1 Introduction
For a given family
$\mathcal {F}$
of graphs, throughout this note we denote
${\mathrm {ex}}(n,\mathcal {F})$
to be the maximum number of edges in an n-vertex graph which does not contain any member in
$\mathcal {F}$
as its subgraph. This number – often referred as the extremal number or Turán number of
$\mathcal {F}$
– is the main subject in the field of extremal graph theory (see [Reference Füredi and Simonovits9]). One of the central, extremely challenging problems in this field asks for the determination of the extremal number
${\mathrm {ex}}(n,\{C_3,C_4\})$
of the family consisting of the triangle
$C_3$
and the 4-cycle
$C_4$
, whose study can be dated back to a paper of Erdős [Reference Erdős5] in 1938.
A relevant object is the Zarankiewicz number
$z(n,C_4)$
of the 4-cycle, that is, the maximum number of edges in an n-vertex bipartite graph without containing a 4-cycle. It is well-known that
${z(n,C_4)=(\frac {n}{2})^{3/2}+o(n^{3/2})}$
(see [Reference Damásdi, Héger and Szőnyi4, Reference Füredi and Simonovits9]). To be more precise, there exists a constant
$c>0$
such that for any positive integer n,
$$ \begin{align} \left(\frac{n}{2}\right)^{3/2}-cn^{4/3}\leq z(n,C_4)\leq \frac{n}{4}\left(\sqrt{2n-3}+1\right)\leq \left(\frac{n}{2}\right)^{3/2}+\frac{1}{4}n, \end{align} $$
where the lower bound follows from [Reference Füredi8] and the upper bound can be found in [Reference Keevash, Sudakov and Verstraëte12] (see its Proposition 1.4).Footnote 1 As a bipartite graph cannot contain a triangle, evidently one can relate these two aforementioned numbers with the following inequality
$$ \begin{align} {\mathrm{ex}}(n,\{C_3,C_4\})\geq z(n,C_4). \end{align} $$
A famous old conjecture of Erdős [Reference Erdős5, Reference Erdős6] (restated in Erdős-Simonovits [Reference Erdős and Simonovits7]) asserts that this lower bound on
${\mathrm {ex}}(n,\{C_3,C_4\})$
essentially is optimal.
Conjecture 1.1 (Erdős [Reference Erdős5, Reference Erdős6], Erdős-Simonovits [Reference Erdős and Simonovits7]).
It holds that
$$ \begin{align*} \lim\limits_{n\to +\infty} \frac{{\mathrm{ex}}(n,\{C_3,C_4\})}{z(n,C_4)}=1. \end{align*} $$
In view of (1.1) and (1.2), this conjecture is equivalent to the upper bound
${\mathrm {ex}}(n,\{C_3,C_4\})\leq (\frac {n}{2})^{3/2}+o(n^{3/2})$
. It is still widely open. The best known upper bound on
${\mathrm {ex}}(n,\{C_3,C_4\})$
remains the following trivial bound that
$$ \begin{align*}{\mathrm{ex}}(n,\{C_3,C_4\})\leq {\mathrm{ex}}(n,C_4)=\frac12 n^{3/2}+O(n).\end{align*} $$
OnFootnote
2
the other hand, Parsons [Reference Parsons14] gave a construction in 1976, showing that for integers
$n=\binom {q}{2}$
where
$q=1$
mod
$4$
is a prime, the inequality (1.2) can be improved to
$$ \begin{align*} {\mathrm{ex}}(n,\{C_3,C_4\})\geq \left(\frac{n}{2}\right)^{3/2}+\frac{3}{8}n\geq z(n,C_4)+\frac{1}{8}n. \end{align*} $$
To the best of our knowledge, no progress has been made since then. A stronger version of Conjecture 1.1 was stated as a problem in the book of Chung and Graham [Reference Chung and Graham3] (see p.41 in Section 3.4).
Problem 1.2 (Chung-Graham [Reference Chung and Graham3]).
Is it true that
${\mathrm {ex}}(n,\{C_3,C_4\})=\left (\frac {n}{2}\right )^{3/2}+O(n)$
?
