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Math-5387-A5 · revision r1 · on Math-5387

gpt pro 5.4 attempt on Math-5387

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Unverified LLM-generated attempt. Automated screening checks form, honesty and scope, not correctness. Status at erdosproblems.com: solved. Claims here do not change the source status.

Manuscript

Contents
  1. Erdős Problem 42 --- Round 4 (Continuation & Gap-Closure)
  2. 1) ROUND-4 OBJECTIVE
  3. 2) Round-3 FOUNDATION USED
  4. 3) NEW INSIGHT / TOOL (ROUND-4)
  5. 4) ATTACK PLAN (ROUND-4)
  6. 5) WORK (ROUND-4)
  7. 5.1. A general obstruction-interval lemma
  8. 5.2. Correction of the Round 3 interval
  9. 5.3. A new explicit obstruction interval for $M=3$
  10. 6) ADVERSARIAL VERIFICATION
  11. 7) FINAL (EXACTLY ONE)
  12. 8) COMPLETION ESTIMATE (MANDATORY)
  13. 9) REFERENCES
  14. References

Erdős Problem 42 --- Round 4 (Continuation & Gap-Closure)

1) ROUND-4 OBJECTIVE

Path (C) (obstruction/correction). Round 3 left two precise issues that can be closed rigorously without restarting:
  • the Round 3 interval consequence from the explicit $M=3$ obstruction was off by one, and
  • the obstruction mechanism itself can be upgraded from isolated values of $N$ to a general interval-obstruction lemma.

Using that lemma, we both correct the Round 3 interval and produce a genuinely stronger explicit obstruction for $M=3$, showing failure for every \[ 86\le N\le 96. \] Consequently, if an eventual threshold $N_0(3)$ exists, then necessarily \[ N_0(3)\ge 97. \]

2) Round-3 FOUNDATION USED

We use the following earlier results as black boxes.
  • Round 1 Lemma 42.2 (small-gap lemma). Any $M$-element set $B\subseteq[1,N]$ has a positive difference at most $\lfloor (N-1)/(M-1)\rfloor$.
  • Round 2 theorem (case $M=2$). The case $M=2$ is completely solved with sharp threshold $N_0(2)=8$.
  • Round 3 Proposition $M=3, N=71$. The set \[ A_{10}:=\{1,2,7,11,24,27,35,42,54,56\} \] is Sidon and satisfies $[1,35]\subseteq D^+(A_{10})$, where $D^+(A):=\{a-a':a,a'\in A,\ a>a'\}$.

3) NEW INSIGHT / TOOL (ROUND-4)

The new ingredient is a general interval-obstruction lemma: once one has a Sidon set $A$ whose positive differences contain an initial interval $[1,L]$, the same set $A$ obstructs not just one value of $N$, but an entire interval of values of $N$.

A second genuinely new ingredient is an explicit 12-element Sidon set \[ A_{12}=\{1,3,7,25,30,41,44,56,69,76,77,86\} \] whose positive differences contain $[1,47]$. This yields a new obstruction interval \[ 86\le N\le 96, \] strictly beyond Round 3.

4) ATTACK PLAN (ROUND-4)

The exact unresolved gap after Round 3 was this: the proof produced a single certified obstruction at $N=71$, together with a stated interval consequence, but that interval consequence was not optimized and in fact missed one endpoint.

We therefore proceed as follows.

  1. Prove a general lemma converting one initial-segment difference obstruction into an interval of forbidden $N$-values.
  2. Apply it to the Round 3 set $A_{10}$, thereby correcting the Round 3 interval.
  3. Exhibit a new explicit Sidon set $A_{12}$ with a longer initial segment of covered differences, and apply the lemma again.

This overcomes the Round 3 limitation by turning a sporadic obstruction into a theorem about entire intervals.

5) WORK (ROUND-4)

5.1. A general obstruction-interval lemma

Lemma (Interval-obstruction lemma): Let $M\ge 2$. Let $A\subseteq[1,N_{\min}]$ be Sidon, and suppose that for some integer $L\ge 1$, \[ [1,L]\subseteq D^+(A). \] Then for every integer $N$ satisfying \[ N_{\min}\le N\le (M-1)L+M-1, \] there is no Sidon set $B\subseteq[1,N]$ of size $M$ such that \[ (A-A)\cap(B-B)=\{0\}. \]
Proof: Fix such an $N$, and let $B\subseteq[1,N]$ with $|B|=M$. By Round 1 Lemma 42.2, the set $B$ has a positive difference \[ d\in(B-B)\cap\left[1,\lfloor\frac{N-1}{M-1}\rfloor\right]. \] Since $N\le (M-1)L+M-1$, we have \[ \lfloor\frac{N-1}{M-1}\rfloor\le L. \] Therefore $d\in[1,L]\subseteq D^+(A)\subseteq A-A$. Hence $d\in (A-A)\cap(B-B)$ with $d\ne 0$, so \[ (A-A)\cap(B-B)\neq \{0\}. \] Thus no such $B$ exists. ∎

5.2. Correction of the Round 3 interval

Round 3 proved that \[ A_{10}=\{1,2,7,11,24,27,35,42,54,56\} \] is Sidon and satisfies $[1,35]\subseteq D^+(A_{10})$. Since $\max A_{10}=56$, Lemma Lemma (Interval-obstruction lemma) with $M=3$, $N_{\min}=56$, and $L=35$ gives: \[ 56\le N\le 2\cdot 35+2=72 \quad\Longrightarrow\quad \text{the }M=3\text{ statement fails at }N. \] So the same Round 3 construction obstructs every \[ 56\le N\le 72. \] This corrects the Round 3 endpoint omission: $N=72$ also fails.

