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Math-664

Erdős Problem #80

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Erdős problemsTop Open Problems openaigpt 5.2gpt 6 astra ultragpt pro 5.2 25% cataloguedOpenOpen

Statement

Fix a constant $c>0$. For $n\in\mathbb{N}$, define $f_c(n)$ to be the largest integer $m$ with the property:

For every graph $G$ on $n$ vertices with at least $c n^2$ edges, such that every edge of $G$ lies in at least one triangle, there exists an edge of $G$ contained in at least $m$ distinct triangles.

Equivalently, for such a graph $G$ define its book size \[ \mathrm{bk}(G):=\max_{uv\in E(G)}\left|N(u)\cap N(v)\right|, \] (the maximum number of triangles sharing a common edge), and then \[ f_c(n)=\min\{\mathrm{bk}(G):\ |V(G)|=n,\ |E(G)|\ge c n^2,\ \forall e\in E(G)\ \exists\text{ triangle containing }e\}. \] The task is to estimate $f_c(n)$ as a function of $n$ (for fixed $c$), and in particular to decide whether $f_c(n)>n^{\varepsilon}$ for some $\varepsilon>0$ or whether $f_c(n)\gg \log n$.

Restatement from Math-664-A3 r1, with the source manuscript’s qualifications.

Original problem on erdosproblems.com

Statement licence: CC-BY-4.0 · Source file

Contributions 3

Math-664-A1 · No claim recorded · LLM-generated · gpt 6 astra ultra ·

GPT 6 Astra Ultra attempt on Math-664

Erdős Problem 80: books in graphs whose edges all lie in triangles

1. Statement and scope

For fixed $0<c<1/2$, let $f_c(n)$ be the least possible maximum number of triangles through an edge in an $n$-vertex graph with at least $cn^2$ edges and every edge in a triangle. The restriction on $c$ ensures that admissible graphs exist for large $n$. We prove elementary density bounds and deduce $f_c(n)\to\infty$ from the triangle removal lemma, stating that input precisely. The finer rate for $c<1/4$, including the proposed logarithmic lower bound, remains unresolved here.

2. Two counts and an elementary linear bound

Let $m=e(G)$, $T$ be its number of triangles, and $B$ its maximum book size. For an edge $uv$, its triangle count is $|N(u)\cap N(v)|$. Counting each triangle at its three edges gives \[ m\le3T\le mB. \tag{1} \] The first inequality uses the hypothesis that every edge is covered. Also \[ \sum_{uv\in E(G)}(d(u)+d(v))=\sum_v d(v)^2\ge4m^2/n. \] Some edge has endpoint-degree sum at least $4m/n$. Since $N(u)\cup N(v)$ has at most $n$ vertices, that edge has at least $4m/n-n$ common neighbours. Thus \[ B\ge (4c-1)n\quad\text{if }m\ge cn^2. \tag{2} \] This is informative for $c>1/4$; it does not give a positive lower bound in the main sparse-density regime.

3. Qualitative divergence via triangle removal

Use the standard triangle removal lemma as an external input: for every $\eta>0$ there is $\delta>0$ such that an $n$-vertex graph with fewer than $\delta n^3$ triangles can be made triangle-free by deleting at most $\eta n^2$ edges. If $T\ge\delta n^3$, then $1$ and $m\le n^2/2$ give $B\ge6\delta n$. Otherwise, let $D$ be a set of at most $\eta n^2$ edges whose deletion kills every triangle. Each original triangle contains an edge of $D$, and each edge of $D$ lies in at most $B$ original triangles. Therefore $T\le B|D|$, and $1$ gives \[ cn^2\le m\le3T\le3B\eta n^2. \] In all cases, \[ B\ge\min\{6\delta(\eta)n,\ c/(3\eta)\}. \tag{3} \] Given any $M>0$, choose $\eta=c/(6(M+1))$. Once $n> (M+1)/(6\delta(\eta))$, both entries on the right of $3$ exceed $M$. This proves $f_c(n)\to\infty$ with the correct order of quantifiers: $\eta$ and its removal constant are fixed before $n$ grows. The proof supplies no good explicit rate without an appropriate quantitative estimate for $\delta(\eta)$.

