Proof. The second part of the lemma follows by the first claim because, say,

$$\begin{eqnarray}\mathop{\sum }_{\ell \in L_{M}}i(\ell )^{3}\leqslant 8M^{3}|L_{M}|.\end{eqnarray}$$

First we show that for all $M\geqslant 2|A|^{2}/p>1$ ,

(15) $$\begin{eqnarray}|L_{M}|\leqslant \frac{4p|A|^{2}}{M^{2}}.\end{eqnarray}$$

The hypothesis $i(\ell )\geqslant 2|A|^{2}/p$ implies that $i(\ell )-|A|^{2}/p\geqslant i(\ell )/2\geqslant M/2$ .

Equation (9) now implies

$$\begin{eqnarray}\frac{M^{2}}{4}|L_{M}|\leqslant \mathop{\sum }_{\ell \in L_{M}}\biggl(i(\ell )-\frac{|A|^{2}}{p}\biggr)^{2}\leqslant \mathop{\sum }_{\text{all}\;\text{lines}\;\ell }\biggl(i(\ell )-\frac{|A|^{2}}{p}\biggr)^{2}\leqslant p|A|^{2}.\end{eqnarray}$$

Thus (15) follows.

To show that $|L_{M}|\ll |A|^{5}/M^{4}$ for $M\geqslant 2|A|^{2}/p$ , we consider two cases. First, suppose that $M<2|A|^{3/2}/p^{1/2}$ . By Lemma 15, we have

$$\begin{eqnarray}|L_{M}|\ll \frac{p|A|^{2}}{M^{2}}<\frac{p|A|^{2}}{M^{2}}\frac{4|A|^{3}}{pM^{2}}=\frac{4|A|^{5}}{M^{4}}.\end{eqnarray}$$

Next, suppose that $M\geqslant 2|A|^{3/2}/p^{1/2}$ ; in this case, we apply Theorem 8. Therefore we must confirm the condition $|A||L_{M}|\leqslant p^{2}$ . The hypothesis $M\geqslant 2|A|^{3/2}/p^{1/2}$ implies $M\geqslant 2|A|^{2}/p$ . Lemma 15 can therefore be applied and in conjunction with the hypothesis $M^{2}\geqslant 4|A|^{3}/p$ gives

$$\begin{eqnarray}|L_{M}|\leqslant \frac{4p|A|^{2}}{M^{2}}\leqslant \frac{p^{2}}{|A|}.\end{eqnarray}$$

Hence, $|A||L_{M}|\leqslant p^{2}$ and Theorem 8 may be applied. It gives

$$\begin{eqnarray}M|L_{M}|\leqslant \mathop{\sum }_{\ell \in L_{M}}i(\ell )={\mathcal{I}}(A\times A,L_{M})\ll |A|^{5/4}|L_{M}|^{3/4}+|A|^{2}+|L_{M}|.\end{eqnarray}$$

If $2\leqslant M\leqslant 2C$ , where $C$ is the implicit constant in Theorem 8, then

$$\begin{eqnarray}|L_{M}|\leqslant |A|^{4}\ll |A|^{5}/M^{4}.\end{eqnarray}$$

On the other hand, if $M>2C$ , it follows that

$$\begin{eqnarray}M|L_{M}|\ll |A|^{2}+|A|^{5/4}|L_{M}|^{3/4}.\end{eqnarray}$$

Hence, using $M\leqslant |A|$ ,

$$\begin{eqnarray}|L_{M}|\ll \frac{|A|^{5}}{M^{4}}+\frac{|A|^{2}}{M}\ll \frac{|A|^{5}}{M^{4}}.\Box\end{eqnarray}$$

Definition.

Let $A\subset \mathbb{F}$ . $T(A)$ is the number of ordered collinear triples in $A\times A$ . That is, ordered triples $(u,v,w)\in (A\times A)\times (A\times A)\times (A\times A)$ such that $u$ , $v$ and $w$ are all incident to the same line. $Q(A)$ is the number of ordered collinear quadruples in $A\times A$ .

Proof of inequality (24).

Using Lemma 9 and the formula $\sum _{\ell }i(\ell )=(p+1)|A|^{2}$ , we have

$$\begin{eqnarray}\displaystyle \mathop{\sum }_{\ell }i^{3}(\ell ) & = & \displaystyle \mathop{\sum }_{\ell }i(\ell )\biggl(i(\ell )-\frac{|A|^{2}}{p}+\frac{|A|^{2}}{p}\biggr)^{2}\nonumber\\ \displaystyle & = & \displaystyle \mathop{\sum }_{\ell }i(\ell )\biggl(i(\ell )-\frac{|A|^{2}}{p}\biggr)^{2}+\frac{2|A|^{2}}{p}\mathop{\sum }_{\ell }i^{2}(\ell )-\frac{|A|^{4}}{p^{2}}\mathop{\sum }_{\ell }i(\ell )\nonumber\\ \displaystyle & = & \displaystyle \mathop{\sum }_{\ell }i(\ell )\biggl(i(\ell )-\frac{|A|^{2}}{p}\biggr)^{2}+\frac{|A|^{6}}{p}+2|A|^{4}-\frac{|A|^{6}}{p^{2}}.\nonumber\end{eqnarray}$$

Thus by (16),

(26) $$\begin{eqnarray}\biggl|T(A)-\frac{|A|^{6}}{p}-2|A|^{4}+\frac{|A|^{6}}{p^{2}}\biggr|=\mathop{\sum }_{\ell }i(\ell )\biggl(i(\ell )-\frac{|A|^{2}}{p}\biggr)^{2}+O(|A|^{4}).\end{eqnarray}$$

Since any line contains at most $|A|$ points of $A\times A$ , we have $i(\ell )\leqslant |A|$ for all  $\ell$ . Thus by Lemma 9,

$$\begin{eqnarray}\mathop{\sum }_{\ell }i(\ell )\biggl(i(\ell )-\frac{|A|^{2}}{p}\biggr)^{2}\leqslant |A|\mathop{\sum }_{\ell }\biggl(i(\ell )-\frac{|A|^{2}}{p}\biggr)^{2}\leqslant p|A|^{3},\end{eqnarray}$$

which proves equation (24). ◻

Proof of Theorem 11.

We prove the following asymptotic estimates for the number of collinear triples and quadruples, from which the theorem follows from $|A|\leqslant p$ :

(27) $$\begin{eqnarray}\biggl|T(A)-\frac{|A|^{6}}{p}-2|A|^{4}+\frac{|A|^{6}}{p^{2}}\biggr|\ll p^{1/2}|A|^{7/2}\end{eqnarray}$$

and

(28) $$\begin{eqnarray}\biggl|Q(A)-\frac{|A|^{8}}{p^{2}}\biggr|\ll |A|^{5}\log |A|.\end{eqnarray}$$

We begin with (27). By equation (26), we have

$$\begin{eqnarray}\biggl|T(A)-\frac{|A|^{6}}{p}-2|A|^{4}+\frac{|A|^{6}}{p^{2}}\biggr|=\mathop{\sum }_{\ell }i(\ell )\biggl(i(\ell )-\frac{|A|^{2}}{p}\biggr)^{2}+O(|A|^{4}).\end{eqnarray}$$

We will show that

(29) $$\begin{eqnarray}\mathop{\sum }_{\ell }i(\ell )\biggl(i(\ell )-\frac{|A|^{2}}{p}\biggr)^{2}\ll p^{1/2}|A|^{7/2}.\end{eqnarray}$$

This will finish the proof of (27), because $|A|\leqslant p$ .