Let us also mention that Allen, Keevash, Sudakov, and Verstraëte [Reference Allen, Keevash, Sudakov and Verstraëte1] (see Conjecture 1.7 therein) made an opposite conjecture that
$\liminf \limits _{n\to +\infty }\frac {{\mathrm {ex}}(n,\{C_3,C_4\})}{z(n,C_4)}>1.$
A general conjecture of Erdős and Simonovits [Reference Erdős and Simonovits7] concerning extremal numbers of families containing bipartite graphs and odd cycles is as follows: Let
$\mathcal {C}$
denote the family of all odd cycles, and let
$\mathcal {C}_k$
denote the family of all odd cycles of length at most k. Then for any finite family
$\mathcal {F}$
containing a bipartite graph, there exists an odd integer k such that
$$ \begin{align*}\lim\limits_{n\to +\infty} \frac{{\mathrm{ex}}(n,\mathcal{F}\cup \mathcal{C}_k)}{{\mathrm{ex}}(n,\mathcal{F}\cup \mathcal{C})}=1.\end{align*} $$
Erdős and Simonovits [Reference Erdős and Simonovits7] confirmed this for
$\mathcal {F}=\{C_4\}$
by showing that
${\mathrm {ex}}(n,\{C_4,C_5\})=\left (\frac {n}{2}\right )^{3/2}+O(n)$
and thus
$\lim \limits _{n\to +\infty }\frac {{\mathrm {ex}}(n,\{C_4,C_5\})}{z(n,C_4)}=1$
. This was strengthened by Keevash, Sudakov, and Verstraëte [Reference Keevash, Sudakov and Verstraëte12], where their main result implies that for all integers
$k\geq 2$
,
$$ \begin{align} {\mathrm{ex}}(n,\{C_4,C_{2k+1}\})=\left(\frac{n}{2}\right)^{3/2}+O(n). \end{align} $$
The main result of this paper is the following theorem, which improves the error term in Parsons’ lower bound on
${\mathrm {ex}}(n,\{C_3,C_4\})$
from
$\Omega (n)$
to
$\Omega (n^{1.25})$
.
Theorem 1.3. There exists an absolute constant
$c>0$
such that for every integer
$n\geq 7$
,
$$ \begin{align*}{\mathrm{ex}}(n,\{C_3,C_4\})\geq z(n,C_4)+c\cdot n^{1.25}.\end{align*} $$
We would like to emphasize that this result works for every integer
$n\geq 7$
, while the construction of Parsons is applicable only for a special form of integers n. Let us also note that when
$n=6$
, both numbers
${\mathrm {ex}}(n,\{C_3,C_4\})$
and
$z(n,C_4)$
are equal to
$6$
.
Using the above bound, one can immediately derive the following corollary that the difference between
${\mathrm {ex}}(n,\{C_3,C_4\})$
and
$\left (\frac {n}{2}\right )^{3/2}$
can be a superlinear term in n, thus showing that the answer to Problem 1.2 is negative.
Corollary 1.4. For integers
$n=2(q^2+q+1)$
where q is a prime power,
$$ \begin{align*}{\mathrm{ex}}(n,\{C_3,C_4\})=\left(\frac{n}{2}\right)^{3/2}+\Omega(n^{1.25}).\end{align*} $$
In particular, this provides a negative answer to Problem 1.2.
We also see from (1.3) that the behavior of
${\mathrm {ex}}(n,\{C_3,C_4\})$
is different from
${\mathrm {ex}}(n,\{C_4,C_{2k+1}\})$
for any integer
$k\geq 2$
, as they vary in their second order terms. We remark (see the last paragraph of Section 2) that using results from number theory on the distribution of primes, the conclusion of Corollary 1.4 in fact holds for almost all integers n. Also, if we assume some reasonable conjecture about prime gaps (that is, there exists a prime in
$[n-o(n^{1/2}),n]$
for every sufficiently large n), then Corollary 1.4 holds for all sufficiently large integers n.
The rest of the paper is organized as follows. In Section 2, we present the proofs of Theorem 1.3 and Corollary 1.4. In the final section, we conclude with a remark by explaining that similar constructions as in Theorem 1.3 are unlikely to give better bounds.
2 The proofs of Theorem 1.3 and Corollary 1.4
In this section, we prove Theorem 1.3 and then use it to infer Corollary 1.4. Before proving Theorem 1.3, we start with a warm-up theorem as follows.
