}_{\mathbf{m}}^{(n)}=\unicode[STIX]{x1D719}_{m_{1}}\cdots \unicode[STIX]{x1D719}_{m_{n}}\quad \text{and}\quad \unicode[STIX]{x1D6FE}_{\mathbf{m}}^{(n)}=\tilde{b}_{n}(\mathbf{m})\quad (\mathbf{m}\in [0,r-1]^{n}),\end{eqnarray}$$

in the natural sense. In order to analyze the quantity $\unicode[STIX]{x1D6FA}$ that will emerge from the application of this lemma, we define

$$\begin{eqnarray}B_{n}=\mathop{\sum }_{\mathbf{m}\in [0,r-1]^{n}}\unicode[STIX]{x1D6FD}_{\mathbf{m}}^{(n)}\tilde{b}_{n}(\mathbf{m})\quad \text{and}\quad A_{n}=\mathop{\sum }_{\mathbf{m}\in [0,r-1]^{n}}\unicode[STIX]{x1D6FD}_{\mathbf{m}}^{(n)}\tilde{a}_{n}(\mathbf{m}),\end{eqnarray}$$

and then put $\unicode[STIX]{x1D6FA}=B_{R}$ . From the iterative formulae (8.2) and (8.3), we obtain for $n\geqslant 1$ the relations

(8.9) $$\begin{eqnarray}\displaystyle B_{n+1} & = & \displaystyle \mathop{\sum }_{m_{n+1}=0}^{r-1}~\mathop{\sum }_{\mathbf{m}\in [0,r-1]^{n}}((k-m_{n+1})\tilde{b}_{n}(\mathbf{m})-m_{n+1}\tilde{a}_{n}(\mathbf{m}))\unicode[STIX]{x1D719}_{m_{1}}\cdots \unicode[STIX]{x1D719}_{m_{n+1}}\nonumber\\ \displaystyle & = & \displaystyle B_{n}\mathop{\sum }_{m=0}^{r-1}(k-m)\unicode[STIX]{x1D719}_{m}-A_{n}\mathop{\sum }_{m=0}^{r-1}m\unicode[STIX]{x1D719}_{m}\end{eqnarray}$$

and

(8.10) $$\begin{eqnarray}A_{n+1}=\mathop{\sum }_{m_{n+1}=0}^{r-1}\mathop{\sum }_{\mathbf{m}\in [0,r-1]^{n}}\tilde{b}_{n}(\mathbf{m})\unicode[STIX]{x1D719}_{m_{1}}\cdots \unicode[STIX]{x1D719}_{m_{n+1}}=B_{n}\mathop{\sum }_{m=0}^{r-1}\unicode[STIX]{x1D719}_{m}.\end{eqnarray}$$

We observe that from (7.1), one has

$$\begin{eqnarray}\mathop{\sum }_{m=0}^{r-1}(k-m)\unicode[STIX]{x1D719}_{m}=\mathop{\sum }_{m=0}^{r-1}\frac{(k-m)(s-r)}{(s-m)(s-m-1)}.\end{eqnarray}$$

The summands on the right-hand side here may be rewritten in the shape

$$\begin{eqnarray}\frac{(r-m)(k-m)}{s-m}-\frac{(r-m-1)(k-m-1)}{s-m-1}-\frac{r-m-1}{s-m-1}.\end{eqnarray}$$

Hence, we obtain

$$\begin{eqnarray}s\mathop{\sum }_{m=0}^{r-1}(k-m)\unicode[STIX]{x1D719}_{m}=kr-s\mathop{\sum }_{l=1}^{r}\frac{r-l}{s-l}=kr-\frac{1}{2}r(r-1)-\unicode[STIX]{x1D6E5},\end{eqnarray}$$

in which $\unicode[STIX]{x1D6E5}$ is defined via (2.3). In addition, one has

$$\begin{eqnarray}\displaystyle s\mathop{\sum }_{m=0}^{r-1}m\unicode[STIX]{x1D719}_{m} & = & \displaystyle sk\mathop{\sum }_{m=0}^{r-1}\unicode[STIX]{x1D719}_{m}-s\mathop{\sum }_{m=0}^{r-1}(k-m)\unicode[STIX]{x1D719}_{m}\nonumber\\ \displaystyle & = & \displaystyle kr-\biggl(kr-\frac{1}{2}r(r-1)-\unicode[STIX]{x1D6E5}\biggr)\nonumber\\ \displaystyle & = & \displaystyle \frac{1}{2}r(r-1)+\unicode[STIX]{x1D6E5}.\nonumber\end{eqnarray}$$

Utilizing the formulae just obtained, we conclude from (8.9) that for $n\geqslant 1$ , one has

$$\begin{eqnarray}sB_{n+1}=(kr-{\textstyle \frac{1}{2}}r(r-1)-\unicode[STIX]{x1D6E5})B_{n}-({\textstyle \frac{1}{2}}r(r-1)+\unicode[STIX]{x1D6E5})A_{n},\end{eqnarray}$$

while from (8.7) and (8.10), one sees that

$$\begin{eqnarray}sA_{n+1}=rB_{n}.\end{eqnarray}$$

Then, on recalling (2.5), we arrive at the iterative relation

(8.11) $$\begin{eqnarray}s^{2}B_{n+2}-s\mathfrak{a}B_{n+1}+\mathfrak{b}B_{n}=0\quad (n\geqslant 1).\end{eqnarray}$$

In addition, also from (2.5) and (8.2), one has the initial data

(8.12) $$\begin{eqnarray}sB_{1}=s\mathop{\sum }_{m=0}^{r-1}((k-m)\tilde{b}_{0}-m\tilde{a}_{0})\unicode[STIX]{x1D719}_{m}=\mathfrak{a}-\mathfrak{b}/(r\sqrt{k})\end{eqnarray}$$

and, from (8.2) and (8.7), one finds that

$$\begin{eqnarray}sA_{1}=s\mathop{\sum }_{m=0}^{r-1}\tilde{b}_{0}\unicode[STIX]{x1D719}_{m}=r.\end{eqnarray}$$

Thus, we see also that

(8.13) $$\begin{eqnarray}\displaystyle s^{2}B_{2} & = & \displaystyle s^{2}\mathop{\sum }_{m=0}^{r-1}((k-m)B_{1}-mA_{1})\unicode[STIX]{x1D719}_{m}=s(\mathfrak{a}B_{1}-\mathfrak{b}A_{1}/r)\nonumber\\ \displaystyle & = & \displaystyle \mathfrak{a}(\mathfrak{a}-\mathfrak{b}/(r\sqrt{k}))-\mathfrak{b}.\end{eqnarray}$$

On recalling (2.6), one finds that the recurrence formula (8.11) has a solution of the shape

$$\begin{eqnarray}s^{n}B_{n}=\unicode[STIX]{x1D70E}_{+}\unicode[STIX]{x1D703}_{+}^{n}+\unicode[STIX]{x1D70E}_{-}\unicode[STIX]{x1D703}_{-}^{n}\quad (n\geqslant 1),\end{eqnarray}$$

where, in view of (8.12) and (8.13), one has

$$\begin{eqnarray}\unicode[STIX]{x1D70E}_{+}\unicode[STIX]{x1D703}_{+}+\unicode[STIX]{x1D70E}_{-}\unicode[STIX]{x1D703}_{-}=sB_{1}=\mathfrak{a}-\mathfrak{b}/(r\sqrt{k})\end{eqnarray}$$

and

$$\begin{eqnarray}\unicode[STIX]{x1D70E}_{+}\unicode[STIX]{x1D703}_{+}^{2}+\unicode[STIX]{x1D70E}_{-}\unicode[STIX]{x1D703}_{-}^{2}=s^{2}B_{2}=\mathfrak{a}(\mathfrak{a}-\mathfrak{b}/(r\sqrt{k}))-\mathfrak{b}.\end{eqnarray}$$

