3.1 Lattices

We use $\mathrm {Vol}(B)$ to denote the volume of a body $B \subseteq \mathbb {R}^d$ . For a lattice $\Gamma \subseteq \mathbb {R}^{d}$ , we recall that the quotient space $\mathbb {R}^d/\Gamma $ (called the fundamental domain) is compact and so $\mathrm {Vol}(\mathbb {R}^d/\Gamma )$ is correctly defined; see also [Reference Tao and Vu30, Sections 3.1 and 3.5] for basic definitions and properties of lattices. In particular, we define the successive minima $\lambda _i$ , $i=1, \ldots , d$ , of B with respect to $\Gamma $ as

$$ \begin{align*}\lambda_i =\inf\{\lambda>0:~\lambda B \text{ contains}\ i\ \text{linearly independent elements of } \Gamma\}, \end{align*} $$

where $\lambda B$ is the homothetic image of B with the coefficient $\lambda $ .

The following is Minkowski’s second theorem. For a proof see [Reference Tao and Vu30, Theorem 3.30].

Lemma 3.1. Suppose $\Gamma \subseteq \mathbb {R}^{d}$ is a lattice of rank d, $B\subseteq \mathbb {R}^{d}$ a symmetric convex body, and let $\lambda _1,\ldots ,\lambda _d$ denote the successive minima of $\Gamma $ with respect to B. Then we have

$$ \begin{align*}\frac{1}{\lambda_1\ldots\lambda_d}\,{\leqslant}\,\frac{d!}{2^d}\frac{ \mathrm {Vol} (B)}{\mathrm {Vol}(\mathbb{R}^d/\Gamma)}. \end{align*} $$

For a proof of the following, see [Reference Betke, Henk and Wills4, Proposition 2.1].

Lemma 3.2. Suppose $\Gamma \subseteq \mathbb {R}^{d}$ is a lattice, $B\subseteq \mathbb {R}^{d}$ a symmetric convex body, and let $\lambda _1,\ldots ,\lambda _d$ denote the successive minima of $\Gamma $ with respect to B. Then we have

$$ \begin{align*}\# \left(\Gamma \cap B\right) \,{\leqslant}\, \prod_{i=1}^{d}\left(\frac{2i}{\lambda_i}+1\right).\end{align*} $$

3.2 Reduction to counting points in lattices

It more convenient to estimate $\mathsf {T}_{2,2}(N;\overline {j},q)$ rather than $\mathsf {T}_{2,2}(N;j,q)$ for the multiplicative inverse $\overline {j}$ of j modulo q, which or course is an equivalent question.

Let ${\mathcal A}$ denote the set

$$ \begin{align*}{\mathcal A}=\{ x\in \mathbb{F}_q^{*} :~ j x^2 \in \{1,\ldots, N\} \} \end{align*} $$

so that

(3.1) $$ \begin{align} \mathsf {T}_{2,2}(N;\overline{j},q)=\sum_{d\in \mathbb{F}_q}({\mathcal A}\circ {\mathcal A})(d)^2, \end{align} $$

where $({\mathcal A}\circ {\mathcal A})(d)$ is defined by Equation (1.4).

If $a_1,a_2\in {\mathcal A}$ satisfy

$$ \begin{align*}a_1-a_2=d,\end{align*} $$

then elementary algebraic manipulations imply

$$ \begin{align*}(a_1^2-a_2^2-d^2)^2=4d^2a_2^2. \end{align*} $$

We have

$$ \begin{align*}ja_1^2-ja_2^2, ja_2^2\in \{-N,\ldots,N\}.\end{align*} $$

Since for any $\lambda , \mu \in \mathbb {F}_q$ the number of solutions to

$$ \begin{align*}ja_1^2-ja_2^2=\lambda, \quad ja_2^2=\mu, \qquad a_1,a_2\in {\mathcal A},\end{align*} $$

is $O(1)$ , we derive from Equation (3.1)

$$ \begin{align*}\mathsf {T}_{2,2}(N;\overline{j},q) \ll \sum_{d\in \mathbb{F}_q}J_0(d)^2, \end{align*} $$

where

If $n,m$ satisfy

then

This implies

(3.2) $$ \begin{align} \mathsf {T}_{2,2}(N;\overline{j},q)\ll \sum_{d\in \mathbb{F}_q}J(d)^2, \end{align} $$

where

(3.3)

Let ${\mathcal L}(d)$ denote the lattice

B the convex body

$$ \begin{align*}B=\{(x,y)\in \mathbb{R}^2 :~ |x| \,{\leqslant}\, 72 N^2,\ |y| \,{\leqslant}\, 12N \}, \end{align*} $$

and let $\lambda _1(d),\, \lambda _2(d)$ denote the first and second successive minima of ${\mathcal L}(d)$ with respect to B.

We now partition summation in Equation (3.2) according to the size of $\lambda _1(d)$ and $\lambda _2(d)$ to get

(3.4) $$ \begin{align} \mathsf {T}_{2,2}(N;\overline{j},q)\ll S_0+S_1+S_2, \end{align} $$

where

$$ \begin{align*}S_0 =\sum_{\substack{d\in \mathbb{F}_q \\ \lambda_1(d)>1}}J(d)^2,\qquad S_1 =\sum_{\substack{d\in \mathbb{F}_q \\ \lambda_1(d){\leqslant}1 \\ \lambda_2(d)>1}}J(d)^2,\qquad S_2 =\sum_{\substack{d\in \mathbb{F}_q \\ \lambda_1(d),\lambda_2(d){\leqslant}1}}J(d)^2. \end{align*} $$

3.3 Concluding the proof

Consider first $S_0$ . If $\lambda _1(d)>1$ , then

$$ \begin{align*}J(d) \,{\leqslant}\, 1,\end{align*} $$

which follows from the fact that for any distinct points $(n_0,m_0)$ , $(n_1.m_1)$ satisfying the conditions in Equation (3.3) we have

$$ \begin{align*}(n_0^2-n_1^2,m_0-m_1)\in {\mathcal L}(d)\cap B.\end{align*} $$

This implies that $J(d)^2 = J(d)$ , and we derive

(3.5) $$ \begin{align} S_0= \sum_{\substack{d\in \mathbb{F}_q \\ \lambda_1(d)>1}}J(d)\ll N^2. \end{align} $$

Consider next $S_1$ . Suppose d satisfies $\lambda _1(d) \,{\leqslant }\, 1$ and $\lambda _2(d)> 1$ . There exists $n_d,m_d$ satisfying the conditions given in Equation (3.3) such that

$$ \begin{align*}J(d)\ll \# \left\{ |m|,|n| \,{\leqslant}\, 6N :~ (n^2-n_d^2,m-m_d)\in {\mathcal L}(d)\cap B\right\}. \end{align*} $$

Since $\lambda _2(d)>1$ , there exists a unique point $(a_d,b_d) \in {\mathcal L}(d)\cap B$ satisfying

$$ \begin{align*}\gcd(a_d,b_d)=1, \quad |a_d| \,{\leqslant}\, 72N^2, \quad |b_d| \,{\leqslant}\, 12N \end{align*} $$

such that

$$ \begin{align*}J(d)\ll \# \left\{ |m|,|n| \,{\leqslant}\, 6N :~ \frac{n^2-n_d^2}{m-m_d}=\frac{a_d}{b_d}\right\}+1. \end{align*} $$

This implies

(3.6) $$ \begin{align} S_1& \,{\leqslant}\, \sum_{d\in \mathbb{F}_q}J(d)\left(\# \left\{ |m|,|n| \,{\leqslant}\, 6N :~ \frac{n^2-n^2_d}{m-m_d}=\frac{a_d}{b_d}\right\}+1 \right) \\ & \,{\leqslant}\, \sum_{(a,b) \in {\mathcal W}} J(a,b)K(a,b)+N^2,\nonumber \end{align} $$

where ${\mathcal W}$ is the following set of all pairs $(a,b)$ satisfying

(3.7) $$ \begin{align} {\mathcal W} = \{(a,b)\in \mathbb{Z}^2:~|a| \,{\leqslant}\, 72N^2, \ |b| \,{\leqslant}\, 12 N, \ \gcd(a,b)=1\}, \end{align} $$

and $K(a,b)$ is defined by

$$ \begin{align*}K(a,b)=\# \left\{(m,n) \in \mathbb{Z}^2:~|m|,|n| \,{\leqslant}\, 6N, \ \frac{n^2-n_{a,b}^2}{m-m_{a,b}}=\frac{a}{b}\right\}, \end{align*} $$

for some choice of integers $m_{a,b},n_{a,b}$ satisfying $|m_{a,b}|,|n_{a,b}| \,{\leqslant }\, 6 N$ and $J(a,b)$ is defined by

$$ \begin{align*}J(a,b)=\# \left\{(m,n) \in \mathbb{Z}^2:~ |m|,|n| \,{\leqslant}\, 6N, \ n^2+(ab^{-1})^2\equiv ab^{-1}m\quad \mod{q}\right\}. \end{align*} $$

Note that

$$ \begin{align*} \sum_{(a,b)\in {\mathcal W}}J(a,b)& \,{\leqslant}\, \#\{(m,n,\lambda) \in \mathbb{Z}^3:~ |m|,|n| \,{\leqslant}\, 6N, \ 1 \,{\leqslant}\, \lambda<q, \\ & \qquad \qquad \qquad\qquad \qquad\quad \lambda^2-\lambda m+n^2 \equiv 0\quad \mod{q} \} \\ &\ll N^2 \end{align*} $$

since after fixing $m,n$ with $O(N^2)$ choices there exists $O(1)$ solutions to

$$ \begin{align*}\lambda^2-\lambda m+n^2 \equiv 0\quad \mod{q}\end{align*} $$

in the remaining variable $\lambda $ . We also have

(3.8) $$ \begin{align} J(a,b)\ll K(a,b)+1. \end{align} $$

Fix some $a,b$ as in the sum in Equation (3.6), and consider $K(a,b)$ . If $n,m$ satisfy

$$ \begin{align*}\frac{n^2-n_{a,b}^2}{m-m_{a,b}}=\frac{a}{b}, \qquad |m|,|n| \,{\leqslant}\, 6 N, \end{align*} $$

then, since $\gcd (a,b)=1$ , we have

(3.9)

and

(3.10)

Furthermore, if one out of m or n is fixed, then the other number is defined in no more than two ways.

