gpt pro 5.4 attempt on Math-811

ROUND-2 OBJECTIVE

We pursue path (C) obstruction/correction. The strongest Round-1 progress isolated the main unresolved issue as a stability problem for deletion-based constructions. In this round I sharpen that point and prove that the currently proposed dyadic deletion argument is not valid as written: there are two independent fatal obstructions.

1. The "capacity-constrained hitting set" lemma used in the January 16, 2026 draft is false.
2. Even if one deletes a hitting set for all currently visible collisions at a dyadic stage $[X,2X)$, that deletion can create new collisions whose maximal value still lies in $[X,2X)$.

Thus the current continuation strategy does not yet yield a proof of Erdős problem #421. The natural corrected statement is a stronger dyadic implication theorem, proved below, which isolates exactly what a future construction must establish.

Round-1 FOUNDATION USED

I use the following Round-1 results without reproving them.

1. Collision normal form. Any collision can be rewritten as an equality of products of two disjoint consecutive blocks \[ \prod_{i=M-\ell+1}^{M} d_i = \prod_{i=N}^{N+k} d_i, \qquad M<N, \] after cancelling overlap.

2. Deletion is non-monotone. Round-1 already proved that deleting terms can create new collisions.

3. Analytic input is not the current bottleneck. Round-1 verified a uniform polylogarithmic point-count for the non-line separated-variable curves in the logarithmic-degree regime, and also proved that affine line components are harmless in the collision region $x<y$, $x,y>0$.

The new work below strengthens item 2 in a way that directly breaks the dyadic stage argument, and separates the true combinatorial obstruction from the already-treated Diophantine point-counting input.

NEW INSIGHT / TOOL (ROUND-2)

The new ingredient is a precise decomposition of the stability issue.

1. Future-scale deletions are harmless. Once a dyadic stage has produced a set with no collisions whose values are $<2X$, later deletions above $2X$ cannot change that fact.

2. Same-scale deletions are dangerous. A deletion inside $[X,2X)$ can create a brand-new collision whose maximal value is still in $[X,2X)$. Hence a one-shot "hit all current collisions in $[X,2X)$" rule is insufficient.

3. The hitting-set lemma used in the draft is false. The lemma is false as a pure combinatorial statement, and the probabilistic proof also contains two separate errors.

This strictly advances the Round-1 analysis: the gap is no longer a vague "prefix-stability problem", but a concrete obstruction with explicit counterexamples and a corrected conditional theorem.

ATTACK PLAN (ROUND-2)

The Round-1 gap was:

Give a rigorous stagewise construction showing that deleting a sparse set in each dyadic interval both preserves local deficiency and eliminates all collisions permanently.

The exact subclaims that must be checked are now:

1. $G1$ Does there exist a low-capacity hitting set $B_X\subset [X,2X)$ for the relevant collision family?
2. $G2$ If such a set is chosen from the current collision family, does deleting it actually guarantee that no collisions with maximal value in $[X,2X)$ remain afterwards?

I prove:

1. a positive lemma showing that later scales are harmless;
2. an explicit same-scale counterexample disproving the naive stagewise stability needed for $G2$;
3. an explicit combinatorial counterexample disproving the hitting-set lemma used for $G1$;
4. a corrected dyadic implication theorem that states the minimal property a future construction would need.

WORK (ROUND-2)

Future-scale deletions do not affect earlier collisions

Lemma 1 (future-scale immutability). Let $A\subset \mathbb N$ be infinite, let $X\ge 1$, and let $B\subset [2X,\infty)$. Let \[ a_1<a_2<\cdots \qquad\text{and}\qquad a'_1<a'_2<\cdots \] be the increasing enumerations of $A$ and $A\setminus B$, respectively. Then the elements of $A$ below $2X$ and the elements of $A\setminus B$ below $2X$ coincide in the same order. Consequently every consecutive block product involving only values $<2X$ is identical in the two sequences. In particular, the existence or nonexistence of collisions all of whose values are $<2X$ is unchanged by deleting $B$.