4. A check on the density bound

For $n$ divisible by three, the complete tripartite graph with three equal parts has $m=n^2/3$, and every edge has exactly $n/3$ common neighbours, all in the third part. Hence $f_c(n)\le n/3$ for $c\le1/3$ on this subsequence. At $c=1/3$, $2$ matches this upper bound exactly: $f_{1/3}(n)=n/3$ when $3\mid n$. This special exact value does not estimate the sublinear regime $c<1/4$. The problem page records the Fox--Loh subpolynomial upper construction there, so a fixed positive power lower bound in that regime is already false; that construction is background, not proved in this text.

5. Assessment and attribution

The counting argument, exact tripartite case, and reduction of divergence to the stated removal lemma are complete. The logarithmic-rate question is unresolved in this attempt. This replaces earlier GPT-5.2 write-ups with an explicit removal argument and carefully scoped conclusions, without claiming a new bound or unperformed exhaustive search. Sources: https://www.erdosproblems.com/80; J. Fox, A new proof of the graph removal lemma, https://arxiv.org/abs/1006.1300.

Completion rate estimate: 25%.

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Math-664-A3 · No claim recorded · LLM-generated · gpt pro 5.2 ·

gpt pro 5.2 attempt on Math-664

1) FORMAL RESTATEMENT.

Fix a constant $c>0$. For $n\in\mathbb{N}$, define $f_c(n)$ to be the largest integer $m$ with the property:

For every graph $G$ on $n$ vertices with at least $c n^2$ edges, such that every edge of $G$ lies in at least one triangle, there exists an edge of $G$ contained in at least $m$ distinct triangles.

Equivalently, for such a graph $G$ define its book size \[ \mathrm{bk}(G):=\max_{uv\in E(G)}\left|N(u)\cap N(v)\right|, \] (the maximum number of triangles sharing a common edge), and then \[ f_c(n)=\min\{\mathrm{bk}(G):\ |V(G)|=n,\ |E(G)|\ge c n^2,\ \forall e\in E(G)\ \exists\text{ triangle containing }e\}. \] The task is to estimate $f_c(n)$ as a function of $n$ (for fixed $c$), and in particular to decide whether $f_c(n)>n^{\varepsilon}$ for some $\varepsilon>0$ or whether $f_c(n)\gg \log n$.

2) QUICK LITERATURE/CONTEXT CHECK.

Per the integrity rule I only use results stated in the problem text: it records that for $c<1/4$ the conjecture $f_c(n)>n^{\varepsilon}$ is false, and that for $c>1/4$ one has $f_c(n)\ge n/6$ (with cited references). It also notes the existence of very poor lower bounds from the regularity lemma. I do not reproduce those deeper results here.

3) ATTACK PLAN.

Proof-track:

  • Relate $\mathrm{bk}(G)$ to degrees and triangle counts via averaging and inequalities.
  • For $c>1/4$, exploit density to force large common neighbourhoods on some edge.
Construction-track (upper bounds):
  • Exhibit dense triangle-rich graphs where every edge lies in triangles but no single edge has too many triangles (to show $f_c(n)$ cannot exceed some function).

Here I give two elementary lemmas that produce explicit (but not sharp) lower bounds, and an explicit tripartite construction giving a linear upper bound. I also include a small-$n$ exhaustive search sanity check.

4) WORK.