Since $\mathfrak{a}=\unicode[STIX]{x1D703}_{+}+\unicode[STIX]{x1D703}_{-}$ and $\mathfrak{b}=\unicode[STIX]{x1D703}_{+}\unicode[STIX]{x1D703}_{-}$ , we therefore deduce that

$$\begin{eqnarray}s^{n}B_{n}=\frac{\unicode[STIX]{x1D703}_{+}^{n+1}-\unicode[STIX]{x1D703}_{-}^{n+1}}{\unicode[STIX]{x1D703}_{+}-\unicode[STIX]{x1D703}_{-}}-\frac{\unicode[STIX]{x1D703}_{+}\unicode[STIX]{x1D703}_{-}}{r\sqrt{k}}\biggl(\frac{\unicode[STIX]{x1D703}_{+}^{n}-\unicode[STIX]{x1D703}_{-}^{n}}{\unicode[STIX]{x1D703}_{+}-\unicode[STIX]{x1D703}_{-}}\biggr).\end{eqnarray}$$

In particular, on recalling (8.1), we find that $s^{R}B_{R}=s_{0}^{R}$ , so that

(8.14) $$\begin{eqnarray}B_{R}=(s_{0}/s)^{R}.\end{eqnarray}$$

Also, therefore, it follows from (2.7) that $s<s_{0}$ and hence also that $B_{R}>1$ .

Returning now to the application of Lemma 8.1, we note first that $\unicode[STIX]{x1D6FA}=B_{R}$ and hence (8.8) yields the relation

$$\begin{eqnarray}\displaystyle & & \displaystyle \mathop{\sum }_{\mathbf{m}\in [0,r-1]^{R}}(X^{-\unicode[STIX]{x1D6EC}}\unicode[STIX]{x27E6}K_{a_{R},b_{R}}^{r,r}(X)\unicode[STIX]{x27E7}+M^{-3sk^{R}b})^{B_{R}/\tilde{b}_{R}(\mathbf{m})}\nonumber\\ \displaystyle & & \displaystyle \quad \gg X^{-(c+1)\unicode[STIX]{x1D6FF}}M^{\unicode[STIX]{x1D6EC}(\unicode[STIX]{x1D713}+(1-r/s)b)+\unicode[STIX]{x1D6FE}}.\nonumber\end{eqnarray}$$

But, in view of (8.4) and (8.14), one has $\tilde{b}_{R}(\mathbf{m})/B_{R}=\unicode[STIX]{x1D70C}_{\mathbf{m}}$ , and thus we find that for some tuple $\mathbf{m}\in [0,r-1]^{R}$ , one has

$$\begin{eqnarray}X^{-\unicode[STIX]{x1D6EC}}\unicode[STIX]{x27E6}K_{a_{R},b_{R}}^{r,r}(X)\unicode[STIX]{x27E7}+M^{-3sk^{R}b}\gg X^{-\unicode[STIX]{x1D70C}_{\mathbf{m}}(c+1)\unicode[STIX]{x1D6FF}}M^{\unicode[STIX]{x1D6EC}\unicode[STIX]{x1D70C}_{\mathbf{m}}(\unicode[STIX]{x1D713}+(1-r/s)b)+\unicode[STIX]{x1D70C}_{\mathbf{m}}\unicode[STIX]{x1D6FE}},\end{eqnarray}$$

whence

(8.15) $$\begin{eqnarray}X^{-\unicode[STIX]{x1D6EC}}\unicode[STIX]{x27E6}K_{a_{R},b_{R}}^{r,r}(X)\unicode[STIX]{x27E7}+M^{-3sk^{R}b}\gg X^{-c^{\prime }\unicode[STIX]{x1D6FF}}M^{\unicode[STIX]{x1D6EC}\unicode[STIX]{x1D713}^{\prime }+\unicode[STIX]{x1D6FE}^{\prime }}.\end{eqnarray}$$

Two further issues remain to be resolved, the first being the removal of the term $M^{-3sk^{R}b}$ on the left-hand side of (8.15). We observe that the relations (8.3) ensure that $\tilde{b}_{R}(\mathbf{m})\leqslant k^{R}$ , and hence (8.4) reveals that $\unicode[STIX]{x1D70C}_{\mathbf{m}}<\tilde{b}_{R}(\mathbf{m})\leqslant k^{R}$ . Our hypothesis on $c$ therefore ensures that

$$\begin{eqnarray}c^{\prime }\unicode[STIX]{x1D6FF}\leqslant k^{R}(c+1)\unicode[STIX]{x1D6FF}\leqslant k^{R}({\textstyle \frac{1}{2}}\unicode[STIX]{x1D703}+\unicode[STIX]{x1D6FF})\leqslant k^{R}\unicode[STIX]{x1D703},\end{eqnarray}$$

so that

$$\begin{eqnarray}X^{-c^{\prime }\unicode[STIX]{x1D6FF}}M^{\unicode[STIX]{x1D6EC}\unicode[STIX]{x1D713}^{\prime }+\unicode[STIX]{x1D6FE}^{\prime }}\geqslant M^{-k^{R}+\unicode[STIX]{x1D70C}_{\boldsymbol{ m}}\unicode[STIX]{x1D6FE}}>M^{-k^{R}-sk^{R}b}.\end{eqnarray}$$

Since

$$\begin{eqnarray}M^{-3sk^{R}b}\leqslant M^{-2sk^{R}-sk^{R}b},\end{eqnarray}$$

it follows from (8.15) that

(8.16) $$\begin{eqnarray}X^{-\unicode[STIX]{x1D6EC}}\unicode[STIX]{x27E6}K_{a_{R},b_{R}}^{r,r}(X)\unicode[STIX]{x27E7}\gg X^{-c^{\prime }\unicode[STIX]{x1D6FF}}M^{\unicode[STIX]{x1D6EC}\unicode[STIX]{x1D713}^{\prime }+\unicode[STIX]{x1D6FE}^{\prime }}.\end{eqnarray}$$

Our final task consists of extracting appropriate constraints on the parameters $a_{R}$ and $b_{R}$ . We note first that from (7.6), one has $a_{0}=\lfloor b/\sqrt{k}\rfloor \leqslant b/\sqrt{k}$ , so that on writing

$$\begin{eqnarray}h_{1}^{\ast }(\mathbf{m})=h_{1}(\mathbf{m})+m_{1}(b/\sqrt{k}-a_{0})\end{eqnarray}$$

and

$$\begin{eqnarray}h_{n}^{\ast }(\mathbf{m})=h_{n}(\mathbf{m})\quad (n>1),\end{eqnarray}$$

we have

$$\begin{eqnarray}h_{1}^{\ast }(\mathbf{m})\leqslant h_{1}(\mathbf{m})+m_{1}\leqslant h_{1}(\mathbf{m})+k<16k^{R}b\end{eqnarray}$$

and

$$\begin{eqnarray}h_{n}^{\ast }(\mathbf{m})=h_{n}(\mathbf{m})\leqslant 16k^{R}b\quad (n>1).\end{eqnarray}$$

Define the sequences $(a_{n}^{\ast })=(a_{n}^{\ast }(\mathbf{m};\mathbf{h}))$ and $(b_{n}^{\ast })=(b_{n}^{\ast }(\mathbf{m};\mathbf{h}))$ by means of the relations

$$\begin{eqnarray}a_{n}^{\ast }=b_{n-1}^{\ast }\quad \text{and}\quad b_{n}^{\ast }=(k-m_{n})b_{n-1}^{\ast }-m_{n}a_{n-1}^{\ast }+h_{n}^{\ast }(\mathbf{m})\quad (1\leqslant n\leqslant R),\end{eqnarray}$$

where

$$\begin{eqnarray}a_{0}^{\ast }=b/\sqrt{k}\quad \text{and}\quad b_{0}^{\ast }=b.\end{eqnarray}$$

Then a comparison with (7.6) and (7.7) reveals that $a_{n}^{\ast }=a_{n}$ and $b_{n}^{\ast }=b_{n}$ for $n\geqslant 1$ . Moreover, it is apparent from (8.2) and (8.3) that $\tilde{a}_{n}b=a_{n}^{\ast }(\mathbf{m};\mathbf{0})$ and $\tilde{b}_{n}b=b_{n}^{\ast }(\mathbf{m};\mathbf{0})$ for $n\geqslant 0$ . We claim that for $n\geqslant 0$ , one has

(8.17) $$\begin{eqnarray}b_{n}^{\ast }(\mathbf{m};\mathbf{0})\leqslant b_{n}^{\ast }(\mathbf{m};\mathbf{h})\leqslant b_{n}^{\ast }(\mathbf{m};\mathbf{0})+16k^{R+n}b.\end{eqnarray}$$

The validity of these inequalities for $n=0$ is immediate from the definition of $b_{n}^{\ast }(\mathbf{m};\mathbf{h})$ . We use an inductive argument to establish (8.17) for $n\geqslant 1$ , though this requires some discussion.