Write Equation (3.9) as

Then we see that there are two integers $a_1,a_2$ satisfying

$$ \begin{align*}a_1a_2=a, \qquad |a_1|,|a_2| \,{\leqslant}\, 12 N \end{align*} $$

such that

Hence, for each fixed pair $(a_1, a_2)$ there are at most

$$ \begin{align*}\frac{N}{\mathrm {lcm}[a_1,a_2]}+1 \ll \frac{N}{|a|} \gcd(a_1,a_2) + 1 \end{align*} $$

possibilities for n. Hence, by a well-known bound

(3.11) $$ \begin{align} \tau(a) = a^{o(1)} \end{align} $$

on the divisor function $\tau (a)$ for $a \ne 0$ , see [Reference Iwaniec and Kowalski19, Equation (1.81)], we have

$$ \begin{align*}K(a,b) \ll \sum_{a_1a_2=a}\left(\frac{N}{\text{lcm}(a_1,a_2)}+1\right)\ll \frac{N}{|a|}\sum_{a_1a_2=a}\gcd(a_1,a_2) + N^{o(1)}. \end{align*} $$

By the Cauchy–Schwarz inequality and Equation (3.11), we now derive

(3.12) $$ \begin{align} K(a,b)^2\ll N^{2+o(1)}\sum_{a_1a_2=a}\frac{\gcd(a_1,a_2)^2}{|a|^2} + N^{o(1)}. \end{align} $$

Similarly, using Equation (3.10) we obtain

(3.13) $$ \begin{align} K(a,b)\ll \frac{N}{|b|}. \end{align} $$

Combining Equations (3.12), (3.13), (3.8) and substituting into Equations (3.6), we see that

$$ \begin{align*} S_1 \,{\leqslant}\, N^{2+o(1)}\sum_{(a,b) \in {\mathcal W}}\sum_{\substack{a_1a_2=a \\ |a_1|,|a_2|{\leqslant}12 N}}\min\left\{\frac{1}{b^2},\frac{\gcd(a_1,a_2)^2}{a^2} \right\} & \\ + \sum_{(a,b)\in {\mathcal W}}J(a,b) & N^{o(1)}. \end{align*} $$

Hence, recalling Equation (3.7), we derive

$$ \begin{align*} S_1 & \,{\leqslant}\, N^{2+o(1)}\sum_{\substack{|a|{\leqslant}72 N^2\\|b|{\leqslant}12 N}}\sum_{\substack{a_1a_2=a \\ |a_1|,|a_2|{\leqslant}12 N}}\min\left\{\frac{1}{b^2},\frac{\gcd(a_1,a_2)^2}{a^2} \right\} + N^{2+o(1)}\\ & \,{\leqslant}\, N^{2+o(1)}\sum_{ a_1,a_2, b{\leqslant}12 N}\min\left\{\frac{1}{b^2},\frac{\gcd(a_1,a_2)^2}{a^2_1a^2_2} \right\} + N^{2+o(1)}\\ & \,{\leqslant}\, N^{2+o(1)}\sum_{e{\leqslant}12N}\sum_{ b{\leqslant}12N}\sum_{\substack{a_1,a_2{\leqslant}12N \\ \gcd(a_1,a_2)=e}} \min\left\{\frac{1}{b^2},\frac{e^2}{a^2_1a^2_2} \right\} + N^{2+o(1)}\\ & \,{\leqslant}\, N^{2+o(1)}\sum_{e{\leqslant}12N}\sum_{ b{\leqslant}12N}\sum_{a_1,a_2{\leqslant}12N/e}\min\left\{\frac{1}{b^2},\frac{1}{a^2_1a^2_2e^2} \right\} + N^{2+o(1)}. \end{align*} $$

Using the bound on the divisor function (3.11) again, we obtain

(3.14) $$ \begin{align} S_1 & \,{\leqslant}\, N^{2+o(1)}\sum_{ b{\leqslant}12N} \sum_{a{\leqslant}12^4 N^2}\min\left\{\frac{1}{b^2},\frac{1}{a^2} \right\}+ N^{2+o(1)} \nonumber\\ & \,{\leqslant}\, N^{2+o(1)}\left(\sum_{ b{\leqslant}12N} \sum_{\substack{ a{\leqslant}b }}\frac{1}{b^2}+ \sum_{a{\leqslant}12^4 N^2}\sum_{\substack{b{\leqslant}a }}\frac{1}{a^2}\right) + N^{2+o(1)}\\ & \,{\leqslant}\, N^{2+o(1)}.\nonumber \end{align} $$

Finally, consider $S_2$ . If d satisfies $\lambda _2(d) \,{\leqslant }\, 1$ , then by Lemmas 3.1 and 3.2

(3.15) $$ \begin{align} \# \left({\mathcal L}(d)\cap B\right)\ll \frac{N^3}{q}. \end{align} $$

In particular, we see that for $N = o(q^{1/3})$ the bound (3.15) implies

$$ \begin{align*}1 \,{\leqslant}\, \# \left({\mathcal L}(d)\cap B\right) = o(1), \end{align*} $$

which means that this case (that is, $\lambda _2(d) \,{\leqslant }\, 1$ ) never occurs for ‘small’ N.

For each $|n| \,{\leqslant }\, 6N$ there exists at most one value of m satisfying Equation (3.3) and for any two pairs $(n_1,m_1), (n_2,m_2)$ satisfying Equation (3.3) we have

This implies

Since for any integer $r\neq 0$ the bound (3.11) on the divisor function implies

$$ \begin{align*}\# \{ |n_1|,|n_2| \,{\leqslant}\, 8N :~ n_1^2-n_2^2=r\} \,{\leqslant}\, N^{o(1)}, \end{align*} $$

we obtain

$$ \begin{align*}J(d)^2 \,{\leqslant}\, \# \left({\mathcal L}(d)\cap B\right) N^{o(1)}.\end{align*} $$

By Equation (3.15)

$$ \begin{align*}J(d)\ll \frac{N^{3/2+o(1)}}{q^{1/2}}, \end{align*} $$

which implies

(3.16) $$ \begin{align} S_2=\sum_{\substack{d\in \mathbb{F}_q \\ \lambda_1(d),\lambda_2(d){\leqslant}1}}J(d)^2\ll \frac{N^{3/2}}{q^{1/2}}\sum_{\substack{d\in \mathbb{F}_q \\ \lambda_1(d),\lambda_2(d){\leqslant}1}}J(d)\ll \frac{N^{7/2+o(1)}}{q^{1/2}}. \end{align} $$

Combining Equations (3.5), (3.14) and (3.16) with Equation (3.4), we derive the desired bound on $\mathsf {T}_{2,2}(N;\overline {j},q))$ .

4.1 Lattices

For a lattice $\Gamma $ and a convex body B, we define the dual lattice $\Gamma ^*$ and dual body $B^*$ by

$$ \begin{align*}\Gamma^*=\{ x\in \mathbb{R}^{d} :~\langle x,y \rangle \in \mathbb{Z} \quad \text{for all} \quad y\in \Gamma\}, \end{align*} $$

and

$$ \begin{align*}B^{*}=\{ x\in \mathbb{R}^{d} :~\langle x,y \rangle \,{\leqslant}\, 1 \quad \text{for all} \quad y\in B\}, \end{align*} $$

respectively.

The following is known as a transference theorem and is due to Mahler [Reference Mahler21] which we present in a form given by Cassels [Reference Cassels8, Chapter VIII, Theorem VI].

Lemma 4.1. Let $\Gamma \subseteq \mathbb {R}^{d}$ be a lattice, $B\subseteq \mathbb {R}^{d}$ a symmetric convex body, and let $\Gamma ^{*}$ and $B^{*}$ denote the dual lattice and dual body. Let $\lambda _1,\ldots ,\lambda _d$ denote the successive minima of $\Gamma $ with respect to B and $\lambda _1^{*},\ldots ,\lambda _d^{*}$ the successive minima of $\Gamma ^{*}$ with respect to $B^{*}$ . For each $1 \,{\leqslant }\, j \,{\leqslant }\, d$ , we have

$$ \begin{align*}\lambda_j \lambda^*_{d-j+1} \,{\leqslant}\, d!.\end{align*} $$

We apply Lemma 4.1 to lattices of a specific type whose dual may be easily calculated. For a proof of the following, see [Reference Bordignon and Kerr6, Lemma 15].

Lemma 4.2. Let $a_1,\ldots ,a_d$ and $q\,{\geqslant }\, 1$ be integers satisfying $\gcd (a_i,q)=1$ , and let ${\mathcal L}$ denote the lattice

Then we have

Our next result should be compared with the case $\nu =3$ of [Reference Bourgain, Garaev, Konyagin and Shparlinski7, Lemma 17]. It is possible to give a more direct variant of [Reference Bourgain, Garaev, Konyagin and Shparlinski7, Lemma 17] to estimate higher-order energies of modular square roots (see the proof of Corollary 4.4 below) although this seems to put tighter restrictions on the size of the parameter N.

Lemma 4.3. Let q be prime, and $L,\, M,\, N$ integers. Let ${\mathcal L}$ denote the lattice

and let B be the convex body

$$ \begin{align*}B=\{ (x,\, y,\, z)\in \mathbb{R}^3 :~ |x| \,{\leqslant}\, L, \ \ |y| \,{\leqslant}\, M, \ \ |z| \,{\leqslant}\, N \}. \end{align*} $$

Let

$$ \begin{align*}K = \# \left({\mathcal L}\cap B\right), \end{align*} $$

and $\lambda _1,\lambda _2$ denote the first and second successive minima of ${\mathcal L}$ with respect to B. Then at least one of the following holds:

1. (i)

   $$ \begin{align*}K <\max\left\{\frac{640LMN}{q},\,1\right\}. \end{align*} $$

2. (ii) $\lambda _1 \,{\leqslant }\, 1$ and $\lambda _2>1$ .

3. (iii) There exists some and $\ell,\, m,\, n\in \mathbb {Z}$ satisfying

   $$ \begin{align*}|\ell| \,{\leqslant}\, \frac{4320MN}{K}, \quad |m| \,{\leqslant}\, \frac{4320LN}{K}, \quad |n| \,{\leqslant}\, \frac{4320LM}{K} \end{align*} $$
   and
   

Proof. Assume that (i) fails. Thus, we have

(4.1) $$ \begin{align} K \,{\geqslant}\, \max\left\{ \frac{640 LMN}{q},\,1\right\}. \end{align} $$

Then $K \,{\geqslant }\, 1$ . Hence, if $\lambda _1 \,{\leqslant }\, \lambda _2 \,{\leqslant }\, \lambda _3$ denote the successive minima of ${\mathcal L}$ with respect to B, then $\lambda _1 \,{\leqslant }\, 1$ . We first show Equation (4.1) implies

$$ \begin{align*}\lambda_3>1. \end{align*} $$

Indeed, otherwise by Lemma 3.2

(4.2) $$ \begin{align} K \,{\leqslant}\, \left(\frac{2}{\lambda_1} + 1\right) \left(\frac{4}{\lambda_2} + 1\right) \left(\frac{6}{\lambda_3} + 1\right) \,{\leqslant}\, \frac{3}{\lambda_1} \frac{5}{\lambda_2} \frac{7}{\lambda_3} = \frac{105}{\lambda_1 \lambda_2 \lambda_3}. \end{align} $$

Since

$$ \begin{align*}\mathrm {Vol}(\mathbb{R}^3/{\mathcal L})=q \qquad\mbox{and}\qquad \mathrm {Vol}(B) = 8LMN, \end{align*} $$

we see from Lemma 3.1 that

(4.3) $$ \begin{align} \frac{1}{\lambda_1 \lambda_2 \lambda_3}\,{\leqslant}\,\frac{3!}{8} \frac{8LMM}{q}= \frac{6LMN}{q}, \end{align} $$

which together with Equation (4.2) contradicts Equation (4.1).

Hence, we have either

(4.4) $$ \begin{align} \lambda_1 \,{\leqslant}\, 1, \qquad \lambda_2, \lambda_3>1, \end{align} $$

or

(4.5) $$ \begin{align} \lambda_1,\, \lambda_2 \,{\leqslant}\, 1, \qquad \lambda_3>1. \end{align} $$

Clearly, Equation (4.4) is the same as (ii).

Next, suppose that we have Equation (4.5). By Lemma 3.2, a similar calculation as before, together with Equation (4.3) gives

(4.6) $$ \begin{align} K \,{\leqslant}\, \frac{7\times15 }{\lambda_1 \lambda_2}=\frac{105\lambda_3}{\lambda_1\lambda_2\lambda_3}. \end{align} $$

Applying Lemma 3.1 and using

$$ \begin{align*}\mathrm {Vol} (B)=8NML, \quad \mathrm {Vol}(\mathbb{R}^{3}/{\mathcal L})=q, \end{align*} $$

we derive from Equation (4.6) that

$$ \begin{align*}K \,{\leqslant}\, \frac{105 \cdot 3! \, \mathrm {Vol} (B) \lambda_3}{2^3\, \mathrm {Vol}(\mathbb{R}^{3}/{\mathcal L})}= \frac{630 NML \lambda_3 }{q}. \end{align*} $$

Let $\lambda _1^{*}$ denote the first successive minima of the dual lattice ${\mathcal L}^{*}$ with respect to the dual body $B^{*}$ . By Lemma 4.1,

$$ \begin{align*}\lambda_3 \,{\leqslant}\, \frac{6}{\lambda_1^{*}}. \end{align*} $$

The above estimates combined with Equation (4.6) implies

$$ \begin{align*}\lambda_1^{*} \,{\leqslant}\, \frac{4320 NML}{qK}. \end{align*} $$

Hence, by the definition of $\lambda _1^{*}$

(4.7) $$ \begin{align} {\mathcal L}^{*}\cap \frac{4320NML}{qK}B^{*}\neq \{(0,0,0)\}. \end{align} $$

Its remains to recall that by Lemma 4.2

and also it is obvious that

$$ \begin{align*}B^{*}=\{ (x,\, y,\, z)\in \mathbb{R}^3 :~ L|x|+M|y|+ N|z| \,{\leqslant}\, 1\}. \end{align*} $$

By Equation (4.7), this implies there exists some and $\ell,\, m,\, n$ satisfying (iii), which completes the proof.