Let $\unicode[STIX]{x1D6E5}>0$ be a parameter, which we will specify later. Split the sum on the left-hand side of (29) into two parts:

$$\begin{eqnarray}\displaystyle \mathop{\sum }_{\ell }i(\ell )\biggl(i(\ell )-\frac{|A|^{2}}{p}\biggr)^{2} & = & \displaystyle \mathop{\sum }_{i(\ell )\leqslant \unicode[STIX]{x1D6E5}}i(\ell )\biggl(i(\ell )-\frac{|A|^{2}}{p}\biggr)^{2}\nonumber\\ \displaystyle & & \displaystyle +\,\mathop{\sum }_{i(\ell )>\unicode[STIX]{x1D6E5}}i(\ell )\biggl(i(\ell )-\frac{|A|^{2}}{p}\biggr)^{2}\nonumber\\ \displaystyle & = & \displaystyle I+\mathit{II}.\nonumber\end{eqnarray}$$

To bound term $I$ , we use Lemma 9:

$$\begin{eqnarray}I=\mathop{\sum }_{i(\ell )\leqslant \unicode[STIX]{x1D6E5}}i(\ell )\biggl(i(\ell )-\frac{|A|^{2}}{p}\biggr)^{2}\leqslant \unicode[STIX]{x1D6E5}\mathop{\sum }_{\ell }\biggl(i(\ell )-\frac{|A|^{2}}{p}\biggr)^{2}=\unicode[STIX]{x1D6E5}p|A|^{2}.\end{eqnarray}$$

To bound term $\mathit{II}$ , we first suppose that $\unicode[STIX]{x1D6E5}\geqslant 2|A|^{2}/p$ , an assumption that we will show is satisfied by our choice of $\unicode[STIX]{x1D6E5}$ . This allows us to write

$$\begin{eqnarray}\mathit{II}=\mathop{\sum }_{i(\ell )>\unicode[STIX]{x1D6E5}}i(\ell )\biggl(i(\ell )-\frac{|A|^{2}}{p}\biggr)^{2}\ll \mathop{\sum }_{i(\ell )>\unicode[STIX]{x1D6E5}}i(\ell )^{3}.\end{eqnarray}$$

Now we use dyadic decomposition and Lemma 10 to bound the right-hand side of the previous equation:

$$\begin{eqnarray}\displaystyle \mathop{\sum }_{i(\ell )>\unicode[STIX]{x1D6E5}}i(\ell )^{3}\ll \mathop{\sum }_{k\geqslant 0}(2^{k}\unicode[STIX]{x1D6E5})^{3}|L_{2^{k}\unicode[STIX]{x1D6E5}}|\ll \mathop{\sum }_{k\geqslant 0}(2^{k}\unicode[STIX]{x1D6E5})^{3}\frac{|A|^{5}}{(\unicode[STIX]{x1D6E5}2^{k})^{4}}\ll \frac{|A|^{5}}{\unicode[STIX]{x1D6E5}}. & & \displaystyle \nonumber\end{eqnarray}$$

(Note that our application of Lemma 10 is justified, since $2^{k}\unicode[STIX]{x1D6E5}\geqslant 2|A|^{2}/p$ for all $k\geqslant 0$ .)

Taken together, we have

$$\begin{eqnarray}\mathop{\sum }_{\ell }i(\ell )\biggl(i(\ell )-\frac{|A|^{2}}{p}\biggr)^{2}\ll \unicode[STIX]{x1D6E5}p|A|^{2}+\frac{|A|^{5}}{\unicode[STIX]{x1D6E5}}.\end{eqnarray}$$

Choosing $\unicode[STIX]{x1D6E5}=|A|^{3/2}/p^{1/2}$ yields

$$\begin{eqnarray}\mathop{\sum }_{\ell }i(\ell )\biggl(i(\ell )-\frac{|A|^{2}}{p}\biggr)^{2}\ll p^{1/2}|A|^{7/2},\end{eqnarray}$$

assuming that $|A|^{3/2}/p^{1/2}\geqslant 2|A|^{2}/p$ . If not, we have $p<4|A|$ , thus by (24), we have

$$\begin{eqnarray}\mathop{\sum }_{\ell }i(\ell )\biggl(i(\ell )-\frac{|A|^{2}}{p}\biggr)^{2}\ll p|A|^{3}\ll p^{1/2}|A|^{7/2}.\end{eqnarray}$$

Thus in any case, we have (29).

Next we prove (28). Arguing as in the proof of (24), by (16), Lemma 9 and the asymptotic formula for $T(A)$ , we have

$$\begin{eqnarray}\displaystyle Q(A) & = & \displaystyle \mathop{\sum }_{\ell }i(\ell )^{4}+O(|A|^{4})\nonumber\\ \displaystyle & = & \displaystyle \mathop{\sum }_{\ell }i(\ell )^{2}\biggl(i(\ell )-\frac{|A|^{2}}{p}\biggr)^{2}-\frac{2|A|^{2}}{p}T(A)\nonumber\\ \displaystyle & & \displaystyle +\,\frac{|A|^{4}}{p^{2}}(|A|^{4}+p|A|^{2})+O(|A|^{4})\nonumber\\ \displaystyle & = & \displaystyle \frac{|A|^{8}}{p^{2}}+O\biggl(\frac{|A|^{6}}{p}\biggr)+O\biggl(\frac{|A|^{11/2}}{p^{1/2}}\biggr)+\mathop{\sum }_{\ell }i(\ell )^{2}\biggl(i(\ell )-\frac{|A|^{2}}{p}\biggr)^{2}.\nonumber\end{eqnarray}$$

Now we estimate the sum, first by removing the portion where $i(\ell )<2|A|^{2}/p$ ,

$$\begin{eqnarray}\mathop{\sum }_{i(\ell )<2|A|^{2}/p}i(\ell )^{2}\biggl(i(\ell )-\frac{|A|^{2}}{p}\biggr)^{2}\leqslant \frac{4|A|^{4}}{p^{2}}p|A|^{2}=\frac{4|A|^{6}}{p},\end{eqnarray}$$

and then by applying the second part of Lemma 10 to each of the $O(\log |A|)$ remaining dyadic sums of the form $\sum _{2^{i}\leqslant i(\ell )\leqslant 2^{i+1}}i(\ell )^{4}$ with $2|A|^{2}/p\leqslant 2^{i}\leqslant |A|$ ,

$$\begin{eqnarray}\mathop{\sum }_{i(\ell )\geqslant 2|A|^{2}/p}i(\ell )^{2}\biggl(i(\ell )-\frac{|A|^{2}}{p}\biggr)^{2}\ll \mathop{\sum }_{i(\ell )\geqslant 2|A|^{2}/p}i(\ell )^{4}\ll |A|^{5}\log |A|.\end{eqnarray}$$

Since the terms $|A|^{6}/p$ and $|A|^{11/2}/p^{1/2}$ are both smaller than $|A|^{5}$ , we have

$$\begin{eqnarray}Q(A)=\frac{|A|^{8}}{p^{2}}+O(|A|^{5}\log |A|).\Box\end{eqnarray}$$

Proof of part (2) of Theorem 6.

$R[A]$ is the support of the function $t(x)$ defined in (17). Note also that

$$\begin{eqnarray}\mathop{\sum }_{x\in \mathbb{F}_{p}}t(x)=|A|^{2}(|A|-1)\end{eqnarray}$$

because each ordered triplet $(a,b,c)\in A\times A\times A$ with $c\neq a$ contributes precisely 1 to the sum.

By the Cauchy–Schwarz inequality we get

$$\begin{eqnarray}|R[A]|=|\text{supp}(t)|\geqslant \frac{(\mathop{\sum }_{x}t(x))^{2}}{\mathop{\sum }_{x}t(x)^{2}}\gg \frac{|A|^{6}}{\mathop{\sum }_{x}t(x)^{2}}.\end{eqnarray}$$

By identity (19) and part (1) of Theorem 11 we have

$$\begin{eqnarray}\mathop{\sum }_{x\in \mathbb{F}_{p}}t(x)^{2}\leqslant T(A)\ll \frac{|A|^{6}}{p}+p^{1/2}|A|^{7/2}.\end{eqnarray}$$

Substituting above yields

(30) $$\begin{eqnarray}|R[A]|\gg \min \bigg\{p,\frac{|A|^{5/2}}{p^{1/2}}\bigg\}.\end{eqnarray}$$

When $|A|\geqslant p^{3/5}$ , $p\leqslant |A|^{5/2}/p^{1/2}$ and the claim follows.◻

Proof of parts (1) and (3) of Theorem 6.

We begin with some preliminaries. We denote $R[A]$ by $R$ . Recall also the functions $t$ and $q$ defined in (17) and (18), respectively:

(31) $$\begin{eqnarray}\displaystyle & \displaystyle t(x)=\biggl|\bigg\{(a,b,c)\in A^{3}:\frac{b-a}{c-a}=x\bigg\}\biggr|, & \displaystyle \nonumber\\ \displaystyle & \displaystyle q(x,y)=\biggl|\bigg\{(a,b,c,d)\in A^{4}:\frac{b-a}{c-a}=x,\frac{d-a}{c-a}=y\bigg\}\biggr|. & \displaystyle\end{eqnarray}$$

It follows from the definitions that

(32) $$\begin{eqnarray}\mathop{\sum }_{x\in R}\mathop{\sum }_{y\in R}q(x,y)=|A|\mathop{\sum }_{x\in R}t(x)=|A|^{3}(|A|-1)\leqslant |A|^{4}.\end{eqnarray}$$

We will exploit a symmetry of the set $R$ , namely that $R=1-R$ . To prove this, simply note that

(33) $$\begin{eqnarray}1-\frac{b-a}{c-a}=\frac{c-b}{c-a}=\frac{b-c}{a-c}.\end{eqnarray}$$

This shows that $t(x)=t(1-x)$ and $q(x,y)=q(1-x,1-y)$ .