Theorem 2.1. For any integer
$n\geq 7$
, it holds that
$$ \begin{align*}{\mathrm{ex}}(n,\{C_3,C_4\})\geq z(n,C_4)+1.\end{align*} $$
Proof. By (1.2) we have
${\mathrm {ex}}(n,\{C_3,C_4\})\geq z(n,C_4)$
. Assume for the sake of contradiction, that
${\mathrm {ex}} (n, \{C_3, C_4\}) = z(n,C_4)$
for some
$n\geq 7$
. Then there exists an n-vertex bipartite
$C_4$
-freeFootnote
3
graph G with
$e(G)={\mathrm {ex}}(n,\{C_3,C_4\})$
. Let
$(X,Y)$
be the bipartition of G with
$|X|\geq |Y|$
. We claim that
-
(P1). Any two vertices in the same part (X or Y) have a unique common neighbor, and
-
(P2). The maximum degree
$\Delta (G)$
is at most three.
If
$u,v\in X$
do not have a common neighbor, then the graph
$G+\{uv\}$
will be
$\{C_3,C_4\}$
-free, a contradiction to that
$e(G)={\mathrm {ex}}(n,\{C_3,C_4\})$
. So (P1) follows. To see (P2), suppose that there is a vertex
$u\in X$
with at least four neighbors
$a,b,c,d\in Y$
. Let
$G'$
be the graph obtained from G by deleting the edges
$ub,uc$
and adding new edges
$ab,bc,cd$
. If there is a 3-cycle or 4-cycle in
$G'$
containing one edge in
$\{ab,bc,cd\}$
, then this cycle contains a vertex
$u'\in X$
which has two neighbors in
$\{a,b,c,d\}$
. It implies that u and
$u'$
have two common neighbors in G, which is impossible. Thus
$G'$
is
$\{C_3,C_4\}$
-free with
$e(G')>e(G)$
. This contradiction proves the above claim.
Let
$\sigma _X$
denote the number of paths of length two with both end-points in X. By (P1) we have
$\sigma _X=\binom {|X|}{2}$
, while (P2) implies that
$\sigma _X=\sum _{y\in Y}\binom {d(y)}{2}\leq 3|Y|$
. As
$|X|\geq |Y|$
, we get
$\binom {|X|}{2}\leq 3|Y|\leq 3|X|$
, showing that
$|X|\leq 7$
. So
$n\leq 14$
. The precise values of
${\mathrm {ex}}(n, \{C_3,C_4\})$
are determined in [Reference Garnick, Kwong and Lazebnik10] for all integers
$n\leq 24$
(see Theorem 3.1). In particular, when
$n\in \{7,8,9,10,13\}$
, the extremal
$\{C_3,C_4\}$
-free graph on n vertices is unique and nonbipartite (see Figures 1 and 2 in [Reference Garnick, Kwong and Lazebnik10]). So it only remains to consider
$n\in \{11,12,14\}$
. For
$n=11$
, we have
$|Y|\leq 5$
and
${\mathrm {ex}}(11,\{C_3,C_4\})=16$
from [Reference Garnick, Kwong and Lazebnik10], so there must be a vertex in Y of degree at least four, a contradiction to (P2). For
$n=12$
, we have
${\mathrm {ex}}(12,\{C_3,C_4\})=18$
from [Reference Garnick, Kwong and Lazebnik10], implying that all vertices have degree three and
$|X|=|Y|=6$
. However, it leads to a contradiction as we should have
$\binom {|X|}{2}=\sigma _X=3|Y|$
in this situation. Lastly, for
$n=14$
, we have
${\mathrm {ex}}(14,\{C_3,C_4\})=23$
from [Reference Garnick, Kwong and Lazebnik10] and thus the extremal graph G must contain a vertex of degree at least four, a final contradiction to (P2). This proves Theorem 2.1.
We are now prepared to present the proof of Theorem 1.3 by extending the above proof.
Proof of Theorem 1.3.
To complete the proof of Theorem 1.3, in view of Theorem 2.1 it suffices to show that for sufficiently large n,
$$ \begin{align} {\mathrm{ex}}(n,\{C_3,C_4\})=z(n,C_4)+\Omega(n^{1.25}). \end{align} $$
Let
$\varepsilon $
be a sufficiently small (but fixed) positive real and let n be any integer which is sufficiently larger than
$1/\varepsilon $
. Let G be any extremal graph of
$z(n,C_4)$
, that is, an n-vertex bipartite
$C_4$
-free graph with
$z(n,C_4)$
edges. Let
$(X,Y)$
be the bipartition of G. In what follows, based on G we will construct a (nonbipartite)
$\{C_3,C_4\}$
-free graph on the same vertex set
$V(G)$
and with
$\Omega (n^{1.25})$
more edges.