Corollary 4.4. Let $\varepsilon>0$ be a fixed real number. For $j\in \mathbb {F}_q^{*}$ , integer $N\ll q$ and $\Delta \,{\geqslant }\, 1$ , let ${\mathcal A},\, {\mathcal D}\subseteq \mathbb {F}_q$ denote the sets

$$ \begin{align*}{\mathcal A}=\{ x\in \mathbb{F}_q^{*} :~ jx^2 \in [1,\, N] \}. \end{align*} $$

and

$$ \begin{align*}{\mathcal D}=\{ d\in \mathbb{F}_q^{*} :~ ({\mathcal A}\circ {\mathcal A})(d)\,{\geqslant}\, \Delta\}. \end{align*} $$

Let K be sufficiently large, and suppose K and $\Delta $ satisfy

(4.8) $$ \begin{align} K\,{\geqslant}\, \left(\frac{N^{15/2}}{\Delta^{12}q^{1/2}}+\frac{N^{5}}{\Delta^{8}q^{1/4}}\right) N^{\varepsilon} \end{align} $$

and

(4.9) $$ \begin{align} \Delta \,{\geqslant}\, \left(\frac{N^{3/2}}{q^{1/2}}+\frac{N^{5/8}}{q^{1/8}}\right) N^{\varepsilon}. \end{align} $$

Let ${\mathcal F}\subseteq \mathbb {F}_q^{*}$ denote the set of f satisfying

(4.10) $$ \begin{align} ({\mathcal D}\circ {\mathcal D})(f)\,{\geqslant}\, K. \end{align} $$

Then either

(4.11) $$ \begin{align} K\ll 1, \end{align} $$

or

$$ \begin{align*}K\#{\mathcal F}\ll \frac{N^{3+o(1)}}{\Delta^4}. \end{align*} $$

Proof. From Equation (4.10)

(4.12) $$ \begin{align} K \,{\leqslant}\, \# \{ (d_1,\, d_2)\in {\mathcal D}^2 :~d_1-d_2=f\}. \end{align} $$

If $d_1,\, d_2\in {\mathcal D}$ satisfy $d_1-d_2=f$ , then

$$ \begin{align*}d_1^2-d_2^2 - f^2 = (d_1-d_2)^2 + 2d_1d_2 - 2d_2^2 - f^2 = 2d_2 (d_1-d_2) = 2d_2 f \end{align*} $$

and some algebraic manipulations show

$$ \begin{align*}(2jd_1^2-2jd_2^2-2jf^2)^2=8j f^2(2j d_2^2). \end{align*} $$

Since $0\not \in {\mathcal D}$ , for each $d\in {\mathcal D}$ , by Equation (4.9) and [Reference Dunn, Kerr, Shparlinski and Zaharescu13, Lemma 6.4] there exists $m_d,\, n_d$ satisfying

(4.13)

Let $I(f)$ count the number of solutions to the congruence

(4.14)

with $d_1,\, d_2\in {\mathcal D}$ . The above and Equation (4.12) imply

(4.15) $$ \begin{align} K \,{\leqslant}\, I(f). \end{align} $$

Rearranging Equation (4.14), we obtain

This implies that $I(f)$ is bounded by the number of solutions to

(4.16)

with $d_1,\, d_2\in {\mathcal D}$ . Let ${\mathcal L}$ denote the lattice

and B the convex body

$$ \begin{align*}B=\left\{ (x,\, y,\, z)\in \mathbb{R}^3:~ |x| \,{\leqslant}\, \frac{CN^6}{\Delta^8}, \ |y| \,{\leqslant}\, \frac{CN^5}{\Delta^8}, \ |z| \,{\leqslant}\, \frac{CN^4}{\Delta^8}\right\} \end{align*} $$

for a suitable absolute constant C. By Equations (4.13) and (4.16)

(4.17) $$ \begin{align}\bigl((n_{d_1}m_{d_2}-n_{d_2}m_{d_1})^2, -4m_{d_1}m_{d_2}(n_{d_1}m_{d_2}+n_{d_2}m_{d_1}),&\\ 4(m_{d_1}m_{d_2})^2\bigr)&\in {\mathcal L}\cap B. \nonumber\end{align} $$

Let $\lambda _1,\, \lambda _2$ denote the first and second successive minima of ${\mathcal L}$ with respect to B. Assuming that $K\,{\geqslant }\, 1$ , we have $\lambda _1 \,{\leqslant }\, 1$ .

Suppose that

$$ \begin{align*}\lambda_1 \,{\leqslant}\, 1, \quad \lambda_2>1. \end{align*} $$

Then there exists some $(a_0,\, b_0,\, c_0)\in {\mathcal L}\cap B$ such that for any $d_1,\, d_2\in {\mathcal D}$ satisfying Equation (4.17) we have

$$ \begin{align*} \left((n_{d_1}m_{d_2}-n_{d_2}m_{d_1})^2, -4m_{d_1}m_{d_2}(n_{d_1}m_{d_2}+n_{d_2}m_{d_1}),\, 4(m_{d_1}m_{d_2})^2\right)& \\ =m(a_0,\, b_0,\, &c_0), \end{align*} $$

for some $m\in \mathbb {Z}$ . Note from Equation (4.13) for each $d_1,\, d_2\in {\mathcal D}$ we have $m_{d_1}m_{d_2}\neq 0$ and hence $c_0\neq 0$ . This implies

$$ \begin{align*} & \left(\frac{n_{d_1}}{m_{d_1}}-\frac{n_{d_2}}{m_{d_2}}\right)^2=\frac{a_0}{c_0},\\ & \frac{n_{d_1}}{m_{d_1}}+\frac{n_{d_2}}{m_{d_2}}=\frac{b_0}{c_0}. \end{align*} $$

Hence,

$$ \begin{align*}K \,{\leqslant}\, \#\biggl\{ (d_1,\,d_2)\in {\mathcal D}\times {\mathcal D} : ~ \frac{n_{d_1}}{m_{d_1}}-\frac{n_{d_2}}{m_{d_2}}&=\pm\left(\frac{a_0}{c_0}\right)^{1/2}, \\& \frac{n_{d_1}}{m_{d_1}}+\frac{n_{d_2}}{m_{d_2}}=\frac{b_0}{c_0} \biggr\} \,{\leqslant}\, 8\end{align*} $$

since once $n_{d_1}/m_{d_1}$ is fixed, due to the coprimality condition in Equation (4.13), $d^2_1$ is uniquely defined and similarly for $d^2_2$ . This implies Equation (4.11).

Suppose next that

(4.18) $$ \begin{align} \lambda_1 \,{\leqslant}\, 1, \quad \lambda_2 \,{\leqslant}\, 1. \end{align} $$

Let $J(\ell,\, m,\, n)$ count the number of solutions to

$$ \begin{align*}m_{1}m_{2}=\ell, \quad n_{1}m_{2}+n_{2}m_{1}=m, \quad n_{1}m_{2}-n_{2}m_{1}=n, \end{align*} $$

with

(4.19) $$ \begin{align} |m_1|, |m_2|\ll \frac{N}{\Delta^2}, \quad |n_1|,|n_2|\ll \frac{N^2}{\Delta^2}, \quad \quad m_1m_2n_1n_2\neq 0 \end{align} $$

so that

(4.20)

for some absolute constant C. We next show that

(4.21) $$ \begin{align} J(\ell,\, m,\, n)=N^{o(1)}. \end{align} $$

Estimates for the divisor function (3.11) imply the number of solutions to

$$ \begin{align*}m_1m_2=\ell, \quad m_1,\, m_2\ \text{satisfying Equation~(4.19)} \end{align*} $$

is at most $N^{o(1)}$ . For each such $m_1,\, m_2$ , there exists at most one solution to the system

$$ \begin{align*}n_1m_2-n_2m_1=n, \quad n_1m_2+n_2m_1=m, \quad n_1,\, n_2\ \text{satisfying Equation~(4.19)}, \end{align*} $$

which establishes Equation (4.21). By Equations (4.15) and (4.20)

and hence

(4.22)

By Equation (4.9), for each $\ell,\, n \in \mathbb {Z}$ , there exists at most one value of $|m|\ll N^5/\Delta ^8$ satisfying

For any $(\ell _1,\, m_1,\, n_1)$ and $(\ell _2,\, m_2,\, n_2)$ satisfying the conditions of Equation (4.22), there exists some $|m|\ll N^5/\Delta ^8$ such that

(4.23)

Define the lattice

and the convex body

$$ \begin{align*} B=\{ (n,\, m,\, \ell)\in \mathbb{R}^3 :~ |n|& \,{\leqslant}\, C_0N^6/\Delta^8, \\ & |m| \,{\leqslant}\, C_0N^5/\Delta^8, \ |\ell| \,{\leqslant}\, C_0N^4/\Delta^8 \}, \end{align*} $$

for a suitable constant $C_0$ . Since for any integer r

$$ \begin{align*}\# \{ n_1,\, n_2\in \mathbb{Z} :~ n_1^2+n_2^2=r\} \,{\leqslant}\, r^{o(1)}, \end{align*} $$

we see that Equation (4.23) implies

$$ \begin{align*}K^2 \,{\leqslant}\, \# \left({\mathcal L}\cap B\right) N^{o(1)}. \end{align*} $$

By Equation (4.8), Equation (4.18) and Lemma 4.3, there exists $(\ell,\, m,\, n)\neq (0,\, 0,\, 0)$ satisfying

(4.24) $$ \begin{align} |\ell| \,{\leqslant}\, \frac{N^{11+o(1)}}{\Delta^{16}K^2}, \qquad |m| \,{\leqslant}\, \frac{N^{10+o(1)}}{\Delta^{16}K^2}, \qquad |n| \,{\leqslant}\, \frac{N^{9+o(1)}}{\Delta^{16}K^2}, \end{align} $$

and

(4.25)

Note we may assume

(4.26) $$ \begin{align} \gcd(\ell,\, m,\, n)=1. \end{align} $$

Recall Equation (4.16)

(4.27)

If $d_1,\, d_2$ satisfy the conditions in Equation (4.27), then by Equation (4.25)

and hence from Equation (4.8), assuming that N is large enough, we derive

(4.28) $$ \begin{align} n(n_{d_1}m_{d_2}-n_{d_2}m_{d_1})^2-4mm_{d_1}m_{d_2}(n_{d_1}&m_{d_2}+n_{d_2}m_{d_1}) \\ &+4\ell (m_{d_1}m_{d_2})^2=0.\nonumber \end{align} $$

Similarly by Equations (4.24) and (4.25) we have and again Equation (4.8) ensures that

$$ \begin{align*}m^2=n\ell. \end{align*} $$

Therefore, Equation (4.28) implies the following equation

$$ \begin{align*}\left(\frac{n_{d_1}}{m_{d_1}}-\frac{n_{d_2}}{m_{d_2}}\right)^2-4\left(\frac{n_{d_1}}{m_{d_1}}+\frac{n_{d_2}}{m_{d_2}}\right)\left(\frac{m}{n}\right)+4\left(\frac{m}{n}\right)^{2}=0. \end{align*} $$

We see that

(4.29) $$ \begin{align} \frac{m}{n}=\frac{1}{2}\left(\frac{n_{d_1}}{m_{d_1}}+\frac{n_{d_2}}{m_{d_2}}\right)\pm \frac{\sqrt{n_{d_1}m_{d_1}n_{d_2}m_{d_2}}}{m_{d_1}m_{d_2}}. \end{align} $$