Let $G\subset R\times R$ denote the set of pairs $(x,y)$ such that $q(x,y)\not =0$ . Applying Cauchy–Schwarz to the inequality in (32) yields

(34) $$\begin{eqnarray}|A|^{8}\ll \biggl(\mathop{\sum }_{(x,y)\in G}q(x,y)\biggr)^{2}\leqslant |G|\biggl(\mathop{\sum }_{x,y}q(x,y)^{2}\biggr).\end{eqnarray}$$

To bound $|G|$ , we use the elementary observation that if $(x,y)\in G$ , then there exist $a,b,c,d\in A$ such that $x=(b-a)/(c-a)$ and $y=(d-a)/(c-a)$ . It follows that $x/y=(b-a)/(d-a)\in R$ . Since $q(x,y)=q(1-x,1-y)$ , we also have $(1-x,1-y)\in G$ , hence $(1-x)/(1-y)\in R$ .

Setting $G_{x}=\{y:(x,y)\in G\}$ , we have

$$\begin{eqnarray}\displaystyle |G| & = & \displaystyle \mathop{\sum }_{x\in R}|G_{x}|\leqslant |R|^{1/2}\biggl(\mathop{\sum }_{x\in R}|G_{x}|^{2}\biggr)^{1/2}\nonumber\\ \displaystyle & = & \displaystyle |R|^{1/2}|\{(x,y,y^{\prime })\in R^{3}:(x,y),(x,y^{\prime })\in G\}|^{1/2}.\nonumber\end{eqnarray}$$

If $(x,y),(x,y^{\prime })\in G$ , then there exist elements $r,r^{\prime }\in R$ such that $x=ry$ and $1-x=r^{\prime }(1-y^{\prime })$ . Thus

$$\begin{eqnarray}\displaystyle & & \displaystyle |\{(x,y,y^{\prime })\in R^{3}:(x,y),(x,y^{\prime })\in G\}|\nonumber\\ \displaystyle & & \displaystyle \quad \leqslant \,|\{(r,r^{\prime },y,y^{\prime })\in R:1-ry=r^{\prime }(1-y^{\prime })\}|.\nonumber\end{eqnarray}$$

The quantity on the right-hand side can be interpreted as the number of incidences between the pairs $(y,y^{\prime })$ in $R\times R$ and the set of $|R|^{2}$ lines of the form

$$\begin{eqnarray}1-r^{\prime }=ry-r^{\prime }y^{\prime }.\end{eqnarray}$$

Applying Theorem 8 yields

$$\begin{eqnarray}|\{(r,r^{\prime },y,y^{\prime })\in R:1-ry=r^{\prime }(1-y^{\prime })\}|\ll \frac{|R|^{7/2}}{p^{1/2}}+|R|^{11/4}.\end{eqnarray}$$

Thus

(35) $$\begin{eqnarray}|G|\ll |R|^{15/8}+\frac{|R|^{9/4}}{p^{1/4}}.\end{eqnarray}$$

The term in the second parentheses of (34) is, by (18), bounded above by the number of collinear quadruples in $A\times A$ . By part (2) of Theorem 11 we have

(36) $$\begin{eqnarray}\mathop{\sum }_{x,y}q(x,y)^{2}\ll \frac{|A|^{8}}{p^{2}}+|A|^{5}\log |A|.\end{eqnarray}$$

The first term dominates only when $|A|^{3}\gg p^{2}\log |A|$ . This range is outside the scope of part (1). As for part (3), part (2) (proved above) guarantees that $|R|\gg p$ and the conclusion follows. So we may assume that the second term dominates. Therefore,

(37) $$\begin{eqnarray}\mathop{\sum }_{x,y}q(x,y)^{2}\ll |A|^{5}\log |A|.\end{eqnarray}$$

Thus by (34), (35) and (37) we have

$$\begin{eqnarray}\frac{|A|^{3}}{\log |A|}\ll \frac{|R|^{9/4}}{p^{1/4}}+|R|^{15/8}.\end{eqnarray}$$

This implies that

• ∙ if $|R|\leqslant p^{2/3}$ , then the second term on the right dominates and

  (38) $$\begin{eqnarray}|R|\gg \frac{|A|^{8/5}}{\log ^{8/15}|A|};\end{eqnarray}$$

• ∙ if $|R|\geqslant p^{2/3}$ , then the first term on the right dominates and

  (39) $$\begin{eqnarray}|R|\gg \frac{p^{1/9}|A|^{4/3}}{\log ^{4/9}|A|}.\end{eqnarray}$$

To conclude the proof of part (1), by (38), either $|R|\gg |A|^{8/5}\log ^{-8/15}|A|$ or $|R|\geqslant p^{2/3}$ . The hypothesis $|A|\leqslant p^{5/12}$ implies that $p^{2/3}\geqslant |A|^{8/5}$ .

To conclude the proof of part (3), we combine the lower bounds in (30) and (39). They are roughly equal when $|A|=p^{11/21}$ . Using (30) for $|A|\geqslant p^{11/21}$ and (39) for $|A|\leqslant p^{11/21}$ gives $|R|\gg |A|^{17/11}\log ^{-4/9}|A|,$ as claimed.◻

Proof. Without loss of generality $0\not \in C$ . Let $l_{b,c}$ be the line with equation $y=b(c^{-1}x-1)$ and let $L=\{l_{b,c}:b\in B,c\in C\}$ , so that $|L|=|B||C|$ . Let $P=AC\times (A-1)B$ . Each line of $L$ is incident to at least $|A|$ points of $P$ .

We will apply Theorem 8 with these sets of points and lines. By assumption we have $|AC||B||C|\leqslant p^{2}$ , and so (13) is satisfied. Thus

$$\begin{eqnarray}\displaystyle |A||B||C| & {\leqslant} & \displaystyle {\mathcal{I}}(P,L)\nonumber\\ \displaystyle & \ll & \displaystyle |AC|^{3/4}|(A-1)B|^{1/2}|B|^{3/4}|C|^{3/4}+|AC||(A-1)B|+|B||C|.\nonumber\end{eqnarray}$$

If the second term dominates, then the desired lower bound follows from

$$\begin{eqnarray}|AC|,|(A-1)B|\gg |A|.\end{eqnarray}$$

Thus we may assume that

$$\begin{eqnarray}|A||B||C|\leqslant {\mathcal{I}}(P,L)\ll |AC|^{3/4}|(A-1)B|^{1/2}|B|^{3/4}|C|^{3/4}+|B||C|.\end{eqnarray}$$

Rearranging this inequality completes the proof. ◻

Proof of Theorem 1.

Recall that Theorem 1 states that for $A\subset \mathbb{F}$ with $|A|^{16}|A-A|^{30}\leqslant p^{25}$ and $|A|<p^{5/12}$ , we have

$$\begin{eqnarray}|(A-A)(A-A)|\gtrsim |A-A|^{4/5}|A|^{32/75}\geqslant |A-A|^{1+1/75}.\end{eqnarray}$$

The second inequality follows from the trivial inequality $|A|\geqslant |A-A|^{1/2}$ , and so it suffices to prove the first bound only.

We use the shorthand $R=R[A]$ and $D=A-A$ . The proof is split into two cases.

Case 1. Suppose that

(40) $$\begin{eqnarray}|RD||D|^{2}\leqslant p^{2}.\end{eqnarray}$$

Apply Lemma 12 with $A=R$ and $B=C=D$ . Note that, since it is assumed that $|RD||D|^{2}\leqslant p^{2}$ , the conditions of Lemma 12 in the positive characteristic case are satisfied. Then, recalling the property that $R-1=-R$ , we have

$$\begin{eqnarray}|RD|^{5}\gg |R|^{4}|D|^{2}.\end{eqnarray}$$

We can apply the bound $|R|\gtrsim |A|^{8/5}$ from Theorem 6. This is valid by our assumption that $|A|<p^{5/12}$ . It follows that

$$\begin{eqnarray}|RD|^{5}\gtrsim |A|^{32/5}|D|^{2}.\end{eqnarray}$$

On the other hand, we can copy arguments from [Reference Shkredov44] to bound the left-hand side of this inequality from above. Indeed, applying Lemma 13 in the multiplicative setting, we have

$$\begin{eqnarray}|RD|\leqslant |DD/D|\leqslant \frac{|DD|^{3}}{|D|^{2}}.\end{eqnarray}$$

Therefore,

$$\begin{eqnarray}|DD|^{15}\geqslant |RD|^{5}|D|^{10}\gtrsim |A|^{32/5}|D|^{12},\end{eqnarray}$$

as required.

