gpt 5.2 attempt on Math-655

OUTPUT 1) FORMAL RESTATEMENT Definitions/Conventions. - For $N\ge 1$, write $A(N):=|A\cap\{1,\dots,N\}|$. - "Triple sums are all distinct aside from trivial coincidences" is interpreted as the $B_3$ condition: for any $a_1\le a_2\le a_3$ and $b_1\le b_2\le b_3$ in $A$, \[a_1+a_2+a_3=b_1+b_2+b_3 \implies (a_1,a_2,a_3)=(b_1,b_2,b_3).\] Equivalently, the multiset $\{a,b,c\}$ is uniquely determined by its sum $a+b+c$.

Question. If $A\subset\mathbb N$ is infinite and satisfies the above $B_3$ condition, must one have \[\liminf_{N\to\infty} \frac{A(N)}{N^{1/3}}=0?\]

2) QUICK LITERATURE/CONTEXT CHECK Web browsing is not available in this session. I only record what is stated in the problem text: - The $h=2$ statement (Sidon case) is attributed to Erdős. - The $h=4$ case is attributed to Nash, and all even $h$ to Chen. I have not verified these citations here.

3) ATTACK PLAN Proof track ideas: 1. Try to adapt Erdős' $h=2$ argument to $h=3$: show that if $A(N)\ge cN^{1/3}$ for all large $N$, then some "local structure" forces a nontrivial equality of 3-sums. 2. Use a density-increment / block-decomposition approach: compare the number of 3-sums in dyadic intervals and force collisions via pigeonhole across intervals. 3. Rephrase in terms of additive energy of order 3 and attempt to show that a uniformly dense $B_3$ sequence would violate a higher-moment inequality.

Disproof/construction track ideas: 1. Attempt to build an infinite $B_3$ sequence with "almost regular" growth by gluing near-optimal finite $B_3$ sets in disjoint ranges (block constructions), aiming for $A(N)\gg N^{1/3}$ for all large $N$. 2. Test greedy constructions computationally to see whether $A(N)/N^{1/3}$ appears bounded away from $0$ or not.

Chosen path in this attempt: prove two basic structural facts (fully) and do a small sanity-check computation.

4) WORK Lemma 1 ($B_3$ implies Sidon for infinite sets). Let $A\subset\mathbb N$ be infinite and satisfy the $B_3$ condition above. Then $A$ is a Sidon set (a $B_2$ set): if $a,b,c,d\in A$ with $a\le b$, $c\le d$ and $a+b=c+d$, then $\{a,b\}=\{c,d\}$.

Proof. Assume $a+b=c+d$ with $\{a,b\}\ne\{c,d\}$. Since $A$ is infinite, choose $e\in A$ with $e\notin\{a,b,c,d\}$. Then \[a+b+e=c+d+e.\] Apply the $B_3$ property to the triples $(a,b,e)$ and $(c,d,e)$ (after sorting each nondecreasingly). Because $e$ occurs in both triples and $e\notin\{a,b,c,d\}$, equality of the multisets $\{a,b,e\}=\{c,d,e\}$ forces $\{a,b\}=\{c,d\}$, contradicting the assumption. $\square$

Lemma 2 (Counting bound in $[1,N]$). Let $N\ge 1$, and let $S\subseteq\{1,\dots,N\}$ be a finite set such that all sums $x+y+z$ with $x\le y\le z$ in $S$ are distinct. Writing $k:=|S|$, one has \[\binom{k+2}{3}\le 3N-2,\] and hence $k \ll N^{1/3}$ (e.g. $k\le (18N)^{1/3}+2$).

Proof. The number of multisets of size $3$ chosen from a $k$-element set is $\binom{k+2}{3}$, and these correspond exactly to the triples $x\le y\le z$ in $S$. By assumption, all the sums $x+y+z$ for such triples are distinct integers. Moreover, since $1\le x,y,z\le N$, each such sum lies in the interval $\{3,4,\dots,3N\}$, which has size $3N-2$. Therefore the number of such triples is at most $3N-2$, i.e. $\binom{k+2}{3}\le 3N-2$. The displayed explicit bound on $k$ follows from $\binom{k+2}{3}=\frac{k(k+1)(k+2)}{6}\ge \frac{k^3}{6}$. $\square$

FAST REALITY CHECK (local computation). I generated the greedy $B_3$ sequence (start with $\emptyset$ and repeatedly adjoin the smallest integer preserving the $B_3$ condition). First 18 terms: \[1,2,5,14,33,72,125,219,376,573,745,1209,1557,2442,3098,4048,5298,6704.\] Counts and ratios (using this finite prefix as a proxy): \[ \begin{array}{c|c|c} N & A(N) & A(N)/N^{1/3} \\\hline 100 & 6 & 1.29\\ 1000 & 11 & 1.10\\ 10000 & 18 & 0.84 \end{array} \] This suggests the ratio might drift downward in the greedy construction, but this is far from evidence for the liminf statement.

5) VERIFICATION - Lemma 1: the only subtlety is the meaning of "trivial coincidences"; under the $B_3$ multiset-uniqueness formalization, adding a distinct $e$ converts a 2-sum collision into a 3-sum collision, forcing triviality. - Lemma 2: counts unordered-with-repetition triples via $\binom{k+2}{3}$, matching the $x\le y\le z$ convention. - Computation: the greedy algorithm enforces distinct $x\le y\le z$ sums; I did not attempt to optimize performance or reach large $N$.

6) FINAL UNRESOLVED

(i) Strongest fully proved partial result: Any infinite $B_3$ set is Sidon (Lemma 1), and for every $N$ one has the global upper bound $|A\cap[1,N]|\ll N^{1/3}$ (Lemma 2).

(ii) First gap: Show (or refute) that $A(N)\le \varepsilon N^{1/3}$ for infinitely many $N$ for every $\varepsilon>0$. I do not see how to force such "density dips" for $h=3$ using only the elementary counting in Lemma 2.

(iii) Top 3 next moves (concrete): 1. Locate/replicate Erdős' proof for $h=2$ in a form that isolates the mechanism producing $\liminf=0$, then test which steps generalize to $h=3$. 2. Attempt a block argument: compare 3-sums coming from $A\cap(2^j,2^{j+1}]$ across many $j$ and force collisions by counting available sum ranges. 3. Search computationally for explicit constructions with unusually regular growth (e.g. block-glued finite $B_3$ sets) that might keep $A(N)/N^{1/3}$ bounded away from $0$.

(iv) What a minimal counterexample would likely look like (structure/parameters): An infinite $B_3$ sequence whose initial segments are near-extremal in every interval, i.e. there exists $c>0$ with $|A\cap[1,N]|\ge cN^{1/3}$ for all sufficiently large $N$, suggesting a highly structured concatenation of near-optimal finite $B_3$ sets with no "sparse" scales.