Hence, from Equations (4.13) and (4.27), there exists some constant C such that

$$ \begin{align*}I(f)& \,{\leqslant}\, \#\biggl\{ \left(m_{d_1},\, m_{d_2},\, n_{d_1},\, n_{d_2} \right)\in \mathbb{Z}^4:\\& \qquad \qquad |m_{d_1}|,\, |m_{d_2}| \,{\leqslant}\, \frac{CN}{\Delta^2}, \ |n_{d_1}|,|n_{d_2}| \,{\leqslant}\, \frac{CN^2}{\Delta^2},\\ & \qquad \qquad \qquad\qquad m_{d_1}m_{d_2}n_{d_1}n_{d_2}\neq 0,\ \text{and~(4.29) holds} \biggr\}.\end{align*} $$

Summing the above over $f\in {\mathcal F}$ , using Equation (4.15) and noting that for each $\ell,\, m,\, n$ satisfying Equation (4.26) there exists $O(1)$ values of f satisfying Equation (4.25), we see that $K\# {\mathcal F}$ is bounded by the number of solutions to the Equation (4.29) with integer variables satisfying

$$ \begin{align*}|m_{d_1}|,\, |m_{d_2}| \,{\leqslant}\, \frac{CN}{\Delta^2}, \qquad |n_{d_1}|,\, |n_{d_2}| \,{\leqslant}\, \frac{CN^2}{\Delta^2}, \qquad n_{d_1}n_{d_2}m_{d_1}m_{d_2}\neq 0. \end{align*} $$

We see from Equation (4.29) that $n_{d_1}m_{d_1}n_{d_2}m_{d_2}=r^2$ for some $r\in \mathbb {Z}$ and hence a bound (3.11) on the divisor function implies

$$ \begin{align*}K\#{\mathcal F} \,{\leqslant}\, N^{o(1)}\#\left\{ \ell \,{\leqslant}\, C^4 \frac{N^6}{\Delta^8} :~\ell=r^2 \ \text{for some}\ r \in \mathbb{Z}\right\} \,{\leqslant}\, \frac{N^{3+o(1)}}{\Delta^4}, \end{align*} $$

which completes the proof.

4.2 Concluding the proof

As in Section 3.2, here we work again with $\mathsf {T}_{4,2}(N;\overline {j},q)$ for the multiplicative inverse $\overline {j}$ of j modulo q rather than with $\mathsf {T}_{4,2}(N;j,q)$ . Let notation be as in Corollary 4.4 so that

$$ \begin{align*}\mathsf {T}_{4,2}(N;\overline{j},q)=\sum_{x\in \mathbb{F}_q}({\mathcal A}\circ {\mathcal A} \circ {\mathcal A} \circ {\mathcal A})(x)^{2}, \end{align*} $$

where we recall that

$$ \begin{align*}{\mathcal A}=\{x\in \mathbb{F}^{*}_q \ : \ jx^2\in [1,N]\}.\end{align*} $$

By Equation (1.5), we may assume that

(4.30) $$ \begin{align} N \,{\leqslant}\, q^{1/3}. \end{align} $$

Applying the dyadic pigeonhole principle, there exist $\Delta _1,\Delta _2 \,{\geqslant }\, 1$ and ${\mathcal D}_1,{\mathcal D}_2\subseteq \mathbb {F}_q$ given by

$$ \begin{align*}{\mathcal D}_j=\{ x\in \mathbb{F}_q :~ \Delta_j \,{\leqslant}\, ({\mathcal A}\circ {\mathcal A})(x)< 2\Delta_j\}, \qquad j=1,2 \end{align*} $$

such that

$$ \begin{align*}\mathsf {T}_{4,2}(N;\overline{j},q) \,{\leqslant}\, N^{o(1)}(\Delta_1\Delta_2)^2E({\mathcal D}_1,{\mathcal D}_2),\end{align*} $$

where

$$ \begin{align*}E({\mathcal D}_1,{\mathcal D}_2)=\sum_{x\in \mathbb{F}_q}({\mathcal D}_1\circ {\mathcal D}_2)(x)^{2}. \end{align*} $$

By the Cauchy–Schwarz inequality,

$$ \begin{align*}E({\mathcal D}_1,{\mathcal D}_2) \,{\leqslant}\, E({\mathcal D}_1)^{1/2}E({\mathcal D}_2)^{1/2}, \end{align*} $$

and hence there exists some $\Delta $ and ${\mathcal D}$ given by

$$ \begin{align*}{\mathcal D}=\{ x\in \mathbb{F}_q :~ \Delta \,{\leqslant}\, ({\mathcal A} \circ {\mathcal A})(x)< 2\Delta\} \end{align*} $$

such that

(4.31) $$ \begin{align} \mathsf {T}_{4,2}(N;\overline{j},q) \,{\leqslant}\, N^{o(1)}\Delta^{4}E({\mathcal D}). \end{align} $$

It is also obvious from Equationi (3.1) that

(4.32) $$ \begin{align} \Delta^2\left(\# {\mathcal D} \right) \,{\leqslant}\, \mathsf {T}_{2,2}(N;\overline{j},q), \end{align} $$

and

(4.33) $$ \begin{align} \# {\mathcal D} \,{\leqslant}\, \Delta \#{\mathcal D} \ll N^2. \end{align} $$

Isolating the diagonal contribution in $E({\mathcal D})$ , we write

$$ \begin{align*}E({\mathcal D})=(\#{\mathcal D})^2+\sum_{f\in \mathbb{F}_q^{*}}({\mathcal D} \circ {\mathcal D})(f)^2. \end{align*} $$

We may assume

(4.34) $$ \begin{align} E({\mathcal D}) \,{\leqslant}\, 2 \sum_{f\in \mathbb{F}_q^{*}}({\mathcal D} \circ {\mathcal D})(f)^2 \end{align} $$

since otherwise we have $E({\mathcal D}) \,{\leqslant }\, 2 (\#{\mathcal D})^2$ and it follows from the bounds (4.31) and (4.32) that

$$ \begin{align*}\mathsf {T}_{4,2}(N;\overline{j},q)\,{\leqslant}\,\Delta^{4}(\#{\mathcal D})^2 N^{o(1)} \,{\leqslant}\, \mathsf {T}_{2,2}(N;\overline{j},q)^2 N^{o(1)}. \end{align*} $$

Now, recalling the condition (4.30) and using Theorem 1.1, we derive

$$ \begin{align*}\mathsf {T}_{4,2}(N;\overline{j},q) \,{\leqslant}\, N^{4+o(1)}. \end{align*} $$

By Equation (4.34) and the dyadic pigeonhole principle, there exists some K and a set ${\mathcal F}\subseteq \mathbb {F}_q^{*}$ given by

$$ \begin{align*}{\mathcal F}=\{ f\in \mathbb{F}_q^{*} :~ K \,{\leqslant}\, ({\mathcal D}\circ {\mathcal D})(f)<2K\} \end{align*} $$

such that

(4.35) $$ \begin{align} E({\mathcal D})\,{\leqslant}\,K^{2}\#{\mathcal F} N^{o(1)}. \end{align} $$

Combining with Equations (4.31) and (4.35) gives

(4.36) $$ \begin{align} \mathsf {T}_{4,2}(N;\overline{j},q)\,{\leqslant}\,\Delta^{4}K^{2}\#{\mathcal F} N^{o(1)}. \end{align} $$

We apply Corollary 4.4 to estimate the right-hand side of Equation (4.36).

We now fix some $\varepsilon> 0$ and suppose first that one of Equation (4.8) or Equation (4.9) does not hold. In particular, assume

(4.37) $$ \begin{align} K < \left(\frac{N^{15/2}}{\Delta^{12}q^{1/2}}+\frac{N^{5}}{\Delta^{8}q^{1/4}}\right) N^{\varepsilon} \end{align} $$

or

(4.38) $$ \begin{align} \Delta < \frac{N^{5/8+\varepsilon}}{q^{1/8}}, \end{align} $$

where we have use the assumption (4.30) to ignore the term $N^{3/2}/q^{1/2}$ in Equation (4.9). If Equation (4.37) holds, then using the trivial bounds

$$ \begin{align*}K \#{\mathcal F} \,{\leqslant}\, (\#{\mathcal D})^2 \qquad\mbox{and}\qquad \Delta \#{\mathcal D} \ll N^2\,, \end{align*} $$

we derive from Equation (4.36)

(4.39) $$ \begin{align} \mathsf {T}_{4,2}(N;\overline{j},q)& \,{\leqslant}\, \Delta^{4}(\#{\mathcal D})^2K N^{o(1)} \,{\leqslant}\, \Delta^2 K N^{4+o(1)} \nonumber\\ & \,{\leqslant}\, \left(\frac{N^{15/2}}{\Delta^{10}q^{1/2}}+\frac{N^{5}}{\Delta^{6}q^{1/4}}\right) N^{4+\varepsilon+o(1)} \nonumber\\ & \,{\leqslant}\, \left(\frac{N^{15/2 }}{q^{1/2}}+\frac{N^{5}}{q^{1/4}}\right)N^{4+\varepsilon+o(1)}\\ & \,{\leqslant}\, \left(\frac{N^{11/2 }}{q^{1/2}}+\frac{N^{3}}{q^{1/4}}\right)N^{6+\varepsilon+o(1)}.\nonumber \end{align} $$

If Equation (4.38) holds, then from Equation (4.36)

(4.40) $$ \begin{align} \mathsf {T}_{4,2}(N;\overline{j},q)& \,{\leqslant}\, N^{o(1)}\Delta^{4}(\# {\mathcal D})^3 \,{\leqslant}\, N^{6+o(1)}\Delta \\ & \,{\leqslant}\, \frac{N^{6+5/8+o(1)}}{q^{1/8}}.\nonumber \end{align} $$

Hence, if one of the conditions (4.8) or (4.9) does not hold then combining Equations (4.39) and (4.40) we obtain

(4.41) $$ \begin{align} \mathsf {T}_{4,2}(N;\overline{j},q) \,{\leqslant}\, \left(\frac{N^{5/8}}{q^{1/8}}+\frac{N^{8}}{q^{1/2}}\right)N^{6+\varepsilon+o(1)}. \end{align} $$

Suppose next that Equations (4.37) and (4.38) both fail and thus both Equation (4.8) and Equation (4.9) hold. By Corollary 4.4, we have either

(4.42) $$ \begin{align} K\ll 1, \end{align} $$

or

(4.43) $$ \begin{align} K \#{\mathcal F} \,{\leqslant}\, \frac{N^{3+o(1)}}{\Delta^4}. \end{align} $$

If Equation (4.42) holds, then from Equation (4.36) and the trivial bound $K \#{\mathcal F} \,{\leqslant }\, (\#{\mathcal D})^2$ , we derive

$$ \begin{align*}\mathsf {T}_{4,2}(N;\overline{j},q) \,{\leqslant}\, \Delta^{4}K^2\#{\mathcal F} N^{o(1)} \,{\leqslant}\, \Delta^{4}K\#{\mathcal F} N^{o(1)} \,{\leqslant}\, \Delta^4(\#{\mathcal D})^2 N^{o(1)}. \end{align*} $$

Now, the bound (4.32) and Theorem 1.1 (under the condition (4.30)) yield

$$ \begin{align*}\mathsf {T}_{4,2}(N;\overline{j},q)\,{\leqslant}\,T_{2,2}(N;j,q)^2 N^{o(1)} \,{\leqslant}\, N^{4+o(1)}. \end{align*} $$

If Equation (4.43) holds, then using Equation (4.33)

(4.44) $$ \begin{align} \mathsf {T}_{4,2}(N;\overline{j},q) \,{\leqslant}\, N^{3+o(1)}K \,{\leqslant}\, N^{3+o(1)}\#{\mathcal D} \,{\leqslant}\, N^{5+o(1)}. \end{align} $$

Combining Equations (4.41) and (4.44), since $\varepsilon>0$ is arbitrary, we complete the proof.