Case 2. Suppose that

(41) $$\begin{eqnarray}|RD||D|^{2}\geqslant p^{2}.\end{eqnarray}$$

Similarly to the previous case, it follows from Lemma 13 that

$$\begin{eqnarray}p^{2}\leqslant |RD||D|^{2}\leqslant |DD/D||D|^{2}\leqslant |DD|^{3}\end{eqnarray}$$

and so

$$\begin{eqnarray}|DD|^{75}\geqslant p^{50}\geqslant |A|^{32}|D|^{60},\end{eqnarray}$$

where the last inequality follows from the assumption that $|A|^{16}|D|^{30}\leqslant p^{25}$ .

Proof of Theorem 2.

Recall that Theorem 2 states that, for any $A\subset \mathbb{F}_{p}$ with $|A|\leqslant p^{25/77}$ ,

$$\begin{eqnarray}|DD/D|\gtrsim |A|^{2+1/25},\end{eqnarray}$$

where again $D=A-A$ . The proof is similar to the proof of Theorem 1. Write $Q=(A-A)/(A-A)$ and $R=R[A]$ . At the outset, apply Lemma 14 to obtain a subset $Q^{\prime }\subset Q$ with

(42) $$\begin{eqnarray}|Q^{\prime }|\gg |Q|\gg |A|^{2}\end{eqnarray}$$

and

(43) $$\begin{eqnarray}\biggl|\frac{Q^{\prime }D}{D}\biggr|\ll \frac{|QD||Q/D|}{|Q|}=\frac{|DD/D|^{2}}{|Q|}.\end{eqnarray}$$

The bound $|Q|\gg |A|^{2}$ is a consequence of (11), a result which is essential for this proof to work. This is valid here because of our assumption that $|A|\leqslant p^{25/77}<p^{1/2}$ .

We will prove, under our assumption on the size of $A$ , that

(44) $$\begin{eqnarray}|RQ^{\prime }|\gtrsim |A|^{2+2/25}.\end{eqnarray}$$

Suppose that (44) holds. Then, in particular, by (43), we have

$$\begin{eqnarray}|A|^{2+2/25}\lesssim |RQ^{\prime }|\leqslant \biggl|\frac{Q^{\prime }D}{D}\biggr|\ll \frac{|DD/D|^{2}}{|Q|}.\end{eqnarray}$$

Rearranging this inequality and applying the bound $|Q|\gg |A|^{2}$ , we obtain the desired result that

$$\begin{eqnarray}|DD/D|\gtrsim |A|^{2+1/25}.\end{eqnarray}$$

It remains to prove (44). We may assume that $|RQ^{\prime }|\leqslant |A|^{2+2/25}$ , as otherwise we are done. Also, we may assume that $|Q|\leqslant |A|^{2+1/25}$ , as otherwise there is nothing to prove. With these two assumptions and the assumption that $|A|\leqslant p^{25/77}$ , we have

$$\begin{eqnarray}|RQ^{\prime }||Q^{\prime }|^{2}\leqslant |A|^{6+4/25}\leqslant p^{2},\end{eqnarray}$$

and so Lemma 12 can be applied to obtain the bound

$$\begin{eqnarray}|RQ^{\prime }|^{5}=|RQ^{\prime }|^{3}|(R-1)Q^{\prime }|^{2}\gg |R|^{4}|Q^{\prime }|^{2}.\end{eqnarray}$$

We can apply the bound $|R|\gtrsim |A|^{8/5}$ from Theorem 6. This is valid by our assumption that $|A|<p^{5/12}$ . In addition, we can apply the bound on $|Q^{\prime }|$ from (42). Putting everything together, it follows that

$$\begin{eqnarray}|RQ^{\prime }|^{5}\gtrsim |A|^{52/5}.\end{eqnarray}$$

This proves (44), and thus completes the proof of Theorem 2. ◻

Proof. We can assume that $|A|>1$ . Put $\unicode[STIX]{x1D6E4}_{\ast }=\unicode[STIX]{x1D6E4}\sqcup \{0\}$ and $R=R[A]$ . Then in light of our condition (47), we have

(48) $$\begin{eqnarray}R\subset (\unicode[STIX]{x1D709}\unicode[STIX]{x1D6E4}\sqcup \{0\})/\unicode[STIX]{x1D709}\unicode[STIX]{x1D6E4}=\unicode[STIX]{x1D6E4}_{\ast }.\end{eqnarray}$$

Suppose that $|\unicode[STIX]{x1D6E4}|\leqslant p^{3/4}$ . Applying (33) and formula (45) with $k=1$ , we obtain

$$\begin{eqnarray}|R|=|R\cap (1-R)|\leqslant |\unicode[STIX]{x1D6E4}_{\ast }\cap (1-\unicode[STIX]{x1D6E4}_{\ast })|\leqslant |\unicode[STIX]{x1D6E4}\cap (1-\unicode[STIX]{x1D6E4})|+2\ll |\unicode[STIX]{x1D6E4}|^{2/3}\leqslant p^{1/2}.\end{eqnarray}$$

Since $|A|\leqslant |R|<p^{1/2}<p^{5/12}$ , we may apply formula (38) of Theorem 6 to get $|R|\gtrsim |A|^{8/5}$ . Hence $|A|^{8/5}\lesssim |\unicode[STIX]{x1D6E4}|^{2/3}$ , and we obtain the required result.

Now let us suppose that $|\unicode[STIX]{x1D6E4}|\geqslant p^{3/4}$ but $|\unicode[STIX]{x1D6E4}|\leqslant p^{5/6}$ . In this case, by formula (46) and the previous calculations, we get

$$\begin{eqnarray}|R|\ll |\unicode[STIX]{x1D6E4}|^{2}/p\leqslant p^{2/3}.\end{eqnarray}$$

Applying Theorem 6 (namely, formula (38)) once more, we obtain $|A|^{8/5}\lesssim |\unicode[STIX]{x1D6E4}|^{2}/p$ , which gives the required bound for the size of $A$ , namely $|A|\lesssim p^{-5/8}|\unicode[STIX]{x1D6E4}|^{5/4}$ .

Now suppose that $|\unicode[STIX]{x1D6E4}|\geqslant p^{5/6}$ . We use the second formula from (31) and note that if $q(x,y)>0$ , then $x-y\in \unicode[STIX]{x1D6E4}_{\ast }$ . By (34)–(36) and inclusion (48), we obtain

$$\begin{eqnarray}\displaystyle & & \displaystyle |A|^{8}\ll \biggl(\frac{|A|^{8}}{p^{2}}+|A|^{5}\log |A|\biggr)\cdot \biggl(\mathop{\sum }_{x,y}R(x)R(y)\unicode[STIX]{x1D6E4}_{\ast }(x-y)\biggr)\nonumber\\ \displaystyle & & \displaystyle \quad \lesssim \,\biggl(\frac{|A|^{8}}{p^{2}}+|A|^{5}\biggr)\cdot \biggl(\mathop{\sum }_{x}\unicode[STIX]{x1D6E4}_{\ast }(x)\unicode[STIX]{x1D6E4}_{\ast }(1-x)\mathop{\sum }_{y}\unicode[STIX]{x1D6E4}_{\ast }(y)\unicode[STIX]{x1D6E4}_{\ast }(1-y)\unicode[STIX]{x1D6E4}_{\ast }(x-y)\biggr).\nonumber\end{eqnarray}$$

It is easy to see that the summand with $x=1$ is negligible in the last inequality. Using formula (46) with $k=2$ and the condition $|\unicode[STIX]{x1D6E4}|\geqslant p^{5/6}$ , we get

$$\begin{eqnarray}|A|^{8}\lesssim \biggl(\frac{|A|^{8}}{p^{2}}+|A|^{5}\biggr)\cdot \frac{|\unicode[STIX]{x1D6E4}|^{5}}{p^{3}}\ll |A|^{5}\cdot \frac{|\unicode[STIX]{x1D6E4}|^{5}}{p^{3}}.\end{eqnarray}$$

It follows that $|A|\lesssim p^{-1}|\unicode[STIX]{x1D6E4}|^{5/3}$ , as claimed.