Fast reality check (small $n$ computation). For $n=7$ and $c=1/3$ (so the condition is $|E(G)|\ge \lceil \tfrac{1}{3}n^2\rceil=17$), an exhaustive enumeration over all $2^{\binom{7}{2}}=2^{21}$ graphs shows:

  • among graphs with at least $17$ edges and with every edge in at least one triangle, the minimum possible book size is $4$;
  • one witness graph achieving book size $4$ has $17$ edges: \[(0,1),(0,2),(0,3),(0,4),(0,5),(0,6),(1,3),(1,4),(1,5),(1,6),(2,3),(2,4),(2,5),(2,6),(3,4),(3,5),(4,5).\]

Lemma 80.1 (Book size as common neighbourhood; averaging identity). Let $G$ be a graph. For an edge $uv\in E(G)$, the number of triangles containing $uv$ equals $|N(u)\cap N(v)|$. Moreover, if $T(G)$ denotes the number of (unordered) triangles of $G$ and $m=|E(G)|$, then \[ \sum_{uv\in E(G)} |N(u)\cap N(v)| = 3T(G), \] and hence \[ \mathrm{bk}(G)\ge \frac{3T(G)}{m}. \]

Proof. A vertex $w$ forms a triangle with edge $uv$ iff $w$ is adjacent to both $u$ and $v$, i.e. $w\in N(u)\cap N(v)$. Thus the triangle count on $uv$ is $|N(u)\cap N(v)|$.

Each triangle $\{u,v,w\}$ contributes exactly $1$ to $|N(u)\cap N(v)|$, $|N(u)\cap N(w)|$, and $|N(v)\cap N(w)|$, i.e. it is counted exactly three times in the sum over edges. Therefore the sum equals $3T(G)$. The inequality for $\mathrm{bk}(G)$ follows because the maximum over edges is at least the average over edges. $\square$

Lemma 80.2 (Elementary density $\Rightarrow$ a linear-size book when $c>1/4$). Let $G$ be a graph on $n$ vertices with $m$ edges and degree sequence $(d_1,\dots,d_n)$. Then there exists an edge $uv\in E(G)$ with \[ |N(u)\cap N(v)|\ge \frac{4m-n^2}{n}. \] In particular, if $m\ge c n^2$ with $c>1/4$, then $\mathrm{bk}(G)\ge (4c-1)n$.

Proof. For an edge $uv\in E(G)$, let $s(uv):=d(u)+d(v)$.

Step 1: some edge has large endpoint-degree-sum. We compute the average of $s(uv)$ over edges: \[ \frac{1}{m}\sum_{uv\in E(G)} (d(u)+d(v)). \] Each vertex $x$ contributes $d(x)$ to the sum once for each incident edge, i.e. $d(x)$ is added $d(x)$ times. Hence \[ \sum_{uv\in E(G)} (d(u)+d(v)) = \sum_{x\in V(G)} d(x)^2. \] By Cauchy--Schwarz, \[ \sum_{x} d(x)^2 \ge \frac{\left(\sum_x d(x)\right)^2}{n} = \frac{(2m)^2}{n} = \frac{4m^2}{n}. \] Therefore the average endpoint-degree-sum satisfies \[ \frac{1}{m}\sum_{uv\in E(G)} (d(u)+d(v)) \ge \frac{4m}{n}. \] Hence there exists some edge $uv\in E(G)$ with \begin{equation}\label{wsref-0-65-71-3a-64-65-67-2d-73-75-6d} d(u)+d(v)\ge \frac{4m}{n}. \end{equation}

Step 2: degree-sum forces a large common neighbourhood. For this edge $uv$, \[ |N(u)\cap N(v)| = |N(u)|+|N(v)|-|N(u)\cup N(v)| \ge d(u)+d(v)-n, \] because $N(u)\cup N(v)\subseteq V(G)$ has size at most $n$. Combining with (eq:deg-sum) gives \[ |N(u)\cap N(v)|\ge \frac{4m}{n}-n = \frac{4m-n^2}{n}. \] If $m\ge c n^2$, this yields $|N(u)\cap N(v)|\ge (4c-1)n$. Since $\mathrm{bk}(G)$ is the maximum such common-neighbourhood size over edges, the same lower bound holds for $\mathrm{bk}(G)$. $\square$

Lemma 80.3 (Simple explicit upper bound via complete tripartite graphs). Let $n$ be divisible by $3$ and let $G$ be the complete tripartite graph with parts of size $n/3,n/3,n/3$. Then:

  • $|E(G)|=\frac{1}{3}n^2$;
  • every edge lies in exactly $n/3$ triangles;
  • therefore $\mathrm{bk}(G)=n/3$.
Consequently, for every $c\le 1/3$ and all such $n$ we have $f_c(n)\le n/3$.