Observe first that, by linearity, the recurrence sequence $b_{n}^{\ast }(\mathbf{m};\mathbf{h})$ is given by the formula

(8.18) $$\begin{eqnarray}b_{n}^{\ast }(\mathbf{m};\mathbf{h})=b_{n}^{\ast }(\mathbf{m};\mathbf{0})+\mathop{\sum }_{l=1}^{R}c_{n,l}(\mathbf{m};\mathbf{h})\quad (0\leqslant n\leqslant R),\end{eqnarray}$$

in which $c_{n,l}(\mathbf{m};\mathbf{h})$ is determined by the relations

$$\begin{eqnarray}c_{l,l}(\mathbf{m};\mathbf{h})=h_{l}^{\ast }(\mathbf{m})\quad \text{and}\quad c_{n,l}(\mathbf{m};\mathbf{h})=0\quad (n<l),\end{eqnarray}$$

with

(8.19) $$\begin{eqnarray}c_{n+1,l}(\mathbf{m};\mathbf{h})=(k-m_{n+1})c_{n,l}(\mathbf{m};\mathbf{h})-m_{n+1}c_{n-1,l}(\mathbf{m};\mathbf{h})\quad (n\geqslant l).\end{eqnarray}$$

We recall our assumption that $r\leqslant k-\lceil 2\sqrt{k}\rceil +2$ , which ensures that

$$\begin{eqnarray}m_{n+1}\leqslant r-1\leqslant k-\lceil 2\sqrt{k}\rceil +1.\end{eqnarray}$$

In view of this upper bound, we are able to prove by induction that

(8.20) $$\begin{eqnarray}c_{n+1,l}(\mathbf{m};\mathbf{h})>\sqrt{k}c_{n,l}(\mathbf{m};\mathbf{h})\end{eqnarray}$$

for each integer $n$ with $n\geqslant l-1$ . In order to confirm this assertion, note first that such is the case when $n=l-1$ . Granted that $c_{n,l}(\mathbf{m};\mathbf{h})>\sqrt{k}c_{n-1,l}(\mathbf{m};\mathbf{h})$ , meanwhile, one deduces from (8.19) that when $n\geqslant l-1$ , one has

(8.21) $$\begin{eqnarray}\displaystyle c_{n+1,l}(\mathbf{m};\mathbf{h}) & {\geqslant} & \displaystyle (\lceil 2\sqrt{k}\rceil -1)c_{n,l}(\mathbf{m};\mathbf{h})-(k-\lceil 2\sqrt{k}\rceil +1)c_{n,l}(\mathbf{m};\mathbf{h})/\sqrt{k}\nonumber\\ \displaystyle & {\geqslant} & \displaystyle ((2\sqrt{k}-1)-(\sqrt{k}-2+1/\sqrt{k}))c_{n,l}(\mathbf{m};\mathbf{h})\nonumber\\ \displaystyle & {>} & \displaystyle \sqrt{k}c_{n,l}(\mathbf{m};\mathbf{h}),\end{eqnarray}$$

establishing this inductive hypothesis. Thus, in particular, one arrives at the lower bound $c_{n,l}(\mathbf{m};\mathbf{h})\geqslant 0$ for all $n$ . On substituting this conclusion into (8.18), we deduce that $b_{n}^{\ast }(\mathbf{m};\mathbf{h})\geqslant b_{n}^{\ast }(\mathbf{m};\mathbf{0})$ for every $n$ , confirming the first of the inequalities in (8.17).

Next, making use of the lower bound $c_{n,l}(\mathbf{m};\mathbf{h})\geqslant 0$ , just obtained, within (8.19), we find that

$$\begin{eqnarray}c_{n+1,l}(\mathbf{m};\mathbf{h})\leqslant kc_{n,l}(\mathbf{m};\mathbf{h})\quad (n\geqslant l),\end{eqnarray}$$

so that

$$\begin{eqnarray}c_{n,l}(\mathbf{m};\mathbf{h})\leqslant k^{n-l}h_{l}^{\ast }(\mathbf{m})\quad (n\geqslant l).\end{eqnarray}$$

We therefore deduce from (8.18) and the bound $h_{n}^{\ast }(\mathbf{m})\leqslant 16k^{R}b$ that

$$\begin{eqnarray}b_{n}^{\ast }(\mathbf{m};\mathbf{h})\leqslant b_{n}^{\ast }(\mathbf{m};\mathbf{0})+16k^{R}b\mathop{\sum }_{l=1}^{n}k^{n-l}\leqslant b_{n}^{\ast }(\mathbf{m};\mathbf{0})+16k^{R+n}b.\end{eqnarray}$$

This confirms the second of the inequalities in (8.17).

We now make use of the inequalities (8.17). Observe first that in view of (8.4), one has

$$\begin{eqnarray}b_{R}=b_{R}^{\ast }(\mathbf{m};\mathbf{h})\leqslant b_{R}^{\ast }(\mathbf{m};\mathbf{0})+16k^{2R}b=\tilde{b}_{R}(\mathbf{m})b+16k^{2R}b=k_{\boldsymbol{ m}}b+16k^{2R}b.\end{eqnarray}$$

In addition, one has

$$\begin{eqnarray}b_{R}=b_{R}^{\ast }(\mathbf{m};\mathbf{h})\geqslant b_{R}^{\ast }(\mathbf{m};\mathbf{0})=\tilde{b}_{R}(\mathbf{m})b=k_{\mathbf{m}}b.\end{eqnarray}$$

Thus, there exists an integer $h_{\mathbf{m}}$ , with $0\leqslant h_{\mathbf{m}}\leqslant 16k^{2R}b$ , for which one has $b_{R}=k_{\mathbf{m}}b+h_{\mathbf{m}}$ .

The argument confirming (8.21) shows also that for each $n\geqslant 0$ , one has

(8.22) $$\begin{eqnarray}b_{n+1}^{\ast }(\mathbf{m};\mathbf{0})>\sqrt{k}b_{n}^{\ast }(\mathbf{m};\mathbf{0}).\end{eqnarray}$$

Then it follows from (8.18) and (8.20) that

$$\begin{eqnarray}b_{R}^{\ast }(\mathbf{m};\mathbf{h})>\sqrt{k}b_{R-1}^{\ast }(\mathbf{m};\mathbf{h}),\end{eqnarray}$$

whence

$$\begin{eqnarray}a_{R}=b_{R-1}=b_{R-1}^{\ast }(\mathbf{m};\mathbf{h})<b_{R}^{\ast }(\mathbf{m};\mathbf{h})/\sqrt{k}=b_{R}/\sqrt{k}.\end{eqnarray}$$

Then we deduce from (8.16) that there exist integers $b^{\prime }=b_{R}$ , $a^{\prime }=a_{R}$ and $h=h_{\mathbf{m}}$ satisfying the conditions

$$\begin{eqnarray}0\leqslant h_{\mathbf{m}}\leqslant 16k^{2R}b,\qquad b^{\prime }=k_{\boldsymbol{ m}}b+h\qquad \text{and}\qquad a^{\prime }\leqslant b^{\prime }/\sqrt{k},\end{eqnarray}$$

and for which

$$\begin{eqnarray}X^{-\unicode[STIX]{x1D6EC}}\unicode[STIX]{x27E6}K_{a^{\prime },b^{\prime }}^{r,r}(X)\unicode[STIX]{x27E7}\gg X^{-c^{\prime }\unicode[STIX]{x1D6FF}}M^{\unicode[STIX]{x1D6EC}\unicode[STIX]{x1D713}^{\prime }+\unicode[STIX]{x1D6FE}^{\prime }}.\end{eqnarray}$$

The desired conclusion (8.6) now follows, with all the associated conditions.