Finally, it is known that if $A+B=\unicode[STIX]{x1D709}\unicode[STIX]{x1D6E4}$ or $A+B=\unicode[STIX]{x1D709}\unicode[STIX]{x1D6E4}\sqcup \{0\}$ , then $|A|\sim |B|\sim |\unicode[STIX]{x1D6E4}|^{1/2+\unicode[STIX]{x1D700}}$ for all $\unicode[STIX]{x1D700}>0$ ; see [Reference Shkredov and Vyugin47, Reference Shparlinski49]. Thus, if $p^{5/6}\leqslant |\unicode[STIX]{x1D6E4}|\leqslant p^{6/7-\unicode[STIX]{x1D700}}$ , we have $|A|\leqslant |\unicode[STIX]{x1D6E4}|^{1/2-\unicode[STIX]{x1D700}/2}$ , which is a contradiction. Another two bounds which were obtained before give us the same for smaller $\unicode[STIX]{x1D6E4}$ . Thus, we have $\unicode[STIX]{x1D709}\unicode[STIX]{x1D6E4}\sqcup \{0\}\neq A-A$ for sufficiently large $\unicode[STIX]{x1D6E4}$ .◻

Proof of (49).

Without loss of generality, we may assume that, say, at most $|P|^{3/4}$ points of $P\subset \mathbb{F}^{2}\setminus \{0\}$ lie on a line, since otherwise a fixed point $u\in P$ outside this line determines at least $|P|^{3/4}$ distinct values of $\unicode[STIX]{x1D714}(u,u^{\prime })$ with $u^{\prime }$ on the line, as $\unicode[STIX]{x1D714}$ is non-degenerate. This is needed just to ensure in the sequel that $P$ or its large subset determines at least four distinct directions.◻

Proof of Lemma 20.

Note that

(53) $$\begin{eqnarray}(a,b,c,d)\sim (a^{\prime },b^{\prime },c^{\prime },d^{\prime })\quad \text{only if }\frac{t_{ab}t_{cd}}{t_{ac}t_{bd}}=\frac{t_{a^{\prime }b^{\prime }}t_{c^{\prime }d^{\prime }}}{t_{a^{\prime }c^{\prime }}t_{b^{\prime }d^{\prime }}}.\end{eqnarray}$$

The quantities on the right-hand side are cross-ratios. To see why, we express points in coordinates like $a=(a_{1},a_{2})$ and by choosing coordinate axes assume that $a_{1},b_{1},c_{1}$ and $d_{1}$ are all non-zero. We then have

$$\begin{eqnarray}\frac{t_{ab}t_{cd}}{t_{ac}t_{bd}}=\frac{(a_{1}b_{2}-a_{2}b_{1})(c_{1}d_{2}-c_{2}d_{1})}{(a_{1}c_{2}-a_{2}c_{1})(b_{1}d_{2}-b_{2}d_{1})}.\end{eqnarray}$$

Now denote by, say, $\unicode[STIX]{x1D6FF}_{a}=a_{2}/a_{1}$ the direction of the line through the origin that passes through $a$ . Dividing both numerator and denominator by $a_{1}b_{1}c_{1}d_{1}$ gives

$$\begin{eqnarray}\frac{t_{ab}t_{cd}}{t_{ac}t_{bd}}=\frac{(a_{1}b_{2}-a_{2}b_{1})(c_{1}d_{2}-c_{2}d_{1})}{(a_{1}c_{2}-a_{2}c_{1})(b_{1}d_{2}-b_{2}d_{1})}=\frac{(\unicode[STIX]{x1D6FF}_{b}-\unicode[STIX]{x1D6FF}_{a})(\unicode[STIX]{x1D6FF}_{d}-\unicode[STIX]{x1D6FF}_{c})}{(\unicode[STIX]{x1D6FF}_{c}-\unicode[STIX]{x1D6FF}_{a})(\unicode[STIX]{x1D6FF}_{d}-\unicode[STIX]{x1D6FF}_{b})}=[\unicode[STIX]{x1D6FF}_{a},\unicode[STIX]{x1D6FF}_{b},\unicode[STIX]{x1D6FF}_{c},\unicode[STIX]{x1D6FF}_{d}].\end{eqnarray}$$

That is, if $\unicode[STIX]{x1D6FF}_{a},\ldots ,\unicode[STIX]{x1D6FF}_{d^{\prime }}\in \mathbb{F}\mathbb{P}^{1}$ are the directions from the origin, corresponding to the points $a,\ldots ,d^{\prime }$ , respectively, then $(a,b,c,d)\sim (a^{\prime },b^{\prime },c^{\prime },d^{\prime })$ only if $[\unicode[STIX]{x1D6FF}_{a},\unicode[STIX]{x1D6FF}_{b},\unicode[STIX]{x1D6FF}_{c},\unicode[STIX]{x1D6FF}_{d}]=[\unicode[STIX]{x1D6FF}_{a^{\prime }},\unicode[STIX]{x1D6FF}_{b^{\prime }},\unicode[STIX]{x1D6FF}_{c^{\prime }},\unicode[STIX]{x1D6FF}_{d^{\prime }}],$ where $[\,]$ stands for the cross-ratioFootnote ¹ . ◻

Proof of Lemma 21.

The equation (55) is the identity

$$\begin{eqnarray}t_{ad}t_{bc}=t_{ac}t_{bd}-t_{ab}t_{cd}.\end{eqnarray}$$

There are many ways to verify this identity: one can do it, for example, by direct calculation using coordinates (both sides equal $a_{1}b_{2}c_{1}d_{2}+a_{2}b_{1}c_{2}d_{1}-a_{1}b_{1}c_{2}d_{2}-a_{2}b_{2}c_{1}d_{1}$ ), trigonometry, or using the cross and scalar product notation, using the fact that

$$\begin{eqnarray}(a\times d)\cdot (b\times c)=-c\cdot (b\times (a\times d))=(a\cdot b)(c\cdot d)-(a\cdot c)(b\cdot d).\end{eqnarray}$$

Now replace $b=(b_{1},b_{2})$ by $b^{\bot }=(-b_{2},b_{1})$ , and $c$ by $c^{\bot }$ .

Now we will show that $\unicode[STIX]{x1D6F7}$ is two-to-one injective on equivalent quadruples with the same direction quadruple $(\unicode[STIX]{x1D6FF}_{a},\unicode[STIX]{x1D6FF}_{b},\unicode[STIX]{x1D6FF}_{c},\unicode[STIX]{x1D6FF}_{d})$ . Suppose that $(a,b,c,d)$ and $(a^{\prime },b^{\prime },c^{\prime },d^{\prime })$ are two such quadruples. Set $a^{\prime }=ka,$ $b^{\prime }=lb$ , $c^{\prime }=mc,$ $d^{\prime }=nd$ with some $k,l,m,n\neq 0$ . Then $t_{ab}=t_{a^{\prime }b^{\prime }}$ means $kl=1$ and $t_{cd}=t_{c^{\prime }d^{\prime }}$ means $mn=1$ . In addition, $t_{ac}=t_{a^{\prime }c^{\prime }}$ means $km=1$ and $t_{bd}=t_{b^{\prime }d^{\prime }}$ means $ln=1$ . So $k,n=x$ and $l,m=x^{-1}$ , for some $x$ .

Informally, given the direction quadruple $(\unicode[STIX]{x1D6FF}_{a},\unicode[STIX]{x1D6FF}_{b},\unicode[STIX]{x1D6FF}_{c},\unicode[STIX]{x1D6FF}_{d})$ , an equivalence class of the quadruple of points $(a,b,c,d)$ supported in the corresponding directions arises by scaling $a,d$ by some factor $x$ and $b,c$ simultaneously by the factor $x^{-1}$ .

So we have $(a^{\prime },d^{\prime })=x(a,d)$ and $(b^{\prime },c^{\prime })=x^{-1}(b,c)$ for some $x\not =0$ . Thus

$$\begin{eqnarray}t_{a^{\prime }d^{\prime }}=x^{2}t_{ad}\quad \text{and}\quad t_{b^{\prime }c^{\prime }}=x^{-2}t_{bc},\end{eqnarray}$$

so if $t_{a^{\prime }d^{\prime }}=t_{ad}$ , then $x=\pm 1$ and $(a,b,c,d)=(a^{\prime },b^{\prime },c^{\prime },d^{\prime })$ .◻

Proof. For the proof of the first two relations see Lemma 4.3 and the formula in the proof of Corollary 4.4 in [Reference Konyagin and Rudnev23]. Inequality (68) follows by applying the Hölder and then the Cauchy–Schwarz inequality. We set $D^{\prime }=A-A$ and $\mathsf{E}_{3/2}(A)=\sum _{d\in A-A}|A_{d}|^{3/2}.$ Then

$$\begin{eqnarray}\mathsf{E}(A)=\mathop{\sum }_{d\in A-A}r_{A-A}(d)r_{A-A}(d)\leqslant \mathsf{E}_{3}^{1/3}\mathsf{E}_{3/2}^{2/3}(A).\end{eqnarray}$$

Thus $E_{3/2}(A)^{2}\geqslant \mathsf{E}^{3}(A)/\mathsf{E}_{3}(A)$ . Substituting this into (66) and eliminating $\mathsf{E}(A)$ via

$$\begin{eqnarray}\mathsf{E}(A)\geqslant \frac{|A|^{4}}{|A+A|}\end{eqnarray}$$

yields (68). ◻

Proof of Theorem 3.

