gpt pro 5.4 attempt on Math-726

Erdős Problem #241

1) ROUND-2 OBJECTIVE

I am pursuing (C) obstruction/correction. The Round-1 file contained a genuine gap: it produced the Bose--Chowla lower bound only on the subsequence $N=q^3-1$ and then informally suggested a passage to all $N$ without proving it. The most promising Round-2 task is therefore to repair that gap and globalize the lower bound to all large $N$.

2) Round-1 FOUNDATION USED

I use the following Round-1 results without re-proving them.

1. The counting upper bound: if $A\subseteq\{1,\dots,N\}$ is $B_3$-distinct and $|A|=m$, then \[ \frac{(m+2)(m+1)m}{6}\le 3N. \] In particular $f(N)\le (18N)^{1/3}+O(1)$.
2. The Bose--Chowla construction for $h=3$: for every prime power $q$ there exists a $B_3$-distinct set of size $q$ inside $\{1,\dots,q^3-1\}$. Equivalently, \[ f(q^3-1)\ge q. \]
3. Monotonicity of $f$: if $M\le N$, then $f(M)\le f(N)$.

3) NEW INSIGHT / TOOL (ROUND-2)

The new ingredient is a rigorous globalization step: combine the prime-power construction with the prime number theorem to find, for every large $N$, a prime $p\le (N+1)^{1/3}$ with $p=(1-o(1))(N+1)^{1/3}$. Monotonicity then lifts the Bose--Chowla construction from $p^3-1$ up to $N$.

4) ATTACK PLAN (ROUND-2)

The exact Round-1 gap was this: the lower bound $f(q^3-1)\ge q$ was proved only on a sparse subsequence, whereas the target asymptotic concerns every $N$. To close that gap, I will:

1. make monotonicity explicit;
2. choose a prime $p$ just below $(N+1)^{1/3}$;
3. embed the Bose--Chowla set at scale $p$ into $\{1,\dots,N\}$;
4. conclude the sharp lower-bound half \[ f(N)\ge (1-o(1))N^{1/3} \] for all large $N$.

This repairs the Round-1 overclaim and isolates the only remaining gap: the sharp upper constant.

5) WORK (ROUND-2)

Proposition 5.1 (monotonicity).

If $M\le N$, then $f(M)\le f(N)$.

Proof. Any $B_3$-distinct set $A\subseteq\{1,\dots,M\}$ is also a subset of $\{1,\dots,N\}$, so the maximal size over $\{1,\dots,N\}$ is at least the maximal size over $\{1,\dots,M\}$.

Theorem 5.2 (global Bose--Chowla lower bound).

For every $\varepsilon>0$ there exists $N_0(\varepsilon)$ such that for all $N\ge N_0(\varepsilon)$, \[ f(N)\ge (1-\varepsilon)N^{1/3}. \] Equivalently, \[ f(N)\ge (1-o(1))N^{1/3} \qquad (N\to\infty). \]

Proof. Fix $\varepsilon>0$, and let \[ x:=(N+1)^{1/3}. \] By the prime number theorem, for all sufficiently large $x$ there exists a prime $p$ with \[ (1-\varepsilon)x\le p\le x. \] For such a prime $p$, Round-1 gives a $B_3$-distinct set of size $p$ inside $\{1,\dots,p^3-1\}$, so \[ f(p^3-1)\ge p. \] Since $p\le x=(N+1)^{1/3}$, we have $p^3-1\le N$. Hence, by monotonicity, \[ f(N)\ge f(p^3-1)\ge p\ge (1-\varepsilon)x>(1-\varepsilon)N^{1/3}. \] This proves the claim.

Corollary 5.3 (the corrected Round-1 theorem).

The lower-bound half of the conjectured asymptotic is true: \[ \liminf_{N\to\infty}\frac{f(N)}{N^{1/3}}\ge 1. \] Together with the literature upper bound recorded in Round-1, \[ f(N)\le \left(\left(\frac{7}{2}\right)^{1/3}+o(1)\right)N^{1/3}, \] we obtain the rigorous bracket \[ 1\le \liminf_{N\to\infty}\frac{f(N)}{N^{1/3}} \le \limsup_{N\to\infty}\frac{f(N)}{N^{1/3}} \le \left(\frac{7}{2}\right)^{1/3}. \] Thus the only unresolved issue is the sharp upper constant.

6) ADVERSARIAL VERIFICATION

1. Was the Round-1 gap real? Yes. A lower bound on the subsequence $N=q^3-1$ does not by itself imply a lower bound for every $N$. The new proof fixes exactly that point.
2. Does the Bose--Chowla set really live in $\{1,\dots,p^3-1\}$? Round-1 produced it in $\{0,\dots,p^3-2\}$; translating by $+1$ preserves triple-sum uniqueness because every triple sum increases by exactly $3$.
3. Is the prime-selection step strong enough? Yes. Any asymptotic prime existence statement of the form "there is a prime in $[(1-\varepsilon)x,x]$ for all large $x$" is sufficient, and this is a standard consequence of the prime number theorem.
4. Any hidden use of the conjecture? No. The argument uses only Round-1's finite-field construction, monotonicity, and primes near $x$.

7) FINAL

UNRESOLVED (BUT STRICTLY ADVANCED).

The Round-1 lower-bound gap has now been closed rigorously: one really does have \[ f(N)\ge (1-o(1))N^{1/3} \] for all large $N$, not merely along a subsequence. So the unresolved content of Problem #241 is now completely concentrated in the upper bound, namely whether \[ f(N)\le (1+o(1))N^{1/3}. \]

8) COMPLETION ESTIMATE

COMPLETION: 55%

9) REFERENCES

1. T. F. Bloom, Erdős Problem #241, current version page on erdosproblems.com, accessed 2026-03-07.
2. Prime number theorem (standard).