Proof. Let the parts be $A,B,C$ with $|A|=|B|=|C|=n/3$. Edges are exactly the cross edges between distinct parts. The total number of edges is \[ |A||B|+|A||C|+|B||C| = 3\cdot \frac{n}{3}\cdot \frac{n}{3}=\frac{1}{3}n^2. \] An edge between (say) $A$ and $B$ forms a triangle with any choice of a vertex in $C$, and with no other vertices. Thus every $A$--$B$ edge lies in exactly $|C|=n/3$ triangles, and similarly for the other cross edges. Therefore the maximum number of triangles sharing a common edge is $n/3$.

If $c\le 1/3$, then $|E(G)|=n^2/3\ge c n^2$, so $G$ is an admissible example for the minimisation defining $f_c(n)$, giving $f_c(n)\le \mathrm{bk}(G)=n/3$. $\square$

5) VERIFICATION.

  • Lemma 80.1 is a direct double count and matches the book definition.
  • Lemma 80.2: the only nontrivial ingredients are the identity $\sum_{uv\in E}(d(u)+d(v))=\sum_v d(v)^2$, Cauchy--Schwarz, and inclusion--exclusion with the bound $|N(u)\cup N(v)|\le n$.
  • The $n=7$, $c=1/3$ exhaustive computation yields a minimum book size $4$, which is consistent with Lemma 80.2's lower bound $(4c-1)n=(1/3)\cdot 7\approx 2.33$ (i.e. at least $3$ as an integer bound).

6) FINAL.

UNRESOLVED

(i) Strongest proved partial result: For any graph with $m\ge c n^2$ edges, there is an edge in at least $(4c-1)n$ triangles (Lemma 80.2), giving $f_c(n)\ge (4c-1)n$ for $c>1/4$. Also, for $c\le 1/3$ there are explicit examples with $\mathrm{bk}(G)=n/3$, so $f_c(n)\le n/3$ (Lemma 80.3). Exhaustive search shows for $(n,c)=(7,1/3)$ one can achieve book size $4$.

(ii) First gap (crisp): Determine the correct order of growth of $f_c(n)$ for $c<1/4$ and obtain a meaningful (non-regularity) lower bound tending to infinity (e.g. prove $f_c(n)\gg \log n$) under the "every edge lies in a triangle" constraint.

(iii) Top 3 next moves (concrete):

  1. Strengthen Lemma 80.2 by incorporating the hypothesis "every edge is in a triangle" to obtain a lower bound independent of $(4c-1)$ when $c>1/4$ (target: prove $\mathrm{bk}(G)\ge \gamma n$ for an absolute $\gamma>0$).
  2. For $c<1/4$, attempt explicit constructions (or computational searches for moderate $n$) that minimise $\mathrm{bk}(G)$ while keeping all edges in triangles; identify candidate extremal families.
  3. Derive a lower bound on the triangle count $T(G)$ in terms of $m$ under the additional constraint that every edge lies in a triangle, then combine with Lemma 80.1 to force a large average common neighbourhood.

(iv) Minimal counterexample structure (to $f_c(n)\gg \log n$): a sequence of graphs $G_n$ with $|E(G_n)|\ge c n^2$, every edge in a triangle, but with $\mathrm{bk}(G_n)=O(\log n)$ (or $n^{o(1)}$). Such graphs would have triangles "spread out" so that no single edge has many common neighbours, suggesting a highly pseudorandom local common-neighbour structure together with a global density constraint.