The first task is prove inequality (3). Apply respectively (67) in the multiplicative setting, Lemmas 23 and 25 as follows:

(69) $$\begin{eqnarray}\displaystyle |A|^{8} & {\leqslant} & \displaystyle \mathsf{E}_{3}^{\times }(A)|A/A|\mathsf{E}^{\times }(A,A/A)\nonumber\\ \displaystyle & \ll & \displaystyle \frac{|A+\unicode[STIX]{x1D706}A|^{15/4}|A/A|\mathsf{E}^{\times }(A,A/A)}{|A|^{3/4}}\log |A|\nonumber\\ \displaystyle & {\lesssim} & \displaystyle \frac{|A+\unicode[STIX]{x1D706}A|^{21/4}|A/A|^{5/2}}{|A|^{5/4}}.\end{eqnarray}$$

When applying Lemmas 23 and 25, we must ensure that the conditions of these lemmas are satisfied in our situation. For Lemma 23, this is immediate, since $|A|^{11}|A+\unicode[STIX]{x1D706}A|\leqslant |A|^{13}\leqslant (p^{5/9})^{13}<p^{8}.$

We now verify the conditions of Lemma 25. We may assume that $|A+\unicode[STIX]{x1D706}A|^{21}|A/A|^{10}\leqslant |A|^{37}$ , or else there is nothing to prove. Similarly, we may assume that $|A/A|\leqslant |A|^{8/5}$ . Therefore, $|A||A+\unicode[STIX]{x1D706}A||A/A|=|A|(|A+\unicode[STIX]{x1D706}A|^{21}|A/A|^{10})^{1/21}|A/A|^{11/21}\leqslant |A|^{378/105}\leqslant p^{2}$ , where the last inequality comes from the assumption that $|A|\leqslant p^{5/9}$ . The other condition in Lemma 25 is trivial in this case.

Rearranging inequality (69), it follows that

$$\begin{eqnarray}|A+\unicode[STIX]{x1D706}A|^{21}|A/A|^{10}\gtrsim |A|^{37}.\end{eqnarray}$$

For the proof of (4), we apply respectively (68) in the multiplicative setting, Lemmas 23 and 25 to obtain

$$\begin{eqnarray}\displaystyle |A|^{14} & {\leqslant} & \displaystyle (\mathsf{E}_{3}^{\times }(A))^{2}|AA|^{3}\mathsf{E}^{\times }(A,AA)\nonumber\\ \displaystyle & {\lesssim} & \displaystyle \frac{|A+\unicode[STIX]{x1D706}A|^{15/2}|AA|^{3}\mathsf{E}^{\times }(A,AA)}{|A|^{3/2}}\nonumber\\ \displaystyle & {\leqslant} & \displaystyle \frac{|A+\unicode[STIX]{x1D706}A|^{9}|AA|^{9/2}}{|A|^{2}}.\nonumber\end{eqnarray}$$

Again, it is necessary to check the conditions of Lemmas 23 and 25. For Lemma 23, this is immediate, since $|A|^{11}|A+\unicode[STIX]{x1D706}A|\leqslant |A|^{13}\leqslant (p^{9/16})^{13}<p^{8}.$ For Lemma 25, we may assume that $|A+\unicode[STIX]{x1D706}A|^{18}|AA|^{9}\leqslant |A|^{32}$ and $|AA|\leqslant |A|^{14/9}$ , or else there is nothing to prove. Therefore $|A||A\,+\,\unicode[STIX]{x1D706}A||AA|=|A|(|A+\unicode[STIX]{x1D706}A|^{18}|AA|^{9})^{1/18}|AA|^{1/2}\leqslant |A|^{64/18}\leqslant p^{2}$ , as required.

It follows that

$$\begin{eqnarray}|A+\unicode[STIX]{x1D706}A|^{18}|AA|^{9}\gtrsim |A|^{32}.\Box\end{eqnarray}$$

Note 26. Using methods from [Reference Shkredov43], one can show that $|A+\unicode[STIX]{x1D706}A|\leqslant M|A|$ implies that $|AA|\gtrsim M^{-C}|A|^{58/37}$ , where $C>0$ is an absolute constant. Indeed, by the previous arguments the inequality $|A+\unicode[STIX]{x1D706}A|\leqslant M|A|$ gives us the required bound for the third energy and the common energy (see Lemmas 23 and 25), and precisely statements of this kind are needed in the eigenvalues approach from [Reference Shkredov43]. Nevertheless, the methods in [Reference Shkredov43] are less elementary (although give slightly better bounds), so we prefer to give a more transparent proof in the current paper.

Proof of Theorem 27.

Let $a\in A$ be arbitrary. Suppose that $|A+\unicode[STIX]{x1D706}A|\leqslant |A|^{1+1/45}$ . Then $|(A+\unicode[STIX]{x1D706}a)+\unicode[STIX]{x1D706}(A+\unicode[STIX]{x1D706}a)|\leqslant |A|^{1+1/45}$ . By Theorem 3, which can be applied because of the assumption that $|A|\leqslant p^{9/16}$ , we have

$$\begin{eqnarray}|(A+\unicode[STIX]{x1D706}A)(A+\unicode[STIX]{x1D706}A)|\geqslant |(A+\unicode[STIX]{x1D706}a)(A+\unicode[STIX]{x1D706}a)|\gtrsim |A|^{14/9-2/45}=|A|^{3/2+1/90}.\end{eqnarray}$$

On the other hand, suppose that $|A+\unicode[STIX]{x1D706}A|\geqslant |A|^{1+1/45}$ . Then by Lemma 28,

$$\begin{eqnarray}|(A+\unicode[STIX]{x1D706}A)(A+\unicode[STIX]{x1D706}A)|\gg |A||A+\unicode[STIX]{x1D706}A|^{1/2}.\end{eqnarray}$$

Therefore,

$$\begin{eqnarray}|(A+\unicode[STIX]{x1D706}A)(A+\unicode[STIX]{x1D706}A)|\gg |A|^{3/2+1/90},\end{eqnarray}$$

as required.

We must verify that we have applied Lemma 28 correctly in the case where  $\mathbb{F}$ has positive characteristic. That is, we must verify the condition that $|A+\unicode[STIX]{x1D706}A||A|^{2}\leqslant p$ . Note that we may assume that $|A+\unicode[STIX]{x1D706}A|\leqslant |A|^{3/2+1/90}$ , otherwise $|(A+\unicode[STIX]{x1D706}A)(A+\unicode[STIX]{x1D706}A)|\geqslant |A+\unicode[STIX]{x1D706}A|\geqslant |A|^{3/2+1/90}$ and there is nothing to prove. Therefore,

$$\begin{eqnarray}|A|^{2}|A+\unicode[STIX]{x1D706}A|\leqslant |A|^{158/45}\leqslant (p^{9/16})^{158/45}<p^{2}.\Box\end{eqnarray}$$

Proof. We follow closely an argument of Jones [Reference Jones20]; see also [Reference Roche-Newton, Arge and Pach31]. Combining (21) and Theorem 11 gives

$$\begin{eqnarray}\mathop{\sum }_{a,b\in A}\mathsf{E}^{\times }(A-a,A-b)=\frac{|A|^{6}}{p}+O(p^{1/2}|A|^{7/2}).\end{eqnarray}$$

Therefore, by averaging, there exist $a,b\in A$ such that $\mathsf{E}^{\times }(A-a,A-b)\leqslant |A|^{4}/p+O(p^{1/2}|A|^{3/2}).$ The Cauchy–Schwarz inequality then gives

$$\begin{eqnarray}|(A-a)(A-b)|\geqslant \frac{|A|^{4}}{\mathsf{E}^{\times }(A-a,A-b)}\gg \min \{p,|A|^{5/2}p^{-1/2}\}.\end{eqnarray}$$

For the last part we may assume that $|A|\leqslant p^{3/5}$ . Next note that if $|A|\leqslant p^{9/16}$ , then Theorem 27 with $\unicode[STIX]{x1D706}=-1$ gives $|(A-A)(A-A)|\gg |A|^{3/2+1/90-\unicode[STIX]{x1D700}}$ . When $p^{9/16}\leqslant |A|\leqslant p^{3/5}$ we just proved

$$\begin{eqnarray}|(A-A)(A-A)|\gg \frac{|A|^{5/2}}{p^{1/2}}\geqslant |A|^{3/2+1/9},\end{eqnarray}$$

which is asymptotically bigger. ◻

Proof. Note that $0\in (A-A)(A-A)(A-A)(A-A)$ .