We claim that there exists a vertex u in G with
$d(u)\leq (1+\varepsilon )\sqrt {n/2}$
and
$|\cup _{x\in N(u)} N(x)|\geq (1-\varepsilon )n/2$
. Note that as G is
$C_4$
-free, for any vertex u, the neighborhoods
$N(x)$
for all vertices
$x\in N(u)$
are pairwise disjoint. To prove this claim, we will proceed to show that
-
(A). there are less than
$n/2$
vertices with degree at least
$(1+\varepsilon )\sqrt { n/2}$
, and -
(B). there are less than
$n/2$
vertices u with
$|\cup _{x\in N(u)} N(x)|\le (1-\varepsilon ) n/2$
.
First let us see that
$(X,Y)$
is almost balanced. By (1.1), we have
$$ \begin{align*}\left(\frac{n}{2}\right)^{3/2}-cn^{4/3}\leq z(n,C_4)=e(G)\leq (|X||Y|)^{3/4}+\max\{|X|,|Y|\}\leq (|X|(n-|X|))^{3/4}+n,\end{align*} $$
where the second last inequality follows by Proposition 3.9 in [Reference Keevash, Sudakov and Verstraëte12]. Solving the above inequality for
$|X|$
, it gives that
$(1-\frac {\varepsilon }{2})\frac {n}{2}\leq |X|, |Y|\leq (1+\frac {\varepsilon }{2})\frac {n}{2}$
. Let
$\sigma $
be the number of paths of length two in G. As G is bipartite and
$C_4$
-free, each pair of vertices from the same part is contained in at most one path of length two. So
$\sigma \leq \binom {|X|}{2}+\binom {|Y|}{2}\leq (1+\frac {\varepsilon ^2}{4})\frac {n^2}{4}$
. Suppose for a contradiction to (A) that there are
$n/2$
vertices with degree at least
$(1+\varepsilon )\sqrt {n/2}$
in G. By Jensen’s inequality, we have
$$ \begin{align*} \sigma &=\sum_{u\in V(G)}\binom{d(u)}{2} \geq \frac n2 \binom{(1+\varepsilon)\sqrt{n/2}}{2}+ \frac n2 \binom{\left(2e(G)-\frac n2(1+\varepsilon)\sqrt {n/2}\right)/(n/2)}{2}\\ &\ge \frac n2 \binom{(1+\varepsilon)\sqrt{n/2}}{2}+ \frac n2 \binom{(1-\varepsilon)\sqrt{n/2}-O(n^{1/3})}{2} =(1+\varepsilon^2) \frac{n^2}4 -o(n^2). \end{align*} $$
This is a contradiction to the above upper bound on
$\sigma $
, thus proving (A). To see (B), suppose on the contrary that there are
$n/2$
vertices u with
$|\cup _{x\in N(u)} N(x)|\le (1-\varepsilon ) n/2$
. Each of these vertices is contained in at most
$(1-\varepsilon ) n/2$
paths of length two as an end-point, while any other vertex is contained in at most
$\max \{|X|,|Y|\}\leq (1+\frac {\varepsilon }{2})\frac {n}{2}$
paths of length two as an end-point. Totally we have
$2\sigma \leq \frac {n}{2}\cdot (1-\varepsilon )\frac {n}{2}+ \frac {n}{2}\cdot (1+\frac {\varepsilon }{2})\frac {n}{2}=\left (2-\frac {\varepsilon }{2}\right )\frac {n^2}{4},$
implying that
$$ \begin{align*} \left(1-\frac{\varepsilon}{4}\right)\frac{n^2}{4}\geq \sigma =\sum_{u\in V(G)}\binom{d(u)}{2} \geq n \binom{ 2e(G)/n}{2}\geq n \binom{\sqrt{n/2}-O(n^{1/3})}{2}=\frac{n^2}{4}-o(n^2), \end{align*} $$
a contradiction. This completes the proof of the claim.