5.3. A new explicit obstruction interval for $M=3$

We now prove a stronger obstruction.
Proposition (New explicit obstruction interval for $M=3$): Let \[ A_{12}:=\{1,3,7,25,30,41,44,56,69,76,77,86\}\subseteq[1,86]. \] Then $A_{12}$ is Sidon and \[ [1,47]\subseteq D^+(A_{12}). \] Consequently, for every integer $N$ with \[ 86\le N\le 96, \] there is no Sidon triple $B\subseteq[1,N]$ such that \[ (A_{12}-A_{12})\cap(B-B)=\{0\}. \] In particular, the Erdős statement fails for $M=3$ at every $N\in[86,96]$.
Proof: Translate by $-1$, and set \[ R:=A_{12}-1=\{0,2,6,24,29,40,43,55,68,75,76,85\}. \] Translation preserves all differences, so it suffices to compute $D^+(R)=D^+(A_{12})$. A direct enumeration gives \[ \begin{aligned} D^+(R)=\{&1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,\\ &49,51,52,53,55,56,61,62,66,68,69,70,73,74,75,76,79,83,85\}. \end{aligned} \] This displayed set contains $47+19=66$ integers, which equals \[ \binom{12}{2}=66. \] Hence all positive differences determined by the 12 marks of $R$ are distinct. For completeness, here is the short implication: if $a_i+a_j=a_k+a_\ell$ with the two pairs written in nondecreasing order and, say, $a_j\ge a_\ell$, then \[ a_j-a_\ell=a_k-a_i. \] Uniqueness of positive differences forces the unordered pairs $(a_i,a_j)$ and $(a_k,a_\ell)$ to coincide, so the sum representation is trivial. Thus $A_{12}$ is Sidon. Moreover, the displayed list visibly contains every integer from $1$ to $47$, i.e. \[ [1,47]\subseteq D^+(A_{12}). \] Finally, $\max A_{12}=86$. Applying Lemma Lemma (Interval-obstruction lemma) with $M=3$, $N_{\min}=86$, and $L=47$, we conclude that every \[ N\in[86,2\cdot 47+2]=[86,96] \] is obstructed by the same set $A_{12}$. ∎
Corollary (New lower bound on any eventual threshold for $M=3$): If there exists a threshold $N_0(3)$ such that for every $N\ge N_0(3)$ and every Sidon set $A\subseteq[1,N]$ there exists a Sidon triple $B\subseteq[1,N]$ with $(A-A)\cap(B-B)=\{0\}$, then necessarily \[ N_0(3)\ge 97. \]
Proof: By Proposition prop:new-interval-86-96, the statement fails for every $N\in[86,96]$. Hence no valid threshold can be less than or equal to $96$. ∎

6) ADVERSARIAL VERIFICATION

  • Endpoint check in Lemma Lemma (Interval-obstruction lemma). For $M=3$ the upper bound becomes $N\le 2L+2$. At the endpoint $N=2L+2$, one has $\lfloor (N-1)/2\rfloor=\lfloor (2L+1)/2\rfloor=L$, so the lemma still applies.

  • Round 3 correction. With $L=35$, the correct upper endpoint is $2L+2=72$, not $71$. There is no hidden obstruction at $N=72$: the same argument works verbatim because $\lfloor 71/2\rfloor=35$.

  • Sidon verification for $A_{12}$. The displayed difference set has exactly $66$ elements, matching $\binom{12}{2}$, so no positive difference repeats. This is enough for the Sidon property.

  • Sharpness of the new interval. The new obstruction stops at $N=96$ because $48\notin D^+(A_{12})$. Thus the present argument does not extend to $N=97$.

  • Interaction with earlier rounds. Nothing in this round affects the complete $M=2$ solution from Round 2. For $M=3$, the new results strictly strengthen Round 3 by replacing one certified value $N=71$ by two certified obstruction intervals, namely $[56,72]$ and $[86,96]$.

7) FINAL (EXACTLY ONE)

UNRESOLVED (BUT STRICTLY ADVANCED).

We proved a new general interval-obstruction lemma, corrected the Round 3 endpoint error by upgrading the old obstruction to all \[ 56\le N\le 72, \] and produced a new explicit 12-element Sidon obstruction showing failure for every \[ 86\le N\le 96 \] when $M=3$. In particular, any eventual threshold for the $M=3$ case must satisfy $N_0(3)\ge 97$.

8) COMPLETION ESTIMATE (MANDATORY)

COMPLETION: 55%

9) REFERENCES

References

  • [Bloom42] T. F. Bloom, Erdős Problem #42, https://www.erdosproblems.com/42, accessed 2026-03-07. (Checked here to confirm the problem numbering and the current public status note for the cases $M=2$ and $M=3$.)
  • [Gusarich42] D. Sedov (Gusarich), Erdős Problem #42 --- The $M=3$ Case, public Lean repository README, https://github.com/Gusarich/erdos42, checked 2026-03-07. (Used only as contextual background on the current state of the $M=3$ case; the new theorem of this round does not rely on its proof.)