Let $\unicode[STIX]{x1D706}\in \mathbb{F}_{p}^{\ast }$ and let $r(\unicode[STIX]{x1D706}):=r_{(A-A)(A-A)(A-A)(A-A)}(\unicode[STIX]{x1D706})$ be the number of ways $\unicode[STIX]{x1D706}$ can be expressed as a product $\unicode[STIX]{x1D706}=(a_{1}-b_{1})(a_{2}-b_{2})(a_{3}-b_{3})(a_{4}-b_{4})$ with all the $a_{i},b_{i}\in A$ . The orthogonality of multiplicative characters, which we denote by $\unicode[STIX]{x1D712}$ , implies (see [Reference Iwaniec and Kowalski19] for details)

$$\begin{eqnarray}\displaystyle r(\unicode[STIX]{x1D706}) & = & \displaystyle \frac{1}{p-1}\nonumber\\ \displaystyle & & \displaystyle \times \,\mathop{\sum }_{\unicode[STIX]{x1D712}}\mathop{\sum }_{a_{1},b_{2},\ldots ,a_{4},b_{4}\in A:a_{i}\neq b_{i}}\unicode[STIX]{x1D712}\biggl(\frac{(a_{1}-b_{1})(a_{2}-b_{2})(a_{3}-b_{3})(a_{4}-b_{4})}{\unicode[STIX]{x1D706}}\biggr)\nonumber\\ \displaystyle & = & \displaystyle \frac{1}{p-1}\mathop{\sum }_{\unicode[STIX]{x1D712}}\unicode[STIX]{x1D712}(\unicode[STIX]{x1D706}^{-1})\nonumber\\ \displaystyle & & \displaystyle \times \,\mathop{\sum }_{a_{1},b_{2},\ldots ,a_{4},b_{4}\in A:a_{i}\neq b_{i}}\unicode[STIX]{x1D712}((a_{1}-b_{1})(a_{2}-b_{2})(a_{3}-b_{3})(a_{4}-b_{4}))\nonumber\\ \displaystyle & = & \displaystyle \frac{1}{p-1}\mathop{\sum }_{\unicode[STIX]{x1D712}}\unicode[STIX]{x1D712}(\unicode[STIX]{x1D706}^{-1})\biggl(\mathop{\sum }_{a_{1},b_{1}\in A:a_{1}\neq b_{1}}\unicode[STIX]{x1D712}(a_{1}-b_{1})\biggr)^{4}.\nonumber\end{eqnarray}$$

The contribution to the sum coming from the principal character $\unicode[STIX]{x1D712}_{0}$ is $(|A|^{4}(|A|-1)^{4})/(p-1)$ . Using the fact that $|\unicode[STIX]{x1D712}(\unicode[STIX]{x1D706})|=1$ for all $\unicode[STIX]{x1D712}$ and applying the triangle inequality, we get

$$\begin{eqnarray}\displaystyle & & \displaystyle \biggl|r(\unicode[STIX]{x1D706})-\frac{|A|^{4}(|A|-1)^{4}}{p-1}\biggr|\nonumber\\ \displaystyle & & \displaystyle \quad \leqslant \frac{1}{p-1}\mathop{\sum }_{\unicode[STIX]{x1D712}\neq \unicode[STIX]{x1D712}_{0}}\biggl|\mathop{\sum }_{a_{1},b_{1}\in A:a_{1}\neq b_{1}}\unicode[STIX]{x1D712}(a_{1}-b_{1})\biggr|^{4}\nonumber\\ \displaystyle & & \displaystyle \quad =\frac{1}{p-1}\mathop{\sum }_{\unicode[STIX]{x1D712}\neq \unicode[STIX]{x1D712}_{0}}\,\mathop{\sum }_{a_{1},\ldots ,a_{4},b_{1},\ldots ,b_{4}\in A}\unicode[STIX]{x1D712}\biggl(\frac{(a_{1}-b_{1})(a_{2}-b_{2})}{(a_{3}-b_{3})(a_{4}-b_{4})}\biggr).\nonumber\end{eqnarray}$$

The orthogonality of characters implies that the right-hand side equals the number of non-zero solutions to $(a_{1}-b_{1})(a_{2}-b_{2})=(a_{3}-b_{3})(a_{4}-b_{4})\neq 0$ with all the $a_{i},b_{i}\in A$ minus the $(|A|^{4}(|A|-1)^{4})/(p-1)$ contribution that comes from $\unicode[STIX]{x1D712}_{0}$ . A typical application of the Cauchy–Schwarz inequality (see [Reference Petridis and Shparlinski29, the proof of Lemma 2.7], for example) bounds the number of non-zero solutions by $|A|^{2}T(A)$ .

Combining all the above with the first part of Theorem 11 yields

$$\begin{eqnarray}\biggl|r(\unicode[STIX]{x1D706})-\frac{|A|^{4}(|A|-1)^{4}}{p-1}\biggr|\leqslant |A|^{2}\biggl(\frac{|A|^{6}}{p}+O(p^{1/2}|A|^{7/2})\biggr)-\frac{|A|^{4}(|A|-1)^{4}}{p-1}.\end{eqnarray}$$

Note now that $(|A|^{4}(|A|-1)^{4})/(p-1)=|A|^{8}/p+O(|A|^{7}/p)$ and so the above becomes

$$\begin{eqnarray}\biggl|r(\unicode[STIX]{x1D706})-\frac{|A|^{4}(|A|-1)^{4}}{p-1}\biggr|=O\biggl(p^{1/2}|A|^{11/2}+\frac{|A|^{7}}{p}\biggr)=O(|A|^{11/2}p^{1/2}).\end{eqnarray}$$

One more application of $(|A|^{4}(|A|-1)^{4})/(p-1)=|A|^{8}/p+O(|A|^{7}/p)$ gives

$$\begin{eqnarray}r(\unicode[STIX]{x1D706})=\frac{|A|^{8}}{p}+O\biggl(|A|^{11/2}p^{1/2}+\frac{|A|^{7}}{p}\biggr)=\frac{|A|^{8}}{p}+O(|A|^{11/2}p^{1/2}).\end{eqnarray}$$

The first term dominates when $|A|\gg p^{3/5}$ , in which range $r(\unicode[STIX]{x1D706})>0$ and so $\unicode[STIX]{x1D706}\in (A-A)(A-A)(A-A)(A-A)$ .◻

Proof. Let $P=(AA\,\cup \,-\unicode[STIX]{x1D6FC}A)\times (AA\,\cup \,-\unicode[STIX]{x1D6FC}A)$ . The idea is to bound the number of collinear quadruples from above and below, much in the same way as Elekes and Ruzsa [Reference Elekes and Ruzsa9] did for collinear triples.

Let $d,d^{\prime }\in A$ be arbitrary and suppose that $(a+\unicode[STIX]{x1D6FC})/(a^{\prime }+\unicode[STIX]{x1D6FC})=(b+\unicode[STIX]{x1D6FC})/(b^{\prime }+\unicode[STIX]{x1D6FC})=(c+\unicode[STIX]{x1D6FC})/(c^{\prime }+\unicode[STIX]{x1D6FC})$ for some $a,\ldots ,c^{\prime }\in A$ . Then observe that the four points $(-\unicode[STIX]{x1D6FC}d,-\unicode[STIX]{x1D6FC}d^{\prime }),(da,d^{\prime }a^{\prime }),(db,d^{\prime }b^{\prime }),(dc,d^{\prime }c^{\prime })$ are collinear. This implies that the number of collinear quadruples in $P$ is at least

$$\begin{eqnarray}|A|^{2}\mathsf{E}_{3}^{\times }(A+\unicode[STIX]{x1D6FC}).\end{eqnarray}$$

Applying Theorem 11 and our assumption that $|AA|\leqslant p^{2/3}$ , we have

$$\begin{eqnarray}|A|^{2}\mathsf{E}_{3}^{\times }(A+\unicode[STIX]{x1D6FC})\leqslant Q(AA\cup -\unicode[STIX]{x1D6FC}A)\ll |AA|^{5}\log |A|.\end{eqnarray}$$

A rearrangement of this inequality completes the proof. ◻

Proof of Theorem 31.

The remainder of the proof of Theorem 31 is very similar to that of Theorem 3, and so we present only the full proof of inequality (70).