It now remains only to confirm the bounds on $k_{\mathbf{m}}$ asserted in the final line of the statement of the lemma. For this, we note that by applying (8.22) in an inductive argument, one obtains from (8.4) the lower bound

$$\begin{eqnarray}k_{\mathbf{m}}=\tilde{b}_{R}(\mathbf{m})=b_{R}^{\ast }(\mathbf{m};\mathbf{0})/b\geqslant (\sqrt{k})^{R}b_{0}^{\ast }(\mathbf{m};\mathbf{0})/b\geqslant 2^{R}.\end{eqnarray}$$

Likewise, though more directly, one has

$$\begin{eqnarray}k_{\mathbf{m}}=\tilde{b}_{R}(\mathbf{m})=b_{R}^{\ast }(\mathbf{m};\mathbf{0})/b\leqslant k^{R}b_{0}^{\ast }(\mathbf{m};\mathbf{0})/b=k^{R}.\end{eqnarray}$$

Thus, $2^{R}\leqslant k_{\mathbf{m}}\leqslant k^{R}$ , and the proof of the lemma is complete.◻

Proof. On considering the underlying Diophantine equations, we deduce from Lemma 2.3 that

$$\begin{eqnarray}K_{a,b}^{r,r}(X)\ll (J_{s+r}(X/M^{a}))^{r/(s+r)}(J_{s+r}(X/M^{b}))^{s/(s+r)},\end{eqnarray}$$

whence

$$\begin{eqnarray}\unicode[STIX]{x27E6}K_{a,b}^{r,r}(X)\unicode[STIX]{x27E7}\ll \frac{X^{\unicode[STIX]{x1D6FF}}((X/M^{a})^{r/(s+r)}(X/M^{b})^{s/(s+r)})^{s+r+\unicode[STIX]{x1D6EC}}}{(X/M^{a})^{r}(X/M^{b})^{s}}\leqslant X^{\unicode[STIX]{x1D6EC}+\unicode[STIX]{x1D6FF}}.\end{eqnarray}$$

This completes the proof of the lemma. ◻

The Proof of  Theorem 2.1.

We prove that when $s$ is the largest integer smaller than $s_{1}$ , then one has $\unicode[STIX]{x1D6EC}\leqslant 0$ , for in such circumstances the conclusion of the lemma follows at once from (2.24). By reference to (2.5) and (2.6), we find that $s<s_{1}=\unicode[STIX]{x1D703}_{+}$ . Thus, since $R$ has been chosen sufficiently large in terms of $s$ , $k$ and $\unicode[STIX]{x1D703}_{+}$ , the hypothesis $s<s_{1}$ ensures that $s<s_{0}$ , where $s_{0}$ is defined by means of (8.1). Assume then that $\unicode[STIX]{x1D6EC}\geqslant 0$ , for otherwise there is nothing to prove. We begin by noting that as a consequence of Lemma 5.1, one finds from (2.21) and (2.23) that there exists an integer $h_{-1}$ with $0\leqslant h_{-1}\leqslant 4B$ such that

$$\begin{eqnarray}\unicode[STIX]{x27E6}J_{s+r}(X)\unicode[STIX]{x27E7}\ll M^{Bs-sh_{-1}/3}\unicode[STIX]{x27E6}K_{0,B+h_{-1}}^{r,r}(X)\unicode[STIX]{x27E7}.\end{eqnarray}$$

We therefore deduce from (2.24) that

(9.1) $$\begin{eqnarray}X^{\unicode[STIX]{x1D6EC}}\ll X^{\unicode[STIX]{x1D6FF}}\unicode[STIX]{x27E6}J_{s+r}(X)\unicode[STIX]{x27E7}\ll X^{\unicode[STIX]{x1D6FF}}M^{Bs-sh_{-1}/3}\unicode[STIX]{x27E6}K_{0,B+h_{-1}}^{r,r}(X)\unicode[STIX]{x27E7}.\end{eqnarray}$$

Next, we define sequences $(\unicode[STIX]{x1D705}_{n})$ , $(h_{n})$ , $(a_{n})$ , $(b_{n})$ , $(c_{n})$ , $(\unicode[STIX]{x1D713}_{n})$ and $(\unicode[STIX]{x1D6FE}_{n})$ , for $0\leqslant n\leqslant N$ , in such a way that

(9.2) $$\begin{eqnarray}2^{R}\leqslant \unicode[STIX]{x1D705}_{n-1}\leqslant k^{R},\qquad 0\leqslant h_{n-1}\leqslant \max \{16k^{2R}b_{n-1},4B\}\end{eqnarray}$$

and

(9.3) $$\begin{eqnarray}X^{\unicode[STIX]{x1D6EC}}M^{\unicode[STIX]{x1D6EC}\unicode[STIX]{x1D713}_{n}}\ll X^{c_{n}\unicode[STIX]{x1D6FF}}M^{-\unicode[STIX]{x1D6FE}_{n}}\unicode[STIX]{x27E6}K_{a_{n},b_{n}}^{r,r}(X)\unicode[STIX]{x27E7}.\end{eqnarray}$$

We note here that the sequences $(a_{n})$ and $(b_{n})$ are not directly related to our earlier use of these letters. Given a fixed choice for the sequences $(a_{n})$ , $(\unicode[STIX]{x1D705}_{n})$ and $(h_{n})$ , the remaining sequences are defined by means of the relations

(9.4) $$\begin{eqnarray}\displaystyle b_{n+1} & = & \displaystyle \unicode[STIX]{x1D705}_{n}b_{n}+h_{n},\end{eqnarray}$$

(9.5) $$\begin{eqnarray}\displaystyle c_{n+1} & = & \displaystyle (s/s_{0})^{R}\unicode[STIX]{x1D705}_{n}(c_{n}+1),\end{eqnarray}$$

(9.6) $$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D713}_{n+1} & = & \displaystyle (s/s_{0})^{R}\unicode[STIX]{x1D705}_{n}(\unicode[STIX]{x1D713}_{n}+(1-r/s)b_{n}),\end{eqnarray}$$

(9.7) $$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D6FE}_{n+1} & = & \displaystyle (s/s_{0})^{R}\unicode[STIX]{x1D705}_{n}\unicode[STIX]{x1D6FE}_{n}.\end{eqnarray}$$

We put

$$\begin{eqnarray}\displaystyle & \displaystyle \unicode[STIX]{x1D705}_{-1}=k^{R},\qquad b_{-1}=0,\qquad a_{0}=0,\qquad b_{0}=B+h_{-1}, & \displaystyle \nonumber\\ \displaystyle & \displaystyle \unicode[STIX]{x1D713}_{0}=0,\qquad c_{0}=1,\qquad \unicode[STIX]{x1D6FE}_{0}={\textstyle \frac{1}{3}}sh_{-1}-Bs, & \displaystyle \nonumber\end{eqnarray}$$

so that both (9.2) and (9.3) hold with $n=0$ as a consequence of our initial choice of $\unicode[STIX]{x1D705}_{-1}$ and $b_{-1}$ , together with (9.1). We prove by induction that for each non-negative integer $n$ with $n<N$ , the sequences $(a_{m})_{m=0}^{n}$ , $(\unicode[STIX]{x1D705}_{m})_{m=0}^{n}$ and $(h_{m})_{m=-1}^{n}$ may be chosen in such a way that

(9.8) $$\begin{eqnarray}\displaystyle & \displaystyle 1\leqslant b_{n}\leqslant (20Rk^{2R}\unicode[STIX]{x1D703})^{-1},\quad \unicode[STIX]{x1D713}_{n}\geqslant 0,\quad \unicode[STIX]{x1D6FE}_{n}\geqslant -sb_{n},\quad 0\leqslant c_{n}\leqslant (2\unicode[STIX]{x1D6FF})^{-1}\unicode[STIX]{x1D703},\quad \quad & \displaystyle\end{eqnarray}$$

(9.9) $$\begin{eqnarray}\displaystyle & \displaystyle 0\leqslant a_{n}\leqslant b_{n}/\sqrt{k},\qquad (k-r+1)b_{n}\geqslant ra_{n}, & \displaystyle\end{eqnarray}$$

and so that (9.2) and (9.3) both hold with $n$ replaced by $n+1$ .