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Math-664-A2 · No claim recorded · LLM-generated · gpt 5.2 ·

gpt 5.2 attempt on Math-664

OUTPUT 1) FORMAL RESTATEMENT Definitions/Conventions. - For a finite simple graph $G=(V,E)$, write $n:=|V|$ and $m:=|E|$. - For an edge $e=uv\in E$, define the book-size at $e$ as \[b_G(uv):=\big|\{w\in V\setminus\{u,v\}: uw\in E,\ vw\in E\}\big|,\] the number of triangles containing the edge $uv$. - The book size of $G$ is \[B(G):=\max_{uv\in E} b_G(uv).\]

For a fixed constant $c>0$, define $f_c(n)$ to be the largest integer $m$ such that every $n$-vertex graph $G$ with 1. $|E(G)|\ge c n^2$, and 2. every edge of $G$ lies in at least one triangle (equivalently $b_G(uv)\ge 1$ for all $uv\in E$), satisfies $B(G)\ge m$. Equivalently, $f_c(n)=\min\{B(G): G \text{ satisfies (1) and (2)}\}$.

Question. Estimate the growth of $f_c(n)$ as $n\to\infty$ (for fixed $c$). In particular, for $c<1/4$ must $f_c(n)$ still grow (e.g. $\gg\log n$)?

2) QUICK LITERATURE/CONTEXT CHECK Web browsing is not available in this session. From the problem text only: - For $c<1/4$, Alon--Trotter gave an upper bound $f_c(n)\ll_c n^{1/2}$. - Szemerédi observed $f_c(n)\to\infty$ via regularity lemma. - For $c>1/4$, a linear lower bound $f_c(n)\ge n/6$ is stated. - For $c<1/4$, Fox--Loh proved a subpolynomial upper bound $f_c(n)\le n^{O(1/\log\log n)}$. I do not re-prove these results here.

3) ATTACK PLAN Proof track ideas: 1. Relate $B(G)$ to degree information via common-neighbor counts. For $c>1/4$ one expects linear books from "supersaturation" of triangles. 2. For $c<1/4$, leverage the extra hypothesis that every edge lies in a triangle (a global covering constraint) to force some growth of $B(G)$.

Disproof track ideas: 1. For $c<1/4$, try to build dense graphs where edges lie in triangles but each edge is in very few triangles, i.e. $B(G)$ grows slowly.

Chosen path in this attempt: derive two elementary lemmas that already imply a linear lower bound when $c>1/4$, and give the basic book/common-neighbor identities.

4) WORK Lemma 1 (Book size equals maximum edge codegree). For an edge $uv\in E(G)$, \[b_G(uv)=|N(u)\cap N(v)|,\] where $N(u)$ is the neighbor set of $u$. In particular, $B(G)=\max_{uv\in E(G)} |N(u)\cap N(v)|$.

Proof. A vertex $w$ forms a triangle $u-v-w-u$ with edge $uv$ iff $w$ is adjacent to both $u$ and $v$, i.e. $w\in N(u)\cap N(v)$. Counting such $w$ gives the identity. $\square$

Lemma 2 (Degree-sum averaging over edges). Let $G$ be a graph with degrees $d(v)$ and $m$ edges. Then \[\frac{1}{m}\sum_{uv\in E(G)}\big(d(u)+d(v)\big) = \frac{1}{m}\sum_{v\in V(G)} d(v)^2 \ \ge\ \frac{4m}{n}.\] Consequently there exists an edge $uv\in E(G)$ with \[d(u)+d(v)\ge \frac{4m}{n}.\]

Proof. For the first identity, each vertex $v$ contributes $d(v)$ to the sum $\sum_{uv\in E}(d(u)+d(v))$ for each incident edge, i.e. $d(v)$ times, so the total contribution is $d(v)^2$. Summing over vertices gives $\sum_{uv\in E}(d(u)+d(v))=\sum_v d(v)^2$.