Proof. Since every element of $B$ is at least $2X$, deleting $B$ removes no element of $A\cap [1,2X)$. Therefore \[ A\cap [1,2X)=(A\setminus B)\cap [1,2X) \] as sets. Because both sequences are increasing enumerations, their values below $2X$ appear in exactly the same order. Hence if \[ a_i,a_{i+1},\dots,a_j < 2X, \] then these same values form a consecutive block in the enumeration of $A\setminus B$ as well, with the same product, and conversely. Thus every consecutive block product formed entirely from values $<2X$ is the same before and after deleting $B$.

Therefore a collision whose participating values are all $<2X$ exists in the sequence enumerating $A$ if and only if it exists in the sequence enumerating $A\setminus B$. $\square$

This lemma shows that the only genuine stability problem is within the current dyadic scale. Later dyadic stages cannot revive a collision already eliminated below $2X$.

A same-scale instability counterexample

The next proposition strengthens Round-1's non-monotonicity example. It shows that deleting inside one dyadic block can create a new collision whose maximal value remains in that very same dyadic block.

Proposition 2 (same-scale instability). There exists a finite strictly increasing sequence with pairwise distinct consecutive-block products such that deleting a single term $b\in [X,2X)$ creates a new collision whose maximal value also lies in $[X,2X)$.

Proof. Take \[ A=(2,8,9,10,12,15). \] Its $\frac{6\cdot 7}{2}=21$ consecutive-block products are: \[ \begin{array}{llllll} 2, & 8, & 9, & 10, & 12, & 15,\\ 16, & 72, & 90, & 120, & 180,\\ 144, & 720, & 1080, & 1800,\\ 1440, & 8640, & 16200,\\ 17280, & 129600,\\ 259200. \end{array} \] When listed in increasing order these are \[ 2<8<9<10<12<15<16<72<90<120<144<180<720<1080<1440<1800<8640<16200<17280<129600<259200, \] so they are pairwise distinct. Hence $A$ has the distinct consecutive products property.

Now delete the element $8$. The resulting sequence is \[ A'=(2,9,10,12,15). \] In $A'$ we have the collision \[ 2\cdot 9\cdot 10 = 180 = 12\cdot 15. \] So the intervals $[1,3]$ and $[4,5]$ in $A'$ have equal products.

The deleted value is $8\in [8,16)$, and the maximal value occurring in the new collision is $15\in [8,16)$. Thus the new collision lies in the same dyadic scale as the deletion. $\square$

Consequence. A dyadic algorithm cannot justify stage $X$ by saying "choose $B_X$ to hit all collisions currently visible with maximal value in $[X,2X)$" and then conclude that no such collisions remain after deleting $B_X$. Proposition 2 gives a direct counterexample to that inference.

The capacity-constrained hitting-set lemma in the 2026 draft is false

The January 16, 2026 draft introduces the following statement at dyadic scale $X$:

If $\mathcal E_X$ is a family of subsets $V(E)\subset [X,2X)$ with $|V(E)|\le L(X)$ and $|\mathcal E_X|=o(X)$, then there exists a hitting set $B_X\subset [X,2X)$ such that every interval $I\subset [X,2X)$ of length $L(X)$ satisfies $|B_X\cap I|\le T(X)/2$.

Here $L(X)=\lceil C_0\log X\rceil$ and $T(X)=\lceil (\log X)^{2/3}\rceil$.

Proposition 3 (the hitting-set lemma is false). For all sufficiently large $X$, the above statement is false.

Proof. Fix a large $X$ and an interval $I\subset [X,2X)$ of length exactly $L(X)$. Since \[ L(X)=\lceil C_0\log X\rceil \quad\text{and}\quad T(X)=\lceil (\log X)^{2/3}\rceil, \] we have $L(X)\ge T(X)$ for all large $X$. Choose $T(X)$ distinct points \[ v_1,\dots,v_{T(X)}\in I. \] Define a family $\mathcal E_X$ by taking $T(X)$ singleton vertex sets: \[ V(E_r):=\{v_r\},\qquad r=1,\dots,T(X). \] Then every $V(E_r)$ has size $1\le L(X)$, and \[ |\mathcal E_X|=T(X)=O((\log X)^{2/3})=o(X). \] So the hypotheses of the stated lemma are satisfied.