5.1 Product polynomials

In the proof of [Reference Shkredov, Shparlinski and Zaharescu26, Lemma 5.1], a certain polynomial in four variables with integer coefficients played a key role. More precisely, it has been found in [Reference Shkredov, Shparlinski and Zaharescu26] that the polynomial

$$ \begin{align*} F(U,V,X,Y) & = 64 UVXY\\ & \qquad - \left(4U V + 4XY - \left(X +Y -U - V\right)^2\right)^2 \, \end{align*} $$

has the following property. Letting $U = u^2$ , $V = v^2$ , $X = x^2$ and $Y = y^2$ , one has that $F(u^2, v^2, x^2, y^2) = 0$ for any $u, v, x, y$ for which $u+v = x+y$ (over any commutative ring). We now proceed to discuss this property in a more general context.

Denote ${\mathcal U}_k = \{ \omega \in \mathbb {C} :~\omega ^k = 1\}$ , and consider the polynomial

$$ \begin{align*}G_k(X_1,X_2, X_3, X_4) = \prod_{\omega_1,\omega_2,\omega_3 \in{\mathcal U}_k} (\omega_1 X_1+ \omega_2 X_2 - \omega_{3} X_{3} - X_4) \end{align*} $$

defined over the cyclotomic field $K_k = \mathbb {Q}\left (\exp (2 \pi i /k)\right )$ . Since the Galois group ${\mathrm {Gal}}(K_k/\mathbb {Q})$ of K is cyclic and any automorphism $\sigma $ of $K_k$ over $\mathbb {Q}$ is a multiplication by some $\omega \in {\mathcal U}_k$ , we see that

$$ \begin{align*} \sigma&\left(G_k(X_1,X_2, X_3, X_4)\right) \\ & = \prod_{\omega_1, \omega_2, \omega_3 \in{\mathcal U}_k} \left(\sigma\left(\omega_1\right) X_1+ \sigma\left(\omega_2\right) X_2 - \sigma\left(\omega_{3}\right) X_{3} - \sigma\left(1\right) X_4\right)\\ & = \prod_{\omega_1, \omega_2, \omega_3 \in{\mathcal U}_k} \left(\omega \omega_1 X_1+ \omega \omega_2 X_2 - \omega \omega_{3} X_{3} -\omega X_4\right)\\ &= \omega^{k^3} \prod_{\omega_1, \omega_2, \omega_3 \in{\mathcal U}_k} \left(\omega_1 X_1+ \omega_2 X_2 - \omega_{3} X_{3} - X_4\right)\\ &= G_k(X_1,X_2, X_3, X_4). \end{align*} $$

Hence, $G_k$ has rational coefficients. Since obviously these coefficients are algebraic integers, we see that $G_k\left (X_1,X_2, X_3, X_4\right ) \in \mathbb {Z}[X_1,X_2, X_3, X_4]$ .

We also see that

$$ \begin{align*} \prod_{\omega_1, \omega_2, \omega_3 \in{\mathcal U}_k}& \left(\omega_1 X_1+ \omega_2 X_2 - \omega_{3} X_{3} - X_4\right)\\ & = \prod_{\omega_1, \omega_2, \omega_3 \in{\mathcal U}_k} \left(\omega_1 X_1+ \omega_1 \omega_2 X_2 - \omega_1 \omega_{3} X_{3} - X_4\right)\\ & = \prod_{\omega_2, \omega_3 \in {\mathcal U}_k} \prod_{\omega_1 \in {\mathcal U}_k} \left(\omega_1 \left(X_1+ \omega_2 X_2 - \omega_{3} X_{3}\right) - X_4\right)\\ & = (-1)^k \prod_{\omega_2, \omega_3 \in {\mathcal U}_k} \left( \left(X_1+ \omega_2 X_2 - \omega_{3} X_{3}\right)^k - X_4^k\right). \end{align*} $$

Therefore, $G_k(X_1,X_2, X_3, X_4)$ is a polynomial in $X_4^k$ . Similarly,

$$ \begin{align*} \prod_{\omega_1, \omega_2, \omega_3 \in{\mathcal U}_k}& \left(\omega_1 X_1+ \omega_2 X_2 - \omega_{3} X_{3} - X_4\right)\\ & = \prod_{ \omega_2, \omega_3 \in{\mathcal U}_k} \prod_{\omega_1 \in {\mathcal U}_k} \left( X_1+ \omega_1^{-1} \left( \omega_2 X_2 - \omega_{3} X_{3} - X_4\right)\right)\\ & = \prod_{\omega_2, \omega_3 \in {\mathcal U}_k} \left( X_1^k+ \left( \omega_{3} X_{3} +X_4- \omega_2 X_2 \right)^k\right). \end{align*} $$

Thus, it is also a polynomial in $X_1^k$ and of course also in $X_2^k$ and $X_3^k$ . Hence, we can write

$$ \begin{align*}G_k(X_1,X_2, X_3, X_4) = F_k\left(X_1^k,X_2^k, X_3^k, X_4^k\right) \end{align*} $$

for some polynomial $F_k\left (X_1,X_2, X_3, X_4\right ) \in \mathbb {Z}[X_1,X_2, X_3, X_4]$ .

Remark 5.1. It is clear that this construction can be extended in several directions, in particular to polynomials $F_{\nu ,k} \in \mathbb {Z}[X_1, \ldots , X_{2\nu }]$ such that

$$ \begin{align*}F_{\nu,k}\left(x_1^k, \ldots, x_{2\nu}^k\right) = 0 \end{align*} $$

whenever $x_1+ \ldots +x_\nu = x_{\nu +1} + \ldots + x_{2\nu }$ .

5.2 The zero set of $F_k(X_1,\ X_2,\ X_3,\ X_4)$

We now need the following bound on the number of integer zeros of $F_k$ in a box. Define by $T_k(N)$ by

$$ \begin{align*} T_k(N)=\# \{ (n_1, n_2, n_3, n_4) \in \mathbb{Z}^4 :~ 1 & \,{\leqslant}\, n_1,n_2,n_3, n_4 \,{\leqslant}\, N,\\ & \qquad F_k(n_1,n_2,n_3,n_4) = 0 \}. \end{align*} $$

Our next result gives a bound for $T_k(N)$ .

Proof. Take a solution $(n_1, n_2, n_3, n_4)$ to $F_k( n_1, n_2, n_3, n_4) = 0$ satisfying $1 \,{\leqslant }\, n_1, n_2, n_3, n_4 \,{\leqslant }\, N$ . Denote by $t_1, t_2, t_3, t_4$ the positive real numbers that are roots of order k of $n_1, n_2, n_3, n_4$ , respectively.

Therefore, there exist roots of unity $\omega _1,\omega _2,\omega _3 \in \mathcal {U}_k$ such that

(5.1) $$ \begin{align} \omega_1 t_1 + \omega_2 t_2 - \omega_3 t_3 - t_4 = 0. \end{align} $$

We now distinguish two cases.

Case 1. At least one of the roots of unity $\omega _1$ , $\omega _2$ , $\omega _3$ is not real. Complex conjugation then provides a second linear equation,

(5.2) $$ \begin{align} \bar \omega_1 t_1 + \bar \omega_2 t_2 - \bar \omega_3 t_3 - t_4 = 0. \end{align} $$

which is different from Equation (5.1). Then using Equations (5.1) and (5.2) to eliminate $t_4$ , one obtains a nontrivial linear equation in $t_1, t_2$ and $t_3$ which obviously has at most $O(N^2)$ solutions, after which $t_4$ is uniquely defined.

Thus, the total number of solutions in Case 1 is $O(N^2)$ .

Case 2. All three of $\omega _1, \omega _2, \omega _3$ are real, that is, $\omega _1, \omega _2, \omega _3 \in \{ -1, 1\}$ , and Equation (5.1) reduces to

(5.3) $$ \begin{align} t_1 \pm t_2 \pm t_3 \pm t_4 = 0. \end{align} $$

We observe that Case 2 also covers the $2N^2 + O(N)$ diagonal solutions.

To treat the nondiagonal solutions, one can now apply results of Besicovitch [Reference Besicovitch3], Mordell [Reference Mordell22], Siegel [Reference Siegel28] or the more recent results of Carr and O’Sullivan [Reference Carr and O’Sullivan9]. For instance, [Reference Carr and O’Sullivan9, Theorem 1.1] shows that a set of real k-th roots of integers that are pairwise linearly independent over the rationals must also be linearly independent. Applying this to the set $t_1, t_2, t_3, t_4$ , which by Equation (5.3) is not linearly independent over $\mathbb {Q}$ , it follows that two of them, for example, $t_1$ and $t_2$ , are linearly dependent over $\mathbb {Q}$ . We derive that there are positive integers $a_1, a_2, b$ such that

$$ \begin{align*}t_1^k = n_1 = b a_1^k \qquad\mbox{and}\qquad t_2^k = n_2 = b a_2^k, \end{align*} $$

where b is not divisible by a k-th power of a prime. That is, $a_1^k$ is the largest k-th power that divides $n_1$ , and $a_2^k$ is the largest k-th power that divides $n_2$ .

Then letting $t_5$ denote the positive k-th root of b, Equation (5.3) becomes

(5.4) $$ \begin{align} (a_1 \pm a_2) t_5 \pm t_3 \pm t_4 = 0. \end{align} $$

Without loss of generality, we can assume that $a_1 \,{\geqslant }\, a_2$ . Hence, for any fixed $1 \,{\leqslant }\, a_2 \,{\leqslant }\, a_1 \,{\leqslant }\, N^{1/k}$ there are at most $N/a_1^k$ possible values for b and thus for $t_5$ . After $a_1$ , $a_2$ and $t_5$ are fixed, there are obviously at most N pairs $(t_3,t_4)$ satisfying Equation (5.4). Hence, the total contribution from such solutions is

$$ \begin{align*}\sum_{ 1{\leqslant}a_2{\leqslant}a_1{\leqslant}N^{1/k}} N^2/a_1^k \,{\leqslant}\, \sum_{ 1{\leqslant}a_1 \,{\leqslant}\, N^{1/k}} N^2/a_1^{k-1} \ll N^2 \end{align*} $$

which concludes the proof.

We remark that the case of $k=2$ can also be included in Lemma 5.2; however, this case is already fully covered by the results of [Reference Shkredov, Shparlinski and Zaharescu26].

5.3 Concluding the proof

Clearly, the congruence

implies that

for the above polynomial $F_k$ . Since $F_k$ is homogeneous, this implies that

Since for a prime $q\sim Q$ , $a \in \mathbb {F}_q$ and $j \in \mathbb {F}_q^*$ , there are at most k solutions to the congruence in variable $z\in \mathbb {F}_q$ , and thus at most $2k$ solution in variable $z \in [1,N]$ (since $N \,{\leqslant }\, Q \,{\leqslant }\, 2q$ ) we have

where, as before, $\overline {j}$ denotes the multiplicative inverse of j modulo q. Changing the order of summation and separating the sum over the variables $U,V,X,Y$ into two parts depending on whether $F_k(U,V,X,Y) = 0$ or not, we derive

$$ \begin{align*} \sum_{\substack{ q \sim Q \\ q~\text{prime}}} & \max_{j \in \mathbb{F}_q^*} \mathsf {E}_k(N;j,q) \ll \mathop{\sum\ldots \sum}_{U,V,X,Y \in [1,N]} \, \sum_{\substack{q \sim Q \\ q~\text{prime} \\ q \mid F_k(U,V,X,Y)}} 1\\ & \ll \frac{Q}{\log Q} \mathop{\sum\ldots \sum}_{\substack{U,V,X,Y \in[1,N]\\ F_k(U,V,X,Y) = 0}} 1 + \mathop{\sum\ldots \sum}_{\substack{U,V,X,Y \in [1,N]\\ F_k(U,V,X,Y) \ne 0} } \sum_{\substack{q \sim Q \\ q~\text{prime} \\ q \mid F_k(U,V,X,Y)}} 1. \end{align*} $$

Recall that $F_k$ is a polynomial with constant coefficients of degree $k^3$ . Hence, $F_k(U,V,X,Y) \ll N^{k^3}$ , and thus trivially has at most $O\left (\log N\right )$ prime divisors. Hence, we derive

$$ \begin{align*}\sum_{\substack{ q \sim Q \\ q~\text{prime}}} \max_{j \in \mathbb{F}_q^*} \mathsf {E}_k(N;j,q) \ll \frac{Q}{\log Q} T_k(N) + N^{4+o(1)} , \end{align*} $$

and applying Lemma 5.2 we conclude the proof.