Let $0\leqslant n<N$ , and suppose that (9.2) and (9.3) both hold for the index $n$ . We have already shown such to be the case for $n=0$ . From (9.2) and (9.4), we find that $b_{n}\leqslant 5(17k^{2R})^{n}B$ , whence, by invoking (2.8), we find that for $0\leqslant n\leqslant N$ , one has

$$\begin{eqnarray}b_{n}\leqslant (20Rk^{2R}\unicode[STIX]{x1D703})^{-1}.\end{eqnarray}$$

It is apparent from (9.5) and (9.6) that $c_{n}$ and $\unicode[STIX]{x1D713}_{n}$ are non-negative for all $n$ . Observe also that since $s\leqslant s_{0}$ and $\unicode[STIX]{x1D705}_{m}\leqslant k^{R}$ , then by iterating (9.5) we obtain the bound

(9.10) $$\begin{eqnarray}c_{n}\leqslant k^{Rn}+k^{R}\biggl(\frac{k^{Rn}-1}{k^{R}-1}\biggr)\leqslant 3k^{Rn}\quad (n\geqslant 0),\end{eqnarray}$$

and by reference to (2.8) we see that $c_{n}\leqslant (2\unicode[STIX]{x1D6FF})^{-1}\unicode[STIX]{x1D703}$ for $0\leqslant n<N$ .

In order to bound $\unicode[STIX]{x1D6FE}_{n}$ , we recall that $s\leqslant s_{0}$ and iterate the relation (9.7) to deduce that

(9.11) $$\begin{eqnarray}\unicode[STIX]{x1D6FE}_{m}=(s/s_{0})^{Rm}\unicode[STIX]{x1D705}_{0}\cdots \unicode[STIX]{x1D705}_{m-1}\unicode[STIX]{x1D6FE}_{0}\geqslant -(s/s_{0})^{Rm}\unicode[STIX]{x1D705}_{0}\cdots \unicode[STIX]{x1D705}_{m-1}Bs.\end{eqnarray}$$

In addition, we find from (9.4) that for $m\geqslant 0$ one has $b_{m+1}\geqslant \unicode[STIX]{x1D705}_{m}b_{m}$ , so that an inductive argument yields the lower bound

$$\begin{eqnarray}b_{m}\geqslant \unicode[STIX]{x1D705}_{0}\cdots \unicode[STIX]{x1D705}_{m-1}b_{0}\geqslant \unicode[STIX]{x1D705}_{0}\cdots \unicode[STIX]{x1D705}_{m-1}B.\end{eqnarray}$$

Hence, we deduce from (9.11) that

$$\begin{eqnarray}\unicode[STIX]{x1D6FE}_{m}\geqslant -(s/s_{0})^{Rm}sb_{m}>-sb_{m}.\end{eqnarray}$$

Assembling this conclusion together with those of the previous paragraph, we have shown that (9.8) holds for $0\leqslant n\leqslant N$ .

At this point in the argument, we may suppose that (9.3), (9.8) and (9.9) hold for the index $n$ . An application of Lemma 8.2 therefore reveals that there exist real numbers $\unicode[STIX]{x1D705}_{n}$ and $h_{n}$ satisfying the constraints implied by (9.2) with $n$ replaced by $n+1$ , for which the upper bound (9.3) holds for some $a_{n}$ with $0\leqslant a_{n}\leqslant b_{n}/\sqrt{k}$ , also with $n$ replaced by $n+1$ . Our hypothesis on $r$ , moreover, ensures that $(k-r+1)b_{n+1}\geqslant ra_{n+1}$ . Hence (9.9) holds also with $n$ replaced by $n+1$ . This completes the inductive step, so that in particular the upper bound (9.3) holds for $0\leqslant n\leqslant N$ .

We now exploit the bound just established. Since we have the upper bound $b_{N}\leqslant 5(17k^{2R})^{N}B\leqslant (2\unicode[STIX]{x1D703})^{-1}$ , it is a consequence of Lemma 9.1 that

$$\begin{eqnarray}\unicode[STIX]{x27E6}K_{a_{N},b_{N}}^{r,r}(X)\unicode[STIX]{x27E7}\ll X^{\unicode[STIX]{x1D6EC}+\unicode[STIX]{x1D6FF}}.\end{eqnarray}$$

By combining this with (9.3) and (9.11), we obtain the bound

(9.12) $$\begin{eqnarray}X^{\unicode[STIX]{x1D6EC}}M^{\unicode[STIX]{x1D6EC}\unicode[STIX]{x1D713}_{N}}\ll X^{\unicode[STIX]{x1D6EC}+(c_{N}+1)\unicode[STIX]{x1D6FF}}M^{\unicode[STIX]{x1D705}_{0}\cdots \unicode[STIX]{x1D705}_{N-1}Bs(s/s_{0})^{RN}}.\end{eqnarray}$$

Meanwhile, an application of (9.10) in combination with (2.8) shows that $X^{(c_{N}+1)\unicode[STIX]{x1D6FF}}<M$ . We therefore deduce from (9.12) that

$$\begin{eqnarray}\unicode[STIX]{x1D6EC}\unicode[STIX]{x1D713}_{N}\leqslant (s/s_{0})^{RN}\unicode[STIX]{x1D705}_{0}\cdots \unicode[STIX]{x1D705}_{N-1}Bs+1.\end{eqnarray}$$

Notice here that $\unicode[STIX]{x1D705}_{n}\geqslant 2^{R}$ and

$$\begin{eqnarray}s/s_{0}\geqslant (s_{1}-1)/s_{0}\geqslant 1-2/s_{0}>{\textstyle \frac{1}{2}}.\end{eqnarray}$$

Hence,

$$\begin{eqnarray}1<(s/s_{0})^{RN}\unicode[STIX]{x1D705}_{0}\cdots \unicode[STIX]{x1D705}_{N-1}Bs,\end{eqnarray}$$

so that

(9.13) $$\begin{eqnarray}\unicode[STIX]{x1D6EC}\unicode[STIX]{x1D713}_{N}\leqslant 2(s/s_{0})^{RN}\unicode[STIX]{x1D705}_{0}\cdots \unicode[STIX]{x1D705}_{N-1}Bs.\end{eqnarray}$$

A further application of the lower bound $b_{n}\geqslant \unicode[STIX]{x1D705}_{0}\cdots \unicode[STIX]{x1D705}_{n-1}B$ leads from (9.6) and the bound $s\leqslant s_{0}$ to the relation