But any hitting set $B_X$ for $\mathcal E_X$ must meet every singleton $\{v_r\}$, hence must contain all the points $v_1,\dots,v_{T(X)}$. Therefore \[ |B_X\cap I|\ge T(X)>T(X)/2, \] which violates the required capacity bound.

Thus the lemma is false. $\square$

Two independent errors in the probabilistic proof of that lemma

The same draft gives a probabilistic proof by choosing one random point $v(E)\in V(E)$ for each $E\in \mathcal E_X$. There are two separate problems.

Proposition 4 (the probability estimate used in the draft is false). In the notation above, the draft's estimate \[ \mathbb P\left(v(E)\in I\right)\le \frac{|I|}{X}=\frac{L(X)}{X} \] is false in general.

Proof. Take the family from Proposition 3. For each $r$ one has $V(E_r)=\{v_r\}\subset I$. Since $v(E_r)$ is chosen uniformly from the singleton $V(E_r)$, it is forced that \[ \mathbb P\left(v(E_r)\in I\right)=1. \] But $L(X)/X<1$ for every sufficiently large $X$. Hence the claimed upper bound fails. $\square$

Proposition 5 (the union-bound asymptotic in the draft is false). Even if one had a tail estimate of the form \[ \mathbb P\left(|B_X\cap I|>T(X)/2\right)\le e^{-cT(X)} \] for each interval $I$ of length $L(X)$, the subsequent claim \[ 2X e^{-cT(X)}\to 0 \qquad (X\to\infty) \] with $T(X)=(\log X)^{2/3}$ is false.

Proof. We have \[ 2X e^{-cT(X)}= \exp\!\left(\log(2X)-c(\log X)^{2/3}\right). \] As $X\to\infty$, the exponent tends to $+\infty$ because $\log X$ dominates $(\log X)^{2/3}$. Therefore \[ 2X e^{-cT(X)}\to \infty, \] not to $0$. $\square$

So the hitting-set step fails at three levels:

1. the combinatorial statement is false,
2. the probability estimate used to prove it is false,
3. the final union-bound asymptotic is false.

Minimal corrected dyadic implication

The previous propositions do not disprove the original Erdős problem. They show that the current stagewise deletion argument is incomplete. The right correction is to formulate exactly what a dyadic construction would actually need to prove.

Theorem 6 (corrected dyadic implication). Let $L(X)$ and $T(X)$ be positive functions on dyadic integers $X=2^t$. Suppose one can choose sets \[ B_X\subset [X,2X)\cap \mathbb Z \qquad (X=2,4,8,\dots) \] such that:

1. (i) for every interval $I\subset [X,2X)$ of length $L(X)$, \[ |B_X\cap I|\le T(X)/2; \]

2. (ii) if \[ A_{\le X}:=\mathbb N\setminus \bigcup_{Y\le X} B_Y, \] then in the increasing enumeration of $A_{\le X}$ there is no collision all of whose values are $<2X$;

3. (iii) $|B_X|=o(X)$ as $X\to\infty$ through dyadic values.

Then the final set \[ A:=\mathbb N\setminus \bigcup_X B_X \] has asymptotic density $1$, and the increasing enumeration of $A$ has pairwise distinct consecutive-block products.

Proof. First we prove density $1$. Write $X_j=2^j$ and $\varepsilon_j:=|B_{X_j}|/X_j$. By hypothesis, $\varepsilon_j\to 0$. Let $N\ge 1$ and choose $m$ with $2^m\le N<2^{m+1}$. Then \[ \#\left(([1,N]\cap \mathbb N)\setminus A\right) \le \sum_{j\le m} |B_{2^j}| =\sum_{j\le m} \varepsilon_j 2^j. \] Fix $\eta>0$. Choose $j_0$ so that $\varepsilon_j\le \eta$ for all $j\ge j_0$. Then \[ \sum_{j\le m} \varepsilon_j 2^j \le \sum_{j<j_0} \varepsilon_j 2^j + \eta \sum_{j=j_0}^{m}2^j \le C(\eta)+2\eta N. \] Dividing by $N$ and letting $N\to\infty$ gives \[ \frac{\#(([1,N]\cap \mathbb N)\setminus A)}{N}\le 2\eta+o(1). \] Since $\eta>0$ is arbitrary, the complement of $A$ has density $0$, so $A$ has density $1$.