6.1 Preliminary discussion

We need some facts about the Gowers norms, introduced in the celebrated work of Gowers [Reference Gowers16, Reference Gowers17] on the first quantitative bound for the famous Szemerédi Theorem [Reference Szemerédi29] about sets avoiding arithmetic progressions of length four and longer. As an important step in the proof, Gowers [Reference Gowers16, Reference Gowers17] observes that there are very random sets having an unexpected number of arithmetic progressions of length $l\,{\geqslant }\, 4$ . An example is, basically, the set

(6.1) $$ \begin{align} {\mathcal A}^{(k)} =\left \{ x \in \mathbb{Z}_N :~ x^k \in \{1,\ldots, c_k N\} \right\} \,, \end{align} $$

where $c_k>0$ is an appropriate constant, depending on $k\,{\geqslant }\, 2$ only (see the beginning of [Reference Gowers17, Section 4] and also [Reference Gowers18]). Then the set ${\mathcal A}^{(k)}$ has an enormous number of arithmetic progressions of length $k+2$ but the expected number of shorter progressions. In Theorem 1.4, we consider the sets ${\mathcal N}^{1/k}$ , where ${\mathcal N}$ is a set with small doubling. Clearly, such sets generalise the construction (6.1). Below, we show that these sets are random in the sense that they all have small additive energy. Actually, we obtain a stronger property that Gowers norms of its characteristic functions are small and thus this has even more parallels to the Gowers construction (6.1). On the other hand, sets ${\mathcal N}^{1/k}$ preserve all essential combinatorial properties of the sets ${\mathcal A}^{(k)}$ . For example, for $k=2$ and any $s\neq 0$ we have for an arbitrary $x\in {\mathcal N}^{1/2} \cap ({\mathcal N}^{1/2} + s)$

$$ \begin{align*}x^2 \in {\mathcal N} \qquad\mbox{and}\qquad (x-s)^2 \in {\mathcal N}. \end{align*} $$

Thus, $2sx - s^2 \in {\mathcal N}-{\mathcal N} $ or $x\in ({\mathcal N}-{\mathcal N} + s^2)/2s$ . Hence, all intersections ${\mathcal N}^{1/2} \cap ({\mathcal N}^{1/2} + s)$ are additively rich sets exactly as in construction (6.1) (we literally use such facts in the proof of Theorem 1.4 below).

6.2 Gowers norms

Now, we are ready to give general definitions. Suppose that G is an abelian group with the group operation $+$ and ${\mathcal A}\subseteq G$ is a finite set. Having a sequence of elements $s_1,\ldots ,s_l \in G$ , we define the set

$$ \begin{align*}{\mathcal A}_{s_1,\ldots,s_l} = {\mathcal A} \cap ({\mathcal A} -s_1) \cap \ldots \cap ({\mathcal A} -s_l). \end{align*} $$

Let $ \| {\mathcal A} \|_{\mathcal {U}^{k}} $ be the Gowers nonnormalised kth-norm [Reference Gowers17] of the characteristic function of ${\mathcal A}$ (in additive form). We have (see, for example, [Reference Shkredov25]):

$$ \begin{align*}\| {\mathcal A} \|_{\mathcal{U}^{k}} = \sum_{x_0,x_1,\ldots,x_k \in G}\, \prod_{\varepsilon \in \{ 0,1 \}^k} {\mathcal A} \left( x_0 + \sum_{j=1}^k \varepsilon_j x_j \right) \,, \end{align*} $$

where $\varepsilon = (\varepsilon _1,\ldots ,\varepsilon _k)$ (we also recall that we use ${\mathcal A}(a)$ for the indicator function of ${\mathcal A}$ ). In particular,

$$ \begin{align*}\| {\mathcal A} \|_{\mathcal{U}^{2}} = \sum_{x_0,x_1,x_2\in G} {\mathcal A}(x_0) {\mathcal A}(x_0 + x_1) {\mathcal A}(x_0 + x_2) {\mathcal A}(x_0 + x_1 + x_2) = E({\mathcal A}) \end{align*} $$

is the additive energy of ${\mathcal A}$ , that is

$$ \begin{align*}E({\mathcal A}) = \# \{(a_1,a_2,a_3,a_4) \in {\mathcal A}^4:~a_1 + a_2 = a_3 + a_4\} , \end{align*} $$

and

$$ \begin{align*}\| {\mathcal A} \|_{\mathcal{U}^{3}} = \sum_{s \in {\mathcal A}-{\mathcal A}} E({\mathcal A}_s) \,. \end{align*} $$

Moreover, the induction property for Gowers norms holds; see [Reference Gowers17]

$$\begin{align*}\| {\mathcal A} \|_{\mathcal{U}^{k+1}} = \sum_{s \in {\mathcal A}-{\mathcal A}} \| {\mathcal A}_s \|_{\mathcal{U}^{k}} \end{align*}$$

and

(6.2) $$ \begin{align} \| {\mathcal A} \|_{\mathcal{U}^k} = \sum_{s_1,\ldots,s_k\in G} \# {\mathcal A}_{\pi(s_1,\ldots,s_k)} \,, \end{align} $$

where $\pi (s_1,\ldots ,s_k)$ is a vector with $2^k$ components, namely,

$$ \begin{align*}\pi(s_1,\ldots,s_k) = \left( \sum_{j=1}^k s_j \varepsilon_j \right)_{\left(\varepsilon_1, \ldots, \varepsilon_k\right) \in \{ 0,1 \}^k} \,. \end{align*} $$

Notice also

(6.3) $$ \begin{align} \| {\mathcal A} \|_{\mathcal{U}^{k+1}} = \sum_{s_1,\ldots,s_k \in G} \left(\# {\mathcal A}_{\pi(s_1,\ldots,s_k)}\right)^2 \,. \end{align} $$

It is proved in [Reference Gowers17] that kth-norms of the characteristic function of any set are connected to each other. It is shown in [Reference Shkredov25] that the connection for the nonnormalised norms does not depend on size of the group G. Here, we formulate a particular case of [Reference Shkredov25, Proposition 35], which relates $\| {\mathcal A} \|_{\mathcal {U}^{k}}$ and $\| {\mathcal A} \|_{\mathcal {U}^{2}}$ .

Lemma 6.1. Let ${\mathcal A}$ be a finite subset of an abelian group G with the group operation $+$ . Then for any integer $k\,{\geqslant }\, 1$ , we have

$$ \begin{align*} \| {\mathcal A} \|_{\mathcal{U}^{k+1}} \,{\geqslant}\, \frac{\| {\mathcal A} \|^{(3k-2)/(k-1)}_{\mathcal{U}^{k}}}{\| {\mathcal A} \|^{2k/(k-1)}_{\mathcal{U}^{k-1}}} \,. \end{align*} $$

Next, we have to relate $\| {\mathcal A} \|_{\mathcal {U}^{k}}$ and $E({\mathcal A})$ ; see [Reference Shkredov25, Remark 36].

Lemma 6.2. Let ${\mathcal A}$ be a finite subset of an abelian group G with the group operation $+$ . Then for any integer $k\,{\geqslant }\, 1$ , we have

$$ \begin{align*} \| {\mathcal A} \|_{\mathcal{U}^{k}} \,{\geqslant}\, E({\mathcal A})^{2^k-k-1}\left( \# {\mathcal A}\right)^{-(3\cdot 2^k -4k -4)} \,. \end{align*} $$

6.3 Concluding the proof

Let ${\mathcal A} = {\mathcal N}^{1/k}$ .

6.3.1 Case $k=3$

Let us start with the case $k=3$ . Below, we can assume that the quantity L is sufficiently small because otherwise the result is trivial.

For any $s\neq 0$ , consider the set ${\mathcal A}_s = {\mathcal A} \cap ({\mathcal A}-s)$ and let $x\in {\mathcal A}_s$ . Then $x^3, (x+s)^3 \in {\mathcal N}$ and hence

$$\begin{align*}3s (x+s/2)^2 + s^3/4 = 3sx^2 + 3s^2 x + s^3 \in {\mathcal N} - {\mathcal N} \,. \end{align*}$$

Put ${\mathcal B}_s ={\mathcal A}_s + s/2$ , so $\# {\mathcal B}_s = \# {\mathcal A}_s$ . Furthermore, let ${\mathcal C}_s = \{x^2:~ x \in {\mathcal B}_s\}$ . Clearly, by the Plünnecke inequality (see [Reference Tao and Vu30, Corollary 6.29]),

$$\begin{align*}\#({\mathcal C}_s + {\mathcal C}_s) \,{\leqslant}\, \#(2{\mathcal N} - 2{\mathcal N})\,{\leqslant}\,L^4 N = L_s \# {\mathcal A}_s \,, \end{align*}$$

where

$$ \begin{align*} L_s = \frac{L^4 N}{\# {\mathcal A}_s}. \end{align*} $$

Then, after applying estimate (1.3) with our restriction $N \,{\leqslant }\, q^{2/3}$ , we obtain

(6.4) $$ \begin{align} E({\mathcal A}_s) & = E({\mathcal B}_s) \ll E_{2}({\mathcal C}_s;q) \\ &\,{\leqslant}\,\left( L^4_s \left(\# {\mathcal A}_s\right)^4/q+ L^2_s \left(\# {\mathcal A}_s\right)^{11/4}\right) q^{o(1)} \,. \nonumber \end{align} $$

We now assume that

(6.5) $$ \begin{align} \#{\mathcal A}_s \,{\geqslant}\, N^{4/5} L^{32/5}. \end{align} $$

We also observe that we can always assume that $L \,{\leqslant }\, N^{1/32}$ as otherwise the result is trivial. Further, to show that the second term in Equation (6.4) dominates the first one, we need to check that

(6.6) $$ \begin{align} L^4_s \left(\# {\mathcal A}_s\right)^4/q \,{\leqslant}\, L^2_s \left(\# {\mathcal A}_s\right)^{11/4} \end{align} $$

or $L^2_s \left ( \#{\mathcal A}_s\right )^{5/4} \,{\leqslant }\, q$ , which in turn is equivalent to $\left ( \#{\mathcal A}_s\right )^{3} \,{\geqslant }\, L^{32} N^8 q^{-4} $ . Since for $L \,{\leqslant }\, N^{1/32}$ and $N \,{\leqslant }\, q^{2/3}$ , we have

$$ \begin{align*}N^{12/5} L^{96/5}\,{\geqslant}\, L^{32} N^8 q^{-4}, \end{align*} $$

we see that under the assumption (6.5) we have Equation (6.6) and hence the bound (6.4) becomes

(6.7) $$ \begin{align} E({\mathcal A}_s)\,{\leqslant}\,L^2_s \left(\# {\mathcal A}_s\right)^{11/4} q^{o(1)}\,{\leqslant}\,L^8 N^2\left( \#{\mathcal A}_s\right)^{3/4} q^{o(1)} \,. \end{align} $$

By the definition of the sets ${\mathcal A}_s$ , we have

(6.8) $$ \begin{align} \sum_{s\in {\mathcal A}-{\mathcal A}} \# {\mathcal A}_s =\left( \# {\mathcal A}\right)^2 \,. \end{align} $$

Furthermore, using the definition of $\mathcal {U}_3$ –norm we write

(6.9) $$ \begin{align} \| {\mathcal A} \|_{\mathcal{U}^3} = \sum_{s \in {\mathcal A}-{\mathcal A}} E({\mathcal A}_s) = \sum_{s :\, \# {\mathcal A}_s{\leqslant}T} E({\mathcal A}_s) + \sum_{s :\, \# {\mathcal A}_s> T} E({\mathcal A}_s). \end{align} $$

First, we observe that

$$ \begin{align*} \sum_{s :\, \# {\mathcal A}_s{\leqslant}T} E({\mathcal A}_s) & = \#\{ (a_1, a_2,a_3, a_4, s) \in {\mathcal A}^4\times\left( {\mathcal A}-{\mathcal A}\right) :\\ & \qquad \qquad \qquad a_1+a_2= a_3+a_4, \ \# {\mathcal A}_s\,{\leqslant}\,T, \\ & \qquad \qquad \qquad \qquad \qquad a_i - s \in {\mathcal A}, \ i =1, \ldots, 4\}\,. \end{align*} $$

Thus, for each of $E({\mathcal A})$ choices of quadruples $(a_1, a_2,a_3, a_4) \in {\mathcal A}^4$ with $a_1+a_2= a_3+a_4$ , there are at most T possibilities for s with $ \# {\mathcal A}_s\,{\leqslant }\,T$ and we derive

(6.10) $$ \begin{align} \sum_{s :\, \# {\mathcal A}_s{\leqslant}T} E({\mathcal A}_s) \,{\leqslant}\, T E({\mathcal A}) \,. \end{align} $$

We now choose

(6.11) $$ \begin{align} T= 27 E({\mathcal A})^{-4/5} L^{32/5} N^{16/5} \end{align} $$

and note that the trivial upper bound $E({\mathcal A}) \,{\leqslant }\, (\# {\mathcal A})^3 \,{\leqslant }\, 27N^3$ implies that $T \,{\geqslant }\, N^{4/5} L^{32/5}$ . Hence, for any s with $\# {\mathcal A}_s> T$ the condition (6.5) is satisfied and so the bound (6.7) holds.