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D713}_{n+1} & = & \displaystyle (s/s_{0})^{R}(\unicode[STIX]{x1D705}_{n}\unicode[STIX]{x1D713}_{n}+\unicode[STIX]{x1D705}_{n}(1-r/s)b_{n})\nonumber\\ \displaystyle & {\geqslant} & \displaystyle (s/s_{0})^{R}\unicode[STIX]{x1D705}_{n}\unicode[STIX]{x1D713}_{n}+(s/s_{0})^{R}\unicode[STIX]{x1D705}_{n}(1-r/s)\unicode[STIX]{x1D705}_{0}\cdots \unicode[STIX]{x1D705}_{n-1}B\nonumber\\ \displaystyle & {\geqslant} & \displaystyle (s/s_{0})^{R}\unicode[STIX]{x1D705}_{n}\unicode[STIX]{x1D713}_{n}+(s/s_{0})^{R(n+1)}(1-r/s)\unicode[STIX]{x1D705}_{0}\cdots \unicode[STIX]{x1D705}_{n}B.\nonumber\end{eqnarray}$$

An inductive argument therefore delivers the lower bound

$$\begin{eqnarray}\unicode[STIX]{x1D713}_{N}\geqslant N(1-r/s)(s/s_{0})^{RN}\unicode[STIX]{x1D705}_{0}\cdots \unicode[STIX]{x1D705}_{N-1}B.\end{eqnarray}$$

Thus, we deduce from (9.13) that

$$\begin{eqnarray}\unicode[STIX]{x1D6EC}\leqslant \frac{2(s/s_{0})^{RN}\unicode[STIX]{x1D705}_{0}\cdots \unicode[STIX]{x1D705}_{N-1}Bs}{N(1-r/s)(s/s_{0})^{RN}\unicode[STIX]{x1D705}_{0}\cdots \unicode[STIX]{x1D705}_{N-1}B}=\frac{2s(1-r/s)^{-1}}{N}.\end{eqnarray}$$

Since we are at liberty to take $N$ as large as we please in terms of $s$ and $k$ , we are forced to conclude that $\unicode[STIX]{x1D6EC}\leqslant 0$ . In view of our opening discussion, this completes the proof of the theorem.◻

Proof. On considering the underlying Diophantine systems, it follows from (2.12) and (2.14) that for some integers $\unicode[STIX]{x1D709}$ and $\unicode[STIX]{x1D702}$ , one has

$$\begin{eqnarray}K_{a,b}^{m+1,r}(X)\leqslant \oint |\mathop{\mathfrak{F}}_{a}^{m}(\unicode[STIX]{x1D736};\unicode[STIX]{x1D709})^{2}\mathfrak{f}_{a}(\unicode[STIX]{x1D736};\unicode[STIX]{x1D709})^{2}\mathfrak{F}_{b}^{r}(\unicode[STIX]{x1D736};\unicode[STIX]{x1D702})^{2}\mathfrak{f}_{b}(\unicode[STIX]{x1D736};\unicode[STIX]{x1D702})^{2s-2m-2}|\,d\unicode[STIX]{x1D736}.\end{eqnarray}$$

Then, by Hölder’s inequality, we obtain

$$\begin{eqnarray}K_{a,b}^{m+1,r}(X)\leqslant I_{1}^{\unicode[STIX]{x1D713}_{1}}I_{2}^{\unicode[STIX]{x1D713}_{2}}I_{3}^{\unicode[STIX]{x1D713}_{3}}I_{4}^{\unicode[STIX]{x1D713}_{4}}I_{5}^{\unicode[STIX]{x1D713}_{5}},\end{eqnarray}$$

where

$$\begin{eqnarray}\displaystyle & \displaystyle I_{1}=\oint |\mathop{\mathfrak{F}}_{a}^{m}(\unicode[STIX]{x1D736};\unicode[STIX]{x1D709})^{4}\mathfrak{f}_{a}(\unicode[STIX]{x1D736};\unicode[STIX]{x1D709})^{2s+2r-4m}|\,d\unicode[STIX]{x1D736}, & \displaystyle \nonumber\\ \displaystyle & \displaystyle I_{2}=\oint |\mathfrak{f}_{a}(\unicode[STIX]{x1D736};\unicode[STIX]{x1D709})|^{2s+2r}\,d\unicode[STIX]{x1D736},\qquad I_{3}=\oint |\mathfrak{f}_{b}(\unicode[STIX]{x1D736};\unicode[STIX]{x1D702})|^{2s+2r}\,d\unicode[STIX]{x1D736}, & \displaystyle \nonumber\\ \displaystyle & \displaystyle I_{4}=\oint |\mathop{\mathfrak{F}}_{a}^{m}(\unicode[STIX]{x1D736};\unicode[STIX]{x1D709})^{2}\mathfrak{F}_{b}^{r}(\unicode[STIX]{x1D736};\unicode[STIX]{x1D702})^{2}\mathfrak{f}_{b}(\unicode[STIX]{x1D736};\unicode[STIX]{x1D702})^{2s-2m}|\,d\unicode[STIX]{x1D736}, & \displaystyle \nonumber\\ \displaystyle & \displaystyle I_{5}=\oint |\mathop{\mathfrak{F}}_{a}^{m}(\unicode[STIX]{x1D736};\unicode[STIX]{x1D709})^{2}\mathfrak{F}_{b}^{r}(\unicode[STIX]{x1D736};\unicode[STIX]{x1D702})^{4}\mathfrak{f}_{b}(\unicode[STIX]{x1D736};\unicode[STIX]{x1D702})^{2s-2r-2m}|\,d\unicode[STIX]{x1D736}, & \displaystyle \nonumber\\ \displaystyle & \displaystyle \unicode[STIX]{x1D713}_{1}=\frac{1}{2s-2m},\qquad \unicode[STIX]{x1D713}_{2}=\frac{s-r}{(s+r)(2s-2m)},\qquad \unicode[STIX]{x1D713}_{3}=\unicode[STIX]{x1D714}_{2}, & \displaystyle \nonumber\\ \displaystyle & \displaystyle \unicode[STIX]{x1D713}_{4}=\frac{s-m-2}{s-m}\quad \text{and}\quad \unicode[STIX]{x1D713}_{5}=\frac{1}{s-m}. & \displaystyle \nonumber\end{eqnarray}$$

A further consideration of the underlying Diophantine systems reveals that $I_{1}\leqslant I_{2}$ and $I_{5}\leqslant I_{4}$ , and thus it follows from Lemma 2.2 and (2.12) that

$$\begin{eqnarray}K_{a,b}^{m+1,r}(X)\ll (J_{s+r}(X/M^{a}))^{\unicode[STIX]{x1D713}_{1}+\unicode[STIX]{x1D713}_{2}}(J_{s+r}(X/M^{b}))^{\unicode[STIX]{x1D714}_{2}}(K_{a,b}^{m,r}(X))^{\unicode[STIX]{x1D713}_{4}+\unicode[STIX]{x1D713}_{5}}.\end{eqnarray}$$

The conclusion of the lemma follows at once. ◻

Proof. By following the argument of the proof of Lemma 7.1, though substituting Lemma 11.1 in place of Lemma 6.1 when $u\leqslant m\leqslant r-1$ , we deduce that

$$\begin{eqnarray}\displaystyle K_{a,b}^{r,r}(X) & \ll & \displaystyle (J_{s+r}(X/M^{b}))^{\unicode[STIX]{x1D719}^{\ast }}(J_{s+r}(X/M^{a}))^{\unicode[STIX]{x1D711}_{0}}(J_{s+r}(X/M^{b}))^{\unicode[STIX]{x1D711}_{1}}\nonumber\\ \displaystyle & & \displaystyle \,\times \mathop{\prod }_{m=0}^{u-1}((M^{b_{m}-a})^{s}I_{b,b_{m}}^{r,r}(X))^{\unicode[STIX]{x1D719}_{m}}.\nonumber\end{eqnarray}$$

It may be useful here to note that

$$\begin{eqnarray}\mathop{\sum }_{m=u}^{r-1}\unicode[STIX]{x1D719}_{m}=(s-r)\mathop{\sum }_{m=u}^{r-1}(s_{m+1}^{-1}-s_{m}^{-1})=(s-r)\biggl(\frac{1}{s-r}-\frac{1}{s-u}\biggr)=\frac{r-u}{s-u}.\end{eqnarray}$$