Let
$u\in X$
be the vertex as claimed. Let
$N(u)=\{u_1,u_2,...,u_t\}$
for some
$t\leq (1+\varepsilon )\sqrt {n/2}$
. For each
$1\leq i\leq t$
, let
$N_i=N(u_i)$
and
$E_i=\{u_ix| x\in N_i\}$
. As pointed out, these
$N_i$
’s are pairwise disjoint and thus
$\sum _{1\leq i\leq t} |N_i|\geq (1-\varepsilon )n/2.$
Let
$G_i$
be an extremal
$\{C_3,C_4\}$
-free graph on the vertex set
$N_i$
. By (1.1), there exists some
$c>0$
such that
$$ \begin{align} e(G_i)\geq \left(|N_i|/2\right)^{3/2}-c\cdot\left(|N_i|/2\right)^{4/3}. \end{align} $$
Let H be obtained from G by deleting all edges in
$E_i$
and adding the graph
$G_i$
into
$N_i$
for every
$1\leq i\leq t$
. We claim that H is
$\{C_3, C_4\}$
-free. Suppose that H contains a triangle say
$abc$
. Then at least one edge (say
$ab$
) must appear in some
$G_i$
(note that
$V(G_i)=N_i\subseteq X$
). As
$G_i$
is
$C_3$
-free, we must have
$c\in Y$
and thus
$ac,bc\in E(G)\cap E(H)$
. But this is a contradiction as the unique common neighbor of
$a,b$
in Y has been destroyed by deleting the edges of
$E_i$
. Now suppose H has a
$C_4$
say
$abcd$
. We may assume that
$ab\in E(G_i)$
. Since
$G_i$
is
$C_4$
-free and
$H[Y]$
is an independent set, we may assume that
$c\in X$
and
$d\in Y$
. As
$N_j$
’s are pairwise disjoint, it is clear that
$c\in N_i$
. Then we get
$ad,cd\in E(G)\cap E(H)$
, which is a contradiction by the same reason. So indeed, H is an n-vertex
$\{C_3, C_4\}$
-free graph. We can estimate the number of edges in H as follows:
$$ \begin{align*} e(H)&=e(G)+\sum_{1\leq i\leq t} (e(G_i)- |E_i|)\geq z(n,C_4)+ \sum_{1\leq i\leq t} \left(\left(|N_i|/2\right)^{3/2}-c\cdot\left(|N_i|/2\right)^{4/3}-|N_i|\right)\\ &\geq z(n,C_4)+\sum_{1\leq i\leq t} (1-o(1))\left(\frac{|N_i|}{2}\right)^{3/2} \ge z(n,C_4)+ (1-o(1))\cdot t\left(\frac{\sum_{1\leq i\leq t}|N_i| /t}{2}\right)^{3/2}\\ & =z(n,C_4)+ \Omega\left(\frac{(\sum_{1\leq i\leq t}|N_i|)^{3/2}}{\sqrt{t}}\right)\geq z(n,C_4)+\Omega(n^{1.25}), \end{align*} $$
where the first inequality follows by (2.2), and the last inequality uses the facts that
$t\leq (1+\varepsilon )\sqrt {n/2}$
and
$\sum _{1\leq i\leq t} |N_i|\geq (1-\varepsilon )n/2.$
This proves (2.1) and thus completes the proof of Theorem 1.3.
Proof of Corollary 1.4.
Let q be a prime power and let
$n=2(q^2+q+1)$
. In this case, it is well-known that a finite projective plane of order q exists and thus
$z(n,C_4)=\frac 12(q+1)n\geq \left (\frac {n}{2}\right )^{3/2}$
(see, e.g., Theorem 1.2 in [Reference Keevash, Sudakov and Verstraëte12]). Therefore, by Theorem 1.3, we have
${\mathrm {ex}}(n,\{C_3,C_4\})=z(n,C_4)+\Omega (n^{1.25})=\left (\frac {n}{2}\right )^{3/2}+\Omega (n^{1.25}).$
In the rest of this section, we present a generalization of Corollary 1.4. Let
$p_n$
denote the n’th prime. It is well-known in number theory (see [Reference Heath-Brown11]) that there exists some constant
$\delta \in (0,1)$
such that for all reals
$x>0$
,
$$ \begin{align} \sum_{p_n\leq x, ~ p_{n+1}-p_n\geq \sqrt{p_n}} (p_{n+1}-p_n)=O(x^{1-\delta}). \end{align} $$
This can imply that for sufficiently small constant
$\varepsilon>0$
and for almost all integers n, there exists a prime in
$[n-\varepsilon \sqrt {n},n]$
. By the proof of [Reference Füredi8], one can then derive that
$z(n,C_4)\geq \left (\frac {n}{2}\right )^{3/2}-O(\varepsilon )\cdot n^{1.25}$
for almost all integers n. Therefore, together with Theorem 1.3, this shows that
${\mathrm {ex}}(n,\{C_3,C_4\})=z(n,C_4)+\Omega (n^{1.25})=\left (\frac {n}{2}\right )^{3/2}+\Omega (n^{1.25})$
holds for almost all integers n.