Hence, by identity (6.8), we obtain

(6.12) $$ \begin{align} \sum_{s :\, \# {\mathcal A}_s> T} E({\mathcal A}_s) &\,{\leqslant}\,L^8 N^2 q^{o(1)} \sum_{s :\, \# {\mathcal A}_s > T}\left( \#{\mathcal A}_s\right)^{3/4} \nonumber\\ &\,{\leqslant}\,L^8 N^2 T^{-1/4} q^{o(1)}\sum_{s :\, \# {\mathcal A}_s> T}\#{\mathcal A}_s \\ &\,{\leqslant}\,L^8 N^2 \cdot N^{2} T^{-1/4} q^{o(1)} = L^8 N^4 T^{-1/4} q^{o(1)}\, .\nonumber \end{align} $$

The value of T in Equation (6.11) is chosen to balance the bounds (6.10) and (6.12) and thus from Equation (6.9) we derive

$$\begin{align*}\| {\mathcal A} \|_{\mathcal{U}^3}\,{\leqslant}\,E({\mathcal A})^{1/5} L^{32/5} N^{16/5} q^{o(1)} \,. \end{align*}$$

Finally, applying Lemma 6.2, we obtain

$$\begin{align*}E({\mathcal A}) \,{\leqslant}\, N^2 \| {\mathcal A} \|^{1/4}_{\mathcal{U}^3}\,{\leqslant}\,L^{8/5} N^{14/5} E({\mathcal A})^{1/20} q^{o(1)} \,, \end{align*}$$

and whence

$$\begin{align*}E({\mathcal A}) \,{\leqslant}\, L^{32/19} N^{56/19} q^{o(1)}\,, \end{align*}$$

which gives the desired result for $k=3$ .

6.3.2 Case $k=4$

Next, we consider the case $k=4$ . Let

$$ \begin{align*} {\mathcal A}_{\pi(s,t)} = {\mathcal A} \cap ({\mathcal A}-s) \cap ({\mathcal A}-t) \cap ({\mathcal A}-s-t), \end{align*} $$

and let $x\in {\mathcal A}_{\pi (s,t)}$ . Then $x^4, (x+s)^4, (x+t)^4, (x+t+s)^4 \in {\mathcal N}$ and hence ${\mathcal N} - {\mathcal N}$ contains

$$\begin{align*}4u x^3 + 6 u^2 x^2 + 4 u^3 x+ u^4, \qquad u \in \{s,t, s+t\}. \end{align*}$$

Subtracting the expressions with s and t from the expression with $s+t$ , we see that $3{\mathcal N}-3{\mathcal N}$ contains $12 st x^2 + 12 (t^2 s + ts^2) x + (t+s)^4-s^4-t^4$ and we can apply a version of previous arguments. Actually, in our particular case $k=4$ one can write exact identity

$$\begin{align*}(x+t+s)^4 + x^4 - (x+s)^4 - (x+t)^4 = 12 st x^2 + 12 (t^2 s + ts^2) x + (t+s)^4-s^4-t^4 \end{align*}$$

and thus even it is enough to consider the set $2{\mathcal N}-2{\mathcal N}$ . In particular, since by the Plünnecke inequality (see [Reference Tao and Vu30, Corollary 6.29])

$$\begin{align*}\#(2{\mathcal N} - 2{\mathcal N})\,{\leqslant}\,L^4 N, \end{align*}$$

the role of $L_s$ is now played by

$$ \begin{align*} L_{s,t} = \frac{L^{8} N}{\# {\mathcal A}_{\pi(s,t)}}. \end{align*} $$

We also set

$$ \begin{align*} T = (E({\mathcal A}) N^{2} L^{16} \| {\mathcal A} \|_{\mathcal{U}^3}^{-1})^{4/5} \end{align*} $$

and note that we have the trivial bound $\| {\mathcal A} \|_{\mathcal {U}^3} \,{\leqslant }\, N E({\mathcal A})$ . We also have

$$ \begin{align*} T \,{\geqslant}\, N^{4/5} L^{64/5}. \end{align*} $$

We now verify that $T^3 \,{\geqslant }\, L^{64} N^8 q^{-4}$ or

$$ \begin{align*} N^{12/5} L^{192/5} \,{\geqslant}\, L^{64} N^8 q^{-4} \end{align*} $$

which is equivalent to $N^{28} L^{128} \,{\leqslant }\, q^{20} $ . Since we can clearly assume that $L \,{\leqslant }\, N^{1/64}$ as otherwise the result is trivial, the last inequality hold under our assumption $N \,{\leqslant }\, q^{2/3}$ .

Hence, similar to the case $k=3$ after simple calculations, one verifies that for $\#{\mathcal A}_{s,t}> T$ , we have $ L_{s,t}^2 \left (\#{\mathcal A}_{\pi (s,t)}\right )^{5/4} \,{\leqslant }\, q$ which in turn is equivalent to

$$ \begin{align*} \left(\#{\mathcal A}_{\pi(s,t)}\right)^3 \,{\geqslant}\, T^3 \,{\geqslant}\, L^{64} N^8 q^{-4}. \end{align*} $$

Therefore, by Equation (1.3), we have

$$ \begin{align*} E({\mathcal A}_{\pi(s,t)}) &\,{\leqslant}\,\left( L^4_{s,t} \left(\# {\mathcal A}_{\pi(s,t)}\right)^4/q+ L^2_{s,t} \left(\# {\mathcal A}_{s,t}\right)^{11/4}\right) q^{o(1)}\\ & \,{\leqslant}\, L^2_{s,t} \left(\# {\mathcal A}_{s,t}\right)^{11/4} q^{o(1)}\\ & = L^{16} N^2 \left(\#{\mathcal A}_{\pi(s,t)}\right)^{3/4} q^{o(1)}. \end{align*} $$

Using Equations (6.2) and (6.3) and the arguments as above, we get

(6.13) $$ \begin{align} \| {\mathcal A} \|_{\mathcal{U}^4} &= \sum_{s,t} E({\mathcal A}_{\pi(s,t)}) \nonumber \\ & \,{\leqslant}\, T \| {\mathcal A} \|_{\mathcal{U}^3} + L^{16} N^2 q^{o(1)} \sum_{(s,t) :\, \#{\mathcal A}_{\pi(s,t)}> T} \#({\mathcal A}_{\pi(s,t)})^{3/4} \nonumber \\ & \,{\leqslant}\,T \| {\mathcal A} \|_{\mathcal{U}^3} + L^{16} N^2 E({\mathcal A}) T^{-1/4} q^{o(1)}\\ & \,{\leqslant}\, L^{64/5} N^{8/5} E^{4/5} ({\mathcal A}) \| {\mathcal A} \|^{1/5}_{\mathcal{U}^3} q^{o(1)}\ \nonumber \end{align} $$

since again we have chosen T to optimise the above bound.

On the other hand, applying Lemma 6.1 and then Lemma 6.2, we derive

(6.14) $$ \begin{align} \| {\mathcal A} \|_{\mathcal{U}^4} \,{\geqslant}\, \frac{\| {\mathcal A} \|^{7/2}_{\mathcal{U}^3}}{\| {\mathcal A} \|_{\mathcal{U}^2}^3} = \frac{\| {\mathcal A} \|^{7/2}_{\mathcal{U}^3}}{E^3({\mathcal A})} \,{\geqslant}\, \| {\mathcal A} \|^{1/5}_{\mathcal{U}^3} \cdot \frac{E^{51/5}({\mathcal A})}{N^{132/5}}. \end{align} $$

Comparing Equations (6.13) and (6.14)

$$\begin{align*}E({\mathcal A}) \,{\leqslant}\, L^{64/47} N^{3-1/47} q^{o(1)} \,, \end{align*}$$

which gives the desired result for $k=4$ .

6.3.3 Case $k\,{\geqslant }\, 5$

Finally, consider the general case, which we treat with a version of Weyl differencing. Now,

$$ \begin{align*}{\mathcal A}_{{s}} = {\mathcal A}_{\pi(s_1,\ldots,s_{k-2})} \end{align*} $$

and let $x\in {\mathcal A}_{\pi (s_1,\ldots ,s_{k-2})}$ . Indeed, we start with ${\mathcal A}_{s_1}$ and reduce the main term in $x^k, (x+s_1)^k \in {\mathcal N}$ deriving that $p_{k-1} (x) \in {\mathcal N}-{\mathcal N}$ , where $\deg p_{k-1}= k-1$ . After that consider $({\mathcal A}_{s_1})_{s_2} = {\mathcal A}_{\pi (s_1,s_2)}$ and reduce degree of the polynomial by one, and so on. We also note that by the Plünnecke inequality (see [Reference Tao and Vu30, Corollary 6.29])

$$\begin{align*}\#\left(2^{k-1}{\mathcal N} - 2^{k-1}{\mathcal N}\right)\,{\leqslant}\,L^{2^k} N, \end{align*}$$

the role of $L_s$ or $L_{s,t}$ is now played by

$$ \begin{align*} L_{{s}} = \frac{L^{2^{k}} N}{\# {\mathcal A}_{\pi({s})}}. \end{align*} $$

We now set

$$ \begin{align*} T = \left(N^{2} L^{2^{k+1}} \| {\mathcal A} \|_{\mathcal{U}^{k-2}} \| {\mathcal A} \|_{\mathcal{U}^{k-1}}^{-1}\right)^{4/5}. \end{align*} $$

Using the same arguments as above, after somewhat tedious calculations to verify all necessary conditions such as

(6.15) $$ \begin{align} N^8 L^{2^{k+3}} q^{-4} \,{\leqslant}\, \left(\#{\mathcal A}_{\pi(s_1,\ldots,s_{k-2})}\right)^3 \end{align} $$

to obtain

$$ \begin{align*} E({\mathcal A}_{\pi(s_1,\ldots,s_{k-2})})\,{\leqslant}\,L^{2^{k+1}} N^2 \left(\#{\mathcal A}_{\pi(s_1,\ldots,s_{k-2})}\right)^{3/4} q^{o(1)}. \end{align*} $$

In particular, to check Equation (6.15) we note that for the above choice of T we have

$$ \begin{align*} T \,{\geqslant}\, N^{4/5}L^{2^{k+3}/5} \end{align*} $$

and then derive

$$\begin{align*}N^8 L^{2^{k+3}} q^{-4} \,{\leqslant}\, N^{12/5} L^{3\cdot 2^{k+3}/5} \,{\leqslant}\, T^3 \end{align*}$$

which is true because $N \,{\leqslant }\, q^{2/3}$ and $L \,{\leqslant }\, N^{1/ 2^{k+3}}$ (which we can assume as otherwise the bound is trivial).