Thus, on recalling the definitions (2.21) to (2.23), we obtain the conclusion of the lemma, just as in the argument delivering Lemma 7.2. ◻

Proof. One may follow the inductive strategy adopted in the proof of Lemma 7.3. The key difference in the present setting is that Lemma 11.2 contains additional factors in the estimate made available for $\unicode[STIX]{x27E6}K_{a,b}^{r,r}(X)\unicode[STIX]{x27E7}$ . However, it follows from Lemma 11.2 that whenever $u\leqslant u(a,b)$ , then

$$\begin{eqnarray}(X/M^{b/2})^{-\unicode[STIX]{x1D6EC}-\unicode[STIX]{x1D6FF}}\unicode[STIX]{x27E6}K_{a,b}^{r,r}(X)\unicode[STIX]{x27E7}\ll (M^{(\unicode[STIX]{x1D6EC}+\unicode[STIX]{x1D6FF})b})^{\unicode[STIX]{x1D6FA}}\mathop{\prod }_{m=0}^{u-1}((X/M^{b/2})^{-\unicode[STIX]{x1D6EC}-\unicode[STIX]{x1D6FF}}\unicode[STIX]{x27E6}I_{b,b_{m}}^{r,r}(X)\unicode[STIX]{x27E7})^{\unicode[STIX]{x1D719}_{m}},\end{eqnarray}$$

where

$$\begin{eqnarray}\unicode[STIX]{x1D6FA}=\frac{1}{2}-\frac{1}{2}\mathop{\sum }_{m=0}^{u-1}\unicode[STIX]{x1D719}_{m}-(\unicode[STIX]{x1D719}^{\ast }+\unicode[STIX]{x1D711}_{1}).\end{eqnarray}$$

But we have

$$\begin{eqnarray}\mathop{\sum }_{m=0}^{u-1}\unicode[STIX]{x1D719}_{m}=(s-r)\mathop{\sum }_{m=0}^{u-1}\biggl(\frac{1}{s-m-1}-\frac{1}{s-m}\biggr)=\frac{u(s-r)}{s(s-u)}\end{eqnarray}$$

and hence $2s(s-u)(s+r)\unicode[STIX]{x1D6FA}$ is equal to

$$\begin{eqnarray}\displaystyle & & \displaystyle -s(s+r)(s-u)-u(s-r)(s+r)+2r(s-u)(s+r)-2r(r-u)s\nonumber\\ \displaystyle & & \displaystyle \qquad =-s^{3}+rs^{2}+urs-ur^{2}=-(s-r)(s^{2}-ur).\nonumber\end{eqnarray}$$

Since we may suppose that $s>r$ , $s>u$ and $s^{2}>ur$ , we find that $\unicode[STIX]{x1D6FA}<0$ and hence

$$\begin{eqnarray}(X/M^{b/2})^{-\unicode[STIX]{x1D6EC}-\unicode[STIX]{x1D6FF}}\unicode[STIX]{x27E6}K_{a,b}^{r,r}(X)\unicode[STIX]{x27E7}\ll \mathop{\prod }_{m=0}^{u-1}((X/M^{b/2})^{-\unicode[STIX]{x1D6EC}-\unicode[STIX]{x1D6FF}}\unicode[STIX]{x27E6}I_{b,b_{m}}^{r,r}(X)\unicode[STIX]{x27E7})^{\unicode[STIX]{x1D719}_{m}}.\end{eqnarray}$$

This relation serves as a substitute for Lemma 7.2 in the proof of Lemma 7.3, the proof of which now applies without serious modification. This completes our account of the proof of Lemma 11.3. ◻

Proof. We prove the asserted inequalities by induction, beginning with the case $n=1$ . From (11.4), we find on the one hand that

$$\begin{eqnarray}\tilde{u} _{1}=r\tilde{b}_{0}/(\tilde{a}_{0}+\tilde{b}_{0})=r/(1+1/l),\end{eqnarray}$$

whilst, on the other, from (11.2), one has

$$\begin{eqnarray}u_{1}=rb_{0}/(a_{0}+b_{0})=r/(1+a_{0}/b)\geqslant r(1+1/l).\end{eqnarray}$$

Thus $\tilde{u} _{1}\leqslant u_{1}$ , confirming the inductive hypothesis when $n=1$ . Suppose next that $n\geqslant 1$ and $\tilde{u} _{n}\leqslant u_{n}$ . One has

(11.7) $$\begin{eqnarray}\frac{b_{n+1}}{a_{n+1}+b_{n+1}}\geqslant \frac{\tilde{b}_{n+1}}{\tilde{a}_{n+1}+\tilde{b}_{n+1}}\end{eqnarray}$$

if and only if $b_{n+1}/a_{n+1}\geqslant \tilde{b}_{n+1}/\tilde{a}_{n+1}$ , which is to say that

$$\begin{eqnarray}((k-m_{n+1})b_{n}-m_{n+1}a_{n}+h_{n+1})/b_{n}\geqslant ((k-m_{n+1})\tilde{b}_{n}-m_{n+1}\tilde{a}_{n})/\tilde{b}_{n}.\end{eqnarray}$$

This lower bound is satisfied if and only if $m_{n+1}\tilde{a}_{n}/\tilde{b}_{n}\geqslant (m_{n+1}a_{n}-h_{n+1})/b_{n}$ . The latter is automatically satisfied when either $m_{n+1}=0$ or $h_{n+1}\geqslant m_{n+1}a_{n}$ , and otherwise it is equivalent to the upper bound

$$\begin{eqnarray}\frac{\tilde{b}_{n}}{\tilde{a}_{n}+\tilde{b}_{n}}\leqslant \frac{b_{n}}{a_{n}+b_{n}-h_{n+1}/m_{n+1}}.\end{eqnarray}$$

However, when $m_{n+1}>0$ , in view of our hypothesis $\tilde{u} _{n}\leqslant u_{n}$ , one has

$$\begin{eqnarray}\frac{\tilde{b}_{n}}{\tilde{a}_{n}+\tilde{b}_{n}}\leqslant \frac{b_{n}}{a_{n}+b_{n}}\leqslant \frac{b_{n}}{a_{n}+b_{n}-h_{n+1}/m_{n+1}}.\end{eqnarray}$$

Thus, we conclude from (11.2), (11.4) and (11.7) that $\tilde{u} _{n+1}\leqslant u_{n+1}$ . This confirms the inductive hypothesis and hence we deduce that $0\leqslant \tilde{u} _{n}\leqslant u_{n}$ for $1\leqslant n\leqslant R$ , as claimed.◻

Proof. We have

(11.13) $$\begin{eqnarray}B_{n+1}=\mathop{\sum }_{m_{1}=0}^{\tilde{v}_{1}-1}\cdots \mathop{\sum }_{m_{n}=0}^{\tilde{v}_{n}-1}\unicode[STIX]{x1D6FD}_{\mathbf{m}}^{(n)}\mathop{\sum }_{m_{n+1}=0}^{\tilde{v}_{n+1}-1}\unicode[STIX]{x1D719}_{m_{n+1}}\tilde{b}_{n+1}(\mathbf{m}).\end{eqnarray}$$

The innermost sum here is

(11.14) $$\begin{eqnarray}\mathop{\sum }_{m_{n+1}=0}^{\tilde{v}_{n+1}-1}\unicode[STIX]{x1D719}_{m_{n+1}}\tilde{b}_{n+1}(\mathbf{m})=\mathfrak{B}(\tilde{v}_{n+1})\tilde{b}_{n}(\mathbf{m})-\mathfrak{A}(\tilde{v}_{n+1})\tilde{a}_{n}(\mathbf{m}),\end{eqnarray}$$

where

$$\begin{eqnarray}\mathfrak{B}(u)=\mathop{\sum }_{m=0}^{u-1}(k-m)\unicode[STIX]{x1D719}_{m}\quad \text{and}\quad \mathfrak{A}(u)=\mathop{\sum }_{m=0}^{u-1}m\unicode[STIX]{x1D719}_{m}.\end{eqnarray}$$