Now suppose, for contradiction, that the increasing enumeration of $A$ has a collision. Let $M$ be the maximal value among all entries occurring in the two equal consecutive blocks. Choose dyadic $X$ with \[ X\le M<2X. \] All values appearing in the collision are then $<2X$. By Lemma 1, deleting elements above $2X$ cannot affect any collision all of whose values are $<2X$. Therefore the same collision already exists in the increasing enumeration of $A_{\le X}$. This contradicts property (ii).

Hence no collision exists in $A$. $\square$

Interpretation. Theorem 6 shows exactly what a dyadic construction must establish. The difficult part is not merely counting current collisions. One must choose $B_X$ so that after the deletion is performed, no collision with maximal value in $[X,2X)$ remains. Proposition 2 shows that hitting only the currently visible collisions is not enough.

ADVERSARIAL VERIFICATION

I now try to break the new work.

1. Does Lemma 1 overlook index changes? No. The statement is about collisions whose values are below $2X$. Deleting only values $\ge 2X$ leaves the list of surviving values below $2X$ unchanged and in the same order, so every consecutive block among those values is unchanged.

2. Could Proposition 2 fail because I missed a duplicate in the original sequence? No. All $21$ consecutive-block products were written explicitly, and when sorted they form a strictly increasing list. Therefore the original sequence has no collision.

3. Could the new collision in Proposition 2 already have existed before deletion in some disguised form? No. The original sequence had no collision at all. After deletion, the equality $2\cdot 9\cdot 10=12\cdot 15$ is a genuinely new equality of consecutive-block products.

4. Does Proposition 3 use a family of vertex sets too special to matter? It does not need to model actual collisions. The draft's lemma was stated for all families of subsets $V(E)\subset [X,2X)$ with $|V(E)|\le L(X)$ and $|\mathcal E_X|=o(X)$. A single counterexample therefore disproves the lemma exactly as stated.

5. Could Proposition 5 be repaired by changing constants? No. For any fixed $c>0$, \[ \log(2X)-c(\log X)^{2/3}\to +\infty, \] so the claimed limit cannot hold with $T(X)=(\log X)^{2/3}$.

6. Does Theorem 6 smuggle in the conclusion? No. It is a genuine implication theorem. Condition (ii) is strictly stronger than the invalid "hit all currently visible collisions" rule, and Proposition 2 proves that such strengthening is genuinely necessary.

FINAL

UNRESOLVED (BUT STRICTLY ADVANCED)

Round-2 does not solve Erdős problem #421, but it makes a mathematically substantive advance:

1. it proves that the currently proposed dyadic deletion proof is not valid as written;
2. it exhibits an explicit same-scale instability counterexample, showing that deleting a hitting set for the current collision family can create new collisions in the same dyadic block;
3. it disproves the draft's capacity-constrained hitting-set lemma by an explicit combinatorial counterexample;
4. it supplies the minimal corrected dyadic implication theorem (Theorem 6), which isolates the exact combinatorial task still required.

So the analytic point-counting side may be in good shape, but the combinatorial deletion mechanism remains unresolved and is now sharply formulated.

COMPLETION ESTIMATE

COMPLETION: 62%

REFERENCES

1. T. F. Bloom, Erdős Problem #421, problem page and discussion thread, accessed March 7, 2026.
2. P. Chojecki, Density-One Sequences with Distinct Consecutive Products, draft dated January 16, 2026.
3. W. Castryck, R. Cluckers, Ph. Dittmann, and K. H. Nguyen, The dimension growth conjecture, polynomial in the degree and without logarithmic factors, Algebra & Number Theory 14 $2020$, no. 8, 2261--2294.