Using the formula (6.2) and Equation (6.3), we obtain

$$ \begin{align*} \| {\mathcal A} \|_{\mathcal{U}^k} & \,{\leqslant}\, T \| {\mathcal A} \|_{\mathcal{U}^{k-1}} + L^{2^{k+1}} N^2 q^{o(1)} \sum_{{s} :\, \#{\mathcal A}_{\pi({s})}> T} \#({\mathcal A}_{\pi({s})})^{3/4} \\ & \,{\leqslant}\, T \| {\mathcal A} \|_{\mathcal{U}^{k-1}} + L^{2^{k+1}} N^{2} \| {\mathcal A}\|_{\mathcal{U}^{k-2}} T^{-1/4} q^{o(1)}\\ & \,{\leqslant}\, L^{2^{k+1} \cdot 4/5} N^{8/5} \| {\mathcal A} \|^{4/5}_{\mathcal{U}^{k-2}} \| {\mathcal A} \|^{1/5}_{\mathcal{U}^{k-1}} q^{o(1)} \end{align*} $$

and hence by induction and Lemma 6.2

$$\begin{align*}E({\mathcal A})^{7\cdot 2^{k-1}-9}\,{\leqslant}\,L^{2^{k+3}} N^{21\cdot 2^{k-1}-28} q^{o(1)}. \end{align*}$$

In other words,

$$\begin{align*}E({\mathcal A})\,{\leqslant}\,L^{2^{k+3}/(7\cdot 2^{k-1}-9)} N^{3-1/(7\cdot 2^{k-1}-9)}q^{o(1)}\,, \end{align*}$$

which completes the proof.

8.1 Preliminaries

Our argument follows the proof of [Reference Dunn, Kerr, Shparlinski and Zaharescu13, Theorem 1.10], the only difference being our use of Corollary 2.1 and Theorem 2.2. We refer the reader to [Reference Dunn, Kerr, Shparlinski and Zaharescu13, Section 7] for more complete details.

Let $\widetilde {S}_q(h,P)$ denote the sum

$$ \begin{align*}\widetilde{S}_{q}(h,P) = \sum_{k=1}^P \Lambda(k) \sum_{\substack{x \in \mathbb{F}_q \\ x^2 =k}} {\mathbf{\,e}}_q(hx). \end{align*} $$

By partial summation, it is sufficient to show

$$ \begin{align*}\widetilde{S}_{q}(h,P) \ll q^{o(1)}(P^{15/16}+q^{1/8}P^{3/4}+q^{1/16}P^{69/80}+q^{13/88}P^{3/4}). \end{align*} $$

Let $J\,{\geqslant }\, 1$ be an integer. Using the Heath–Brown identity and a smooth partition of unity as in [Reference Dunn, Kerr, Shparlinski and Zaharescu13, Section 1.7], there exist some

$$ \begin{align*}\mathbf{V}=(M_1,\ldots , M_J,N_1,\ldots ,N_J) \in [1/2,2P]^{2J} \end{align*} $$

$2J$ -tuple of parameters satisfying

$$ \begin{align*}N_1 \,{\geqslant}\, \ldots \,{\geqslant}\, N_J, \quad M_1,\ldots ,M_J \,{\leqslant}\, P^{1/J},\quad P \ll Q \ll P, \end{align*} $$

(implied constants are allowed to depend on J),

(8.1) $$ \begin{align} Q = \prod_{i=1}^J M_i \prod_{j =1}^JN_j, \end{align} $$

and

• • the arithmetic functions $m_i \mapsto \gamma _i(m_i)$ are bounded and supported in $[M_i/2,2M_i]$ ;

• • the smooth functions $x_i \mapsto V_i(x)$ have support in $[1/2,2]$ and satisfy

  $$ \begin{align*}V_i^{(j)}(x) \ll q^{j \varepsilon} \end{align*} $$
  for all integers $j \,{\geqslant }\, 0$ , where the implied constant may depend on j and $\varepsilon $

such that defining

$$ \begin{align*} \Sigma(\mathbf{V})=\sum_{m_1, \ldots, m_J=1}^{\infty} &\gamma_1(m_1)\cdots \gamma_J(m_J) \sum_{n_1,\ldots , n_J=1}^\infty \\ \\& V_1 \left( \frac{n_1}{N_1} \right) \cdots V_J \left( \frac{n_J}{N_J} \right) \sum_{\substack{x \in \mathbb{F}_q \\ x^2=m_1 \cdots m_J n_1 \cdots n_J}} {\mathbf{\,e}}_q(hx), \end{align*} $$

we have

$$ \begin{align*}\widetilde{S}_{q}(h,P)\ll P^{o(1)}\Sigma(\mathbf{V}). \end{align*} $$

We proceed on a case-by-case basis depending on the size of $N_1$ . We first note a general estimate for the multilinear sums. Let ${\mathcal I},{\mathcal J}\subseteq \{1,\ldots ,J\}$ , and write

$$ \begin{align*}M=\prod_{i\in {\mathcal I}}M_i\prod_{j\in {\mathcal J}}N_j, \quad N=Q/M. \end{align*} $$

Grouping variables in $\Sigma (\mathbf {V})$ according to ${\mathcal I},{\mathcal J}$ , there exists $\alpha ,\beta $ satisfying

$$ \begin{align*}\|\alpha\|_{\infty}, \|\beta\|_{\infty}=Q^{o(1)} \end{align*} $$

such that

$$ \begin{align*}\Sigma(\mathbf{V})=\sum_{\substack{m{\leqslant}2^J M \\ n{\leqslant}2^J N}}\alpha(m)\beta(n)\sum_{\substack{x\in \mathbb{F}_q \\ x^2=mn}}{\mathbf{\,e}}_q(hx). \end{align*} $$

By Corollary 2.1,

(8.2) $$ \begin{align} & \Sigma(\mathbf{V}) \nonumber \\ &\quad\,{\leqslant}\,q^{1/8+o(1)}P^{3/4}\left(\frac{P^{3/16}}{q^{1/16}M^{3/16}}+1\right)\left(\frac{M^{3/16}}{q^{1/16}}+1\right) \\ &\quad \,{\leqslant}\, q^{o(1)}\left(P^{15/16}+\frac{q^{1/16}P^{15/16}}{M^{3/16}}+q^{1/16}P^{3/4}M^{3/16}+q^{1/8}P^{3/4} \right). \nonumber \end{align} $$

We proceed on a case by case basis depending on the size of $N_1$ . Let $P^{1/2}\,{\geqslant }\, H\,{\geqslant }\, P^{\varepsilon }$ be some parameters and take

$$ \begin{align*}J = {\left\lceil{\log P/\log H}\right\rceil}. \end{align*} $$

8.2 Small $N_1$

Suppose first $N_1 \,{\leqslant }\, H$ , then arguing as in [Reference Dunn, Kerr, Shparlinski and Zaharescu13, Equation (7.13)] we can choose two arbitrary sets ${\mathcal I}, {\mathcal J} \subseteq \{1, \ldots , J\}$ such that for

$$ \begin{align*}M = \prod_{i\in {\mathcal I}} M_i \prod_{j \in {\mathcal J}} N_j \qquad\mbox{and}\qquad N = Q/M, \end{align*} $$

where Q is given by Equation (8.1) and we have

$$ \begin{align*}P^{1/2} \ll M \ll H^{1/2}P^{1/2}. \end{align*} $$

Hence, by Equation (8.2)

(8.3) $$ \begin{align}\Sigma(\mathbf{V}) \,{\leqslant}\, q^{o(1)}\left(P^{15/16}+q^{1/16}P^{27/32}H^{3/32}+q^{1/8}P^{3/4} \right).\end{align} $$

8.3 Medium $N_1$

Let L be a parameter satisfying $H \,{\leqslant }\, L$ , and suppose next that

$$ \begin{align*}H \,{\leqslant}\, N_1 \,{\leqslant}\, L. \end{align*} $$

We may also suppose

$$ \begin{align*}H \,{\leqslant}\, N_2 \,{\leqslant}\, N_1 \,{\leqslant}\, L,\end{align*} $$

as otherwise we may argue as before to obtain the bound (8.3). In this case, we define $M,N$ as

$$ \begin{align*}N=\prod_{i=1}^{J}M_i\prod_{j=3}^{J}N_j \quad \text{and} \quad M=N_1N_2 \end{align*} $$

so that

$$ \begin{align*}H^2 \,{\leqslant}\, M \,{\leqslant}\, L ^2. \end{align*} $$

By Equation (8.2)

(8.4) $$ \begin{align}\Sigma(\mathbf{V}) \,{\leqslant}\, q^{o(1)}\biggl(P^{15/16}& +\frac{q^{1/16}P^{15/16}}{H^{3/8}}\\& \quad +q^{1/16}P^{3/4}L^{3/8}+q^{1/8}P^{3/4} \biggr). \nonumber\end{align} $$

8.4 Large $N_1$

Let R be a parameter to be chosen later and satisfying $R\,{\geqslant }\, c P^{1/2}$ for some sufficiently large constant $c> 0$ . Suppose next that

$$ \begin{align*}L^2 \,{\leqslant}\, N_1 \,{\leqslant}\, R.\end{align*} $$

Taking $M=N_1$ as above, we derive from Equation (8.2)

(8.5) $$ \begin{align}\Sigma(\mathbf{V} )\,{\leqslant}\,q^{o(1)}\biggl(P^{15/16}& +\frac{q^{1/16}P^{15/16}}{L^{3/8}}\\&\quad +q^{1/16}P^{3/4}R^{3/16}+q^{1/8}P^{3/4} \biggr). \nonumber \end{align} $$

8.5 Very large $N_1$

Finally, consider when $N_1\,{\geqslant }\, R$ . We now intend to apply Theorem 2.2 with $P/N_1\ll M\ll P/N_1$ and $N = N_1$ , where we notice that the condition $R\,{\geqslant }\, c P^{1/2}$ ensures that $M< N$ , provided that c is large enough. Choosing $r=2$ , we obtain

$$ \begin{align*} \Sigma(\mathbf{V}) &\,{\leqslant}\,q^{3/8+o(1)}(P/N_1)^{3/4} N_1^{1/4} \left(1+ \frac{P^{1/2}}{q^{3/8}}\right)\\ & = q^{3/8+o(1)}P^{3/4} N_1^{-1/2} \left(1+ \frac{P^{1/2}}{q^{3/8}}\right). \end{align*} $$

Using the assumption $P \,{\leqslant }\, q^{3/4}$ , we obtain

(8.6) $$ \begin{align} \Sigma(\mathbf{V})\,{\leqslant}\,q^{3/8+o(1)} \frac{P^{3/4}}{R^{1/2}}. \end{align} $$

8.6 Optimisation

Combining all previous bounds (8.3), (8.4), (8.5) and (8.6) results in

$$ \begin{align*} \widetilde{S}_{q}(h,P)& \,{\leqslant}\, q^{o(1)}(P^{15/16}+q^{1/8}P^{3/4})\\ & \qquad \qquad+q^{o(1)}\left(q^{1/16}P^{27/32}H^{3/32}+\frac{q^{1/16}P^{15/16}}{H^{3/8}}\right) \\ & \qquad \qquad \qquad +q^{o(1)}\left( q^{1/16}P^{3/4}L^{3/8}+\frac{q^{1/16}P^{15/16}}{L^{3/8}}\right) \\ & \qquad \qquad \qquad \qquad +q^{o(1)}\left(q^{1/16}P^{3/4}R^{3/16}+q^{3/8+o(1)}\frac{P^{3/4}}{R^{1/2}}\right). \end{align*} $$

Taking parameters

$$ \begin{align*}H=P^{1/5}, \quad L=P^{1/4}, \quad R=q^{5/11}, \end{align*} $$

gives

$$ \begin{align*}\widetilde{S}_{q}(h,P) \,{\leqslant}\, q^{o(1)}(P^{15/16}+q^{1/8}P^{3/4}+q^{1/16}P^{69/80}+q^{13/88}P^{3/4}), \end{align*} $$

which completes the proof.