3 A remark
In the above construction, we take an extremal graph of
$z(n,C_4)$
, choose vertex-disjoint subsets of size roughly
$q=\sqrt {n/2}$
, and then for each of these subsets A, add
$\Omega (q^{3/2})$
edges into A and delete fewer edges incident with A to make a
$\{C_3, C_4\}$
-free graph with more than
$z(n,C_4)$
edges. One may ask whether one can take larger subsets (say of size
$n^{1/2+\varepsilon }$
for any
$\varepsilon>0$
) and add/delete edges using similar operations to get a denser
$\{C_3, C_4\}$
-free graph. We illustrate in the following example that it is unlikely to give better constructions.
For the purpose of our presentation, let G be an extremal graph of
$z(n,C_4)$
with bipartite
$(X,Y)$
and let
$n=2q^2$
be an integer with
$q\in \mathbb {R}^+$
such that
$$ \begin{align} \text{except }O(q)\text{ vertices, every vertex in }G\text{ has degree at least }q-o(q). \end{align} $$
Let
$\delta \in (0,1)$
be any real. Consider any set
$A\subseteq X$
of size
$q^{1+\delta }$
.
We will show that it is impossible to construct a
$\{C_3,C_4\}$
-free graph H obtained from G by adding
$c |A|^{3/2}$
edges into A and deleting any subset
$E^*$
of edges such that
$e(H)>e(G)$
. Suppose for a contradiction that such H does exist. Since
$e(H)>e(G)$
, we have
$|E^*|<c |A|^{3/2}=o(q^{2+\delta })$
. It is easy to see that the size of
$|X|$
or
$|Y|$
is
$(1+o(1))q^2$
. By (3.1), the number of edges between A and Y in G is at least
$(1-o(1))q^{2+\delta }$
. Then the induced bipartite subgraph
$H[A,Y]$
of H with parts A and Y has at least
$(1-o(1))q^{2+\delta }-|E^*|\geq (1-o(1))q^{2+\delta }$
edges. Let
$\sigma _1$
be the number of paths of lengths two in
$H[A,Y]$
with both ends in A. Then
$$ \begin{align*} \sigma_1=\sum_{v\in Y}\binom{d_{H[A,Y]}(v)}{2}\geq |Y|\binom{e(H[A,Y])/|Y|}{2}=(1-o(1))q^{2+2\delta}/2=(1-o(1))\binom{|A|}{2}. \end{align*} $$
As
$H[A]$
has at least
$c|A|^{3/2}$
edges, the number
$\sigma _2$
of paths of length two in
$H[A]$
is
$$ \begin{align*} \sigma_2=\sum_{v\in A}\binom{d_{H[A]}(v)}{2}\geq |A|\binom{2e(H[A])/|A|}{2}\geq c^2\binom{|A|}{2}, \end{align*} $$
Therefore, in total H contains
$\sigma _1+\sigma _2\geq (1+c^2-o(1))\binom {|A|}{2}>\binom {|A|}{2}$
paths of length two with both ends in A, which leads to a copy of
$C_4$
in H, a contradiction.
We point out that when
$n=2(t^2+t+1)$
for any prime power t, the extremal graph of
$z(n, C_4)$
is regular and thus satisfies (3.1), so (at least) for these infinitely many integers n, our construction cannot be improved using the above operations. In fact, to make the above arguments work, all we need here is the property that
$e(G[A,Y])\geq (1-o(1))q^{2+\delta }$
(which follows by (3.1)), and one can show that for general n, almost all subsets A of size
$q^{1+\delta }$
in one part of the extremal graph of
$z(n,C_4)$
satisfy this property.
As a side note, Keevash et al. proved in [Reference Keevash, Sudakov and Verstraëte12] (see Theorem 5.1) that any (nearly) extremal graph of
$z(n,C_4)$
satisfies the pseudorandomness property. We would like to conjecture that any extremal graph of
$z(n,C_4)$
satisfies (3.1).
Acknowledgments
We would like to thank Lilu Zhao for his helpful instruction in number theory. We are also grateful to Chunqiu Fang for the stimulating discussion and Guorong Gao for bringing the result on (2.3) to our attention.
Competing interest
The authors have no competing interests to declare.
Financial support
Research supported by the National Key Research and Development Program of China 2020YFA071310 and the National Natural Science Foundation of China grant 12125106.
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