Apply respectively (67) in the multiplicative setting, Lemmas 32 and 33 as follows:

(72) $$\begin{eqnarray}\displaystyle |A|^{8} & {\leqslant} & \displaystyle \mathsf{E}_{3}^{\times }(A+\unicode[STIX]{x1D6FC})\biggl|\frac{A+\unicode[STIX]{x1D6FC}}{A+\unicode[STIX]{x1D6FC}}\biggr|\mathsf{E}^{\times }\biggl(A+\unicode[STIX]{x1D6FC},\frac{A+\unicode[STIX]{x1D6FC}}{A+\unicode[STIX]{x1D6FC}}\biggr)\nonumber\\ \displaystyle & \ll & \displaystyle \frac{|AA|^{5}|(A+\unicode[STIX]{x1D6FC})/(A+\unicode[STIX]{x1D6FC})|\mathsf{E}^{\times }(A+\unicode[STIX]{x1D6FC},(A+\unicode[STIX]{x1D6FC})/(A+\unicode[STIX]{x1D6FC}))}{|A|^{2}}\log |A|\nonumber\\ \displaystyle & \ll & \displaystyle \frac{|AA|^{13/2}|(A+\unicode[STIX]{x1D6FC})/(A+\unicode[STIX]{x1D6FC})|^{5/2}}{|A|^{5/2}}\log |A|.\end{eqnarray}$$

Rearranging this inequality gives inequality (70). However, when applying Lemmas 32 and 33 in the finite characteristic setting, we must ensure that the relevant conditions are met.

We may assume that $|AA|^{13}|(A+\unicode[STIX]{x1D6FC})/(A+\unicode[STIX]{x1D6FC})|^{5}\leqslant |A|^{21}$ , or else there is nothing to prove. Therefore, we may assume that $|AA|\leqslant |A|^{16/13}$ and $|(A+\unicode[STIX]{x1D6FC})/(A+\unicode[STIX]{x1D6FC})|\leqslant |A|^{8/5}$ . The conditions of Lemma 32 hold, since $|AA|\leqslant |A|^{16/13}\leqslant p^{2/3}$ , where the last inequality uses the condition $|A|\leqslant p^{13/24}$ .

The first condition of Lemma 33 holds, since

$$\begin{eqnarray}\displaystyle |A||AA|\biggl|\frac{A+\unicode[STIX]{x1D6FC}}{A+\unicode[STIX]{x1D6FC}}\biggr| & = & \displaystyle |A|\biggl(|AA|^{13}\biggl|\frac{A+\unicode[STIX]{x1D6FC}}{A+\unicode[STIX]{x1D6FC}}\biggr|^{5}\biggr)^{1/13}\biggl|\frac{A+\unicode[STIX]{x1D6FC}}{A+\unicode[STIX]{x1D6FC}}\biggr|^{8/13}\nonumber\\ \displaystyle & {\leqslant} & \displaystyle |A|^{1+21/13+64/65}\nonumber\\ \displaystyle & {\leqslant} & \displaystyle p^{2},\nonumber\end{eqnarray}$$

where the last inequality uses the condition $|A|\leqslant p^{13/24}$ . Finally, the second condition of Lemma 33 holds trivially in this application.◻

Proof. We will apply Theorem 34 where $G$ is the multiplicative group  $\mathbb{F}^{\ast }$ . Thus we require bounds on the third multiplicative energy of $A$ and the multiplicative energy of $A$ with another set $B$ .

By Lemma 25 we have

$$\begin{eqnarray}\mathsf{E}^{\times }(A,B)\ll \frac{|A+A|^{3/2}}{|A|^{1/2}}|B|^{3/2},\end{eqnarray}$$

so we may take $D_{2}=M^{3/2}$ whenever $M|A|^{2}|B|\leqslant p^{2}$ . Since we may assume that $|B|\leqslant 4|A|^{4}/\mathsf{E}^{\times }(A)$ , and the theorem is trivial if $\mathsf{E}^{\times }(A)<4M^{51/26}|A|^{32/13}$ , without loss of generality,

$$\begin{eqnarray}|B|\leqslant \frac{|A|^{20/13}}{M^{51/26}},\end{eqnarray}$$

hence $M|A|^{2}|B|\leqslant p^{2}$ whenever

$$\begin{eqnarray}\frac{|A|^{46/13}}{M^{25/26}}\leqslant p^{2},\end{eqnarray}$$

which follows from the assumption $|A|\leqslant p^{13/23}M^{25/92}$ .

By Lemma 23,

$$\begin{eqnarray}\mathsf{E}_{3}^{\times }(A)\ll \frac{|A+A|^{15/4}}{|A|^{3/4}}\log |A|\end{eqnarray}$$

whenever $|A|<p^{2/3}/M^{1/12}$ , so we may take $D_{1}=M^{15/4}\log |A|$ in this range.

The desired bound follows from plugging $D_{1}$ and $D_{2}$ into Theorem 34.

To check the constraint, first note that $M^{51/26}|A|^{32/13}>M^{3/2}|A|^{5/2}$ unless $M<|A|^{1/12}$ . For $M<|A|^{1/12}$ , we have

$$\begin{eqnarray}p^{13/23}M^{25/92}\leqslant \frac{p^{2/3}}{M^{1/12}}.\Box\end{eqnarray}$$

Proof. We use the expression

$$\begin{eqnarray}T(A)=\mathop{\sum }_{a,b\in A}\mathsf{E}^{\times }(A+a,A+b).\end{eqnarray}$$

By Cauchy–Schwarz, $\mathsf{E}^{\times }(A+a,A+b)\leqslant \mathsf{E}^{\times }(A+a)^{1/2}\mathsf{E}^{\times }(A+b)^{1/2}$ , so

$$\begin{eqnarray}T(A)\leqslant \biggl(\mathop{\sum }_{a\in A}\mathsf{E}^{\times }(A+a)^{1/2}\biggr)^{2}.\end{eqnarray}$$

By a second application of Cauchy–Schwarz, we have

$$\begin{eqnarray}T(A)\leqslant |A|\mathop{\sum }_{a\in A}\mathsf{E}^{\times }(A+a).\end{eqnarray}$$

Now, since $|A+A|=M|A|$ , we have $|(A+a)+(A+a)|=M|A|$ , so by Theorem 35,

$$\begin{eqnarray}\mathsf{E}^{\times }(A+a)\lesssim M^{51/26}|A|^{32/13}.\end{eqnarray}$$

Thus

$$\begin{eqnarray}T(A)\lesssim |A|^{2}\cdot M^{51/26}|A|^{32/13},\end{eqnarray}$$

which is the claimed bound. ◻

Proof. By Lemma 33 we have

$$\begin{eqnarray}\mathsf{E}^{\times }(A+\unicode[STIX]{x1D6FC},B)\ll \frac{|AA|^{3/2}}{|A|^{1/2}}|B|^{3/2},\end{eqnarray}$$

so we may take $D_{2}=M^{3/2}$ whenever $M|A|^{2}|B|\leqslant p^{2}$ . Since we may assume that $|B|\leqslant 4|A|^{4}/\mathsf{E}^{\times }(A+\unicode[STIX]{x1D6FC})$ , and the theorem is trivial if $\mathsf{E}^{\times }(A+\unicode[STIX]{x1D6FC})<4M^{33/13}|A|^{32/13}$ , without loss of generality,

$$\begin{eqnarray}|B|\leqslant \frac{|A|^{20/13}}{M^{33/13}},\end{eqnarray}$$

hence $M|A|^{2}|B|\leqslant p^{2}$ whenever

$$\begin{eqnarray}\frac{|A|^{46/13}}{M^{20/13}}\leqslant p^{2},\end{eqnarray}$$

which reduces to the assumption $|A|\leqslant p^{13/23}M^{10/23}$ .

By Lemma 32,

$$\begin{eqnarray}\mathsf{E}_{3}^{\times }(A+\unicode[STIX]{x1D6FC})\ll \frac{|AA|^{5}}{|A|^{2}}\log |A|\end{eqnarray}$$

whenever $|A|<p^{2/3}/M$ , so we may take $D_{1}=M^{5}\log |A|$ in this range.

The desired bound follows from plugging $D_{1}$ and $D_{2}$ into Theorem 34.

To check the constraint, first note that $M^{33/13}|A|^{32/13}>M^{3/2}|A|^{5/2}$ unless $M<|A|^{1/27}$ . For $M<|A|^{1/27}$ , we have

$$\begin{eqnarray}p^{13/23}M^{10/23}\leqslant \frac{p^{2/3}}{M}.\Box\end{eqnarray}$$