For the inequality, apply Cauchy--Schwarz: \[\sum_v d(v)^2 \ge \frac{1}{n}\left(\sum_v d(v)\right)^2=\frac{(2m)^2}{n}=\frac{4m^2}{n}.\] Divide by $m$ to obtain the displayed bound. The existence of an edge with degree-sum at least the average is immediate. $\square$

Lemma 3 (From large degree-sum to a large book). For any edge $uv\in E(G)$, \[|N(u)\cap N(v)| \ge d(u)+d(v)-(n-2).\] In particular, if $d(u)+d(v)\ge n-1$ then $uv$ lies in at least one triangle.

Proof. We have \[|N(u)\cap N(v)| = |N(u)|+|N(v)|-|N(u)\cup N(v)|=d(u)+d(v)-|N(u)\cup N(v)|.\] Since $N(u)\cup N(v)\subseteq V\setminus\{u,v\}$, its size is at most $n-2$. Substituting gives the lower bound. $\square$

Corollary 4 (A clean linear lower bound when $m>n^2/4$). If $G$ has $n$ vertices and $m$ edges, then \[B(G)\ge \max\!\left(0,\frac{4m}{n}-(n-2)\right).\] In particular, if $m\ge c n^2$ with $c>1/4$, then for all sufficiently large $n$, \[B(G)\ge (4c-1)n.\]

Proof. By Lemma 2, there exists an edge $uv$ with $d(u)+d(v)\ge 4m/n$. Then Lemma 3 gives \[b_G(uv)=|N(u)\cap N(v)|\ge \frac{4m}{n}-(n-2).\] Taking the maximum over edges yields the first inequality, and the second follows by substituting $m\ge c n^2$ and absorbing the $+2$ into $(4c-1)n$ for $n$ large. $\square$

FAST REALITY CHECK - Corollary 4 shows that for $c>1/4$ one gets an elementary linear book-size lower bound (without using the extra hypothesis that every edge lies in a triangle). - For $c<1/4$, Corollary 4 becomes vacuous, so the real difficulty is to use the hypothesis that every edge lies in a triangle to force $B(G)\to\infty$.

5) VERIFICATION - Lemma 2 identity $\sum_{uv\in E}(d(u)+d(v))=\sum_v d(v)^2$ holds by grouping contributions by endpoint. - Lemma 3 is a union bound on neighborhoods, with the only subtlety being $|N(u)\cup N(v)|\le n-2$. - Corollary 4 is a clean pipeline: average degree-sum $\Rightarrow$ large degree-sum edge $\Rightarrow$ large intersection.

6) FINAL UNRESOLVED

(i) Strongest fully proved partial result: An elementary argument shows that any graph with $m$ edges satisfies \[B(G)\ge \max\!\left(0,\frac{4m}{n}-(n-2)\right)\] and hence if $m\ge c n^2$ with $c>1/4$ then $B(G)\ge (4c-1)n$ for large $n$ (Corollary 4).

(ii) First gap: Use the additional constraint "every edge lies in a triangle" to prove any meaningful lower bound on $B(G)$ when $c<1/4$ (e.g. $B(G)\to\infty$ or $B(G)\gg\log n$).

(iii) Top 3 next moves (concrete): 1. Translate the constraint "every edge lies in a triangle" into a lower bound on the average edge-codegree $\frac{1}{m}\sum_{uv\in E}|N(u)\cap N(v)|$ that is stronger than $1$ when $m=cn^2$. 2. Try to prove that if all edge-codegrees are $\le B$, then the edge set must be close to bipartite (few triangles), contradicting that every edge is in a triangle. 3. Implement a small optimization search (for fixed $n$ and edge density $c$) to minimize $B(G)$ under the "all edges in triangles" constraint, to guess the true growth for representative $c<1/4$.

(iv) What a minimal counterexample would likely look like (structure/parameters): For some fixed $c<1/4$, a family of graphs $G_n$ with $|E|=\Theta(n^2)$ where every edge lies in at least one triangle, but every edge lies in at most $n^{o(1)}$ triangles, i.e. the graph locally has triangles everywhere but no edge participates in many.

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