We observe that when $u$ is a positive integer, one has

$$\begin{eqnarray}\displaystyle \mathfrak{B}(u) & = & \displaystyle \mathop{\sum }_{m=0}^{u-1}\frac{(k-m)(s-r)}{(s-m)(s-m-1)}\nonumber\\ \displaystyle & = & \displaystyle \mathop{\sum }_{m=0}^{u-1}\biggl(\frac{(r-m)(k-m)}{s-m}-\frac{(r-m-1)(k-m-1)}{s-m-1}-\frac{r-m-1}{s-m-1}\biggr),\nonumber\end{eqnarray}$$

whence

$$\begin{eqnarray}\displaystyle s\mathfrak{B}(u) & = & \displaystyle kr-\frac{s(r-u)(k-u)}{s-u}-s\mathop{\sum }_{m=1}^{u}\frac{r-m}{s-m}\nonumber\\ \displaystyle & = & \displaystyle u(k+r-u)-\frac{u(r-u)(k-u)}{s-u}-\frac{1}{2}r(r-1)\nonumber\\ \displaystyle & & \displaystyle +\,\frac{1}{2}(r-u-1)(r-u)-\mathop{\sum }_{m=1}^{u}\frac{m(r-m)}{s-m}.\nonumber\end{eqnarray}$$

Thus, we conclude that

(11.15) $$\begin{eqnarray}s\mathfrak{B}(u)=ku-\frac{1}{2}u(u-1)-\frac{u(r-u)(k-u)}{s-u}-\mathop{\sum }_{m=1}^{u}\frac{m(r-m)}{s-m}.\end{eqnarray}$$

One finds in like manner that

(11.16) $$\begin{eqnarray}\displaystyle s\mathfrak{A}(u) & = & \displaystyle sk\mathop{\sum }_{m=0}^{u-1}\unicode[STIX]{x1D719}_{m}-s\mathop{\sum }_{m=0}^{u-1}(k-m)\unicode[STIX]{x1D719}_{m}\nonumber\\ \displaystyle & = & \displaystyle \frac{ku(s-r)}{s-u}-\biggl(ku-\frac{1}{2}u(u-1)-\frac{u(r-u)(k-u)}{s-u}\nonumber\\ \displaystyle & & \displaystyle -\,\mathop{\sum }_{m=1}^{u}\frac{m(r-m)}{s-m}\biggr)\nonumber\\ \displaystyle & = & \displaystyle \frac{1}{2}u(u-1)-\frac{u^{2}(r-u)}{s-u}+\mathop{\sum }_{m=1}^{u}\frac{m(r-m)}{s-m}.\end{eqnarray}$$

Write

$$\begin{eqnarray}v=\tilde{v}_{n+1},\qquad h=r-v=k-l-v\quad \text{and}\quad \widetilde{\unicode[STIX]{x1D6E5}}=\mathop{\sum }_{m=1}^{v}\frac{m(r-m)}{s-m}.\end{eqnarray}$$

Then we see from (11.15) that $s\mathfrak{B}(v)$ is equal to

$$\begin{eqnarray}\displaystyle & & \displaystyle k(k-l-h)-\frac{1}{2}(k-l-h)(k-l-h-1)-\frac{h(l+h)(k-l-h)}{s-k+l+h}-\widetilde{\unicode[STIX]{x1D6E5}}\nonumber\\ \displaystyle & & \displaystyle \quad =\frac{1}{2}k(k+1)-\frac{1}{2}(l+h)(l+h+1)-\frac{h(l+h)(k-l-h)}{s-k+l+h}-\widetilde{\unicode[STIX]{x1D6E5}}.\nonumber\end{eqnarray}$$

Also, from (11.16), we find that $s\mathfrak{A}(v)$ is equal to

$$\begin{eqnarray}\displaystyle & & \displaystyle \frac{1}{2}(k-l-h)(k-l-h-1)-\frac{h(k-l-h)^{2}}{s-k+l+h}+\widetilde{\unicode[STIX]{x1D6E5}}\nonumber\\ \displaystyle & & \displaystyle \quad =\frac{1}{2}k(k-1)-(l+h)k+\frac{1}{2}(l+h)(l+h+1)-\frac{h(k-l-h)^{2}}{s-k+l+h}+\widetilde{\unicode[STIX]{x1D6E5}}.\nonumber\end{eqnarray}$$

On substituting these formulae into (11.14), we find that

(11.17) $$\begin{eqnarray}s\mathop{\sum }_{m_{n+1}=0}^{\tilde{v}_{n+1}-1}\unicode[STIX]{x1D719}_{m_{n+1}}\tilde{b}_{n+1}(\mathbf{m})=\mathfrak{a}_{0}\tilde{b}_{n}(\mathbf{m})-\mathfrak{b}_{0}\tilde{a}_{n}(\mathbf{m})-\unicode[STIX]{x1D6F6},\end{eqnarray}$$

where

$$\begin{eqnarray}\displaystyle \mathfrak{a}_{0} & = & \displaystyle {\textstyle \frac{1}{2}}k(k+1)-{\textstyle \frac{1}{2}}l(l+1)-\widetilde{\unicode[STIX]{x1D6E5}},\nonumber\\ \displaystyle \mathfrak{b}_{0} & = & \displaystyle {\textstyle \frac{1}{2}}k(k-1)-lk+{\textstyle \frac{1}{2}}l(l+1)+\widetilde{\unicode[STIX]{x1D6E5}}\nonumber\end{eqnarray}$$

and

(11.18) $$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D6F6} & = & \displaystyle \biggl(lh+\frac{1}{2}h(h+1)\biggr)(\tilde{a}_{n}+\tilde{b}_{n})+\frac{h(l+h)(k-l-h)}{s-k+l+h}\tilde{b}_{n}\nonumber\\ \displaystyle & & \displaystyle -\,\biggl(hk+\frac{h(k-l-h)^{2}}{s-k+l+h}\biggr)\tilde{a}_{n}.\end{eqnarray}$$

We next seek to estimate the expression $\unicode[STIX]{x1D6F6}$ , this requiring us to obtain an upper bound for $h$ . We begin with a proof that $\tilde{v}_{n}(\mathbf{m})\geqslant k(1-2/l)$ for $1\leqslant n\leqslant R$ . Note first that, since we assume $k$ to be sufficiently large and $l=\lceil k^{1/3}\rceil$ , one finds from (11.4) that

$$\begin{eqnarray}\tilde{u} _{1}=r\tilde{b}_{0}/(\tilde{a}_{0}+\tilde{b}_{0})=(k-l)/(1+1/l)>k-2k/l+1.\end{eqnarray}$$

Meanwhile, when $n\geqslant 1$ , the condition

$$\begin{eqnarray}m_{n}\leqslant \tilde{v}_{n}\leqslant \tilde{u} _{n}=r\tilde{b}_{n-1}/(\tilde{a}_{n-1}+\tilde{b}_{n-1})\end{eqnarray}$$

that follows from (11.4) ensures that

$$\begin{eqnarray}\tilde{b}_{n}=k\tilde{b}_{n-1}-m_{n}(\tilde{a}_{n-1}+\tilde{b}_{n-1})\geqslant (k-r)\tilde{b}_{n-1}=l\tilde{b}_{n-1},\end{eqnarray}$$

whence

$$\begin{eqnarray}\tilde{u} _{n+1}=r\tilde{b}_{n}/(\tilde{a}_{n}+\tilde{b}_{n})=r/(1+\tilde{b}_{n-1}/\tilde{b}_{n})\geqslant r/(1+1/l)>k-2k/l+1.\end{eqnarray}$$