gpt pro 5.4 attempt on Math-647, version 5

Round-10 Objective

We pursue (C) obstruction/correction. Round 9 proved that the full difficulty of the problem survives inside an arbitrarily high-density branch: by an affine padding construction, any instance can be embedded into one with \[ \inf_k d(A^{(k)})\ge 1-\varepsilon. \] The exact gap left after Round 9 is whether the remaining problem becomes easier once the finite-stage densities are not merely bounded away from $0$, but are almost constant from the very beginning.

The new goal of this round is to show that this does not simplify the problem. We prove that for every $\varepsilon>0$, the original Erdős Problem 25 is equivalent to its restriction to instances whose stage-density sequence \[ \delta_k:=d(A^{(k)}) \] lies inside an interval of length at most $\varepsilon$, and in fact satisfies \[ 1-\varepsilon<\delta:=\lim_{k\to\infty}\delta_k<1, \qquad 0\le \delta_k-\delta\le \varepsilon\quad (k\ge 1). \] Thus the full difficulty already appears in the branch where the stage densities are uniformly $\varepsilon$-close to their limit and remain arbitrarily close to $1$.

This rules out an entire class of strategies based on "eventual stabilization" or "small total variation" of the stage-density sequence.

Round-9 Foundation Used

We rely on the following vetted Round 9 results.

• Stage-density monotonicity (Round 2). For every instance, \[ \delta_k:=d(A^{(k)}) \] exists, and \[ 1\ge \delta_1\ge \delta_2\ge \cdots\ge 0, \] so $\delta_k\to\delta\in[0,1]$.

• Round 9 affine padding transformation. Given an arbitrary instance $(m_j,b_j)$ and an integer $M\ge 3$, define the transformed instance by \[ n_1:=M,\quad a_1:=0, \qquad n_{j+1}:=Mm_j,\quad a_{j+1}:=Mb_j+1\pmod{Mm_j}. \] If $B$ is the original set and $A_M$ the transformed set, then Round 9 established: \begin{align} A_M &= \{1\} \cup \{n\in\mathbb N:\ n\equiv r\pmod M\ \text{for some }2\le r\le M-1\} \cup \{Mt+1:\ t\in B\}, \tag{R9.1}\\ d(A_M^{(k+1)}) &= \frac{M-2}{M}+\frac1M\,d(B^{(k)}), \tag{R9.2}\\ \underline{\delta}_{\log}(A_M) &= \frac{M-2}{M}+\frac1M\underline{\delta}_{\log}(B), \tag{R9.3}\\ \overline{\delta}_{\log}(A_M) &= \frac{M-2}{M}+\frac1M\overline{\delta}_{\log}(B). \tag{R9.4} \end{align} In particular, $A_M$ has a logarithmic density if and only if $B$ does.

New Insight / Tool (Round-10)

The new observation is that the Round 9 padding transformation does more than force high density: it compresses the entire stage-density sequence by the factor $1/M$.

If \[ \beta_k:=d(B^{(k)}), \qquad \beta_0:=1, \] and \[ \alpha_k:=d(A_M^{(k)}), \qquad \alpha:=\lim_{k\to\infty}\alpha_k, \] then Round 9 gives the exact affine formula \[ \alpha_k=\frac{M-2}{M}+\frac{\beta_{k-1}}{M} \qquad (k\ge 1), \] so \[ \alpha=\frac{M-2}{M}+\frac{\beta}{M}, \qquad \beta:=\lim_{k\to\infty}\beta_k. \] Therefore, \[ 0\le \alpha_k-\alpha=\frac{\beta_{k-1}-\beta}{M}\le \frac1M \qquad (k\ge 1). \] Hence the transformed instance has:

• stage densities confined to an interval of width at most $1/M$,
• total variation at most $1/M$,
• limit lying in the interval $[1-2/M,1-1/M]$.

So the whole problem can be embedded into instances whose stage-density sequence is uniformly arbitrarily close to a limit arbitrarily close to $1$.

Attack Plan (Round-10)

The exact gap after Round 9 was this: although the full problem reduces to a high-density branch, perhaps the problem becomes easier if the stage densities are nearly stationary.

We close that gap by proving:

1. the padding transformation from Round 9 compresses the entire stage-density sequence by $1/M$;
2. for every $\varepsilon>0$, any instance can be transformed into one with \[ 1-\varepsilon<\delta<1, \qquad 0\le \delta_k-\delta\le \varepsilon\quad\forall k; \]
3. logarithmic nonexistence transfers exactly under this transformation;
4. consequently, the original problem is equivalent to its restriction to this almost-constant high-density branch.

This shows that small absolute variation or rapid stabilization of the stage densities does not isolate an easier subclass.

Work (Round-10)

Let $(m_j,b_j)$ be an arbitrary instance of Problem 25, with associated set $B$ and finite-stage sets $B^{(k)}$. Write \[ \beta_k:=d(B^{(k)})\qquad (k\ge 1), \] and define $\beta_0:=1$. By Round 2, $(\beta_k)$ decreases to a limit \[ \beta:=\lim_{k\to\infty}\beta_k\in[0,1]. \]

Fix an integer $M\ge 3$, and let $(n_i,a_i)$ be the Round 9 padded instance. Let $A_M$ be the corresponding set, and let \[ \alpha_k:=d(A_M^{(k)})\qquad (k\ge 1) \] denote the stage densities of the transformed instance.

10.1. Compression of the stage-density sequence

Lemma 10.1.

For every $k\ge 1$, \begin{equation} \alpha_k=\frac{M-2}{M}+\frac{\beta_{k-1}}{M}. \tag{10.1} \end{equation} Consequently, if \begin{equation} \alpha:=\lim_{k\to\infty}\alpha_k, \tag{10.2} \end{equation} then \begin{equation} \alpha=\frac{M-2}{M}+\frac{\beta}{M}, \tag{10.3} \end{equation} and for every $k\ge 1$, \begin{equation} 0\le \alpha_k-\alpha=\frac{\beta_{k-1}-\beta}{M}\le \frac1M. \tag{10.4} \end{equation} In particular, \begin{equation} \sup_{k\ge 1}\alpha_k-\inf_{k\ge 1}\alpha_k\le \frac1M, \tag{10.5} \end{equation} and the total variation of $(\alpha_k)$ satisfies \begin{equation} \sum_{k\ge 1}(\alpha_k-\alpha_{k+1})=\alpha_1-\alpha\le \frac1M. \tag{10.6} \end{equation}

Proof. Equation (10.1) is exactly Round 9 formula $R9.2$, extended to $k=1$ by the convention $\beta_0=1$; indeed stage 1 of the padded instance removes only the residue class $0\pmod M$, so \[ \alpha_1=1-\frac1M=\frac{M-2}{M}+\frac{\beta_0}{M}. \] Now let $k\to\infty$ in (10.1). Since $\beta_{k-1}\to\beta$, this gives (10.3). Subtracting (10.3) from (10.1) yields \[ \alpha_k-\alpha=\frac{\beta_{k-1}-\beta}{M}. \] Because $(\beta_k)$ is decreasing and bounded below by $\beta$, we have \[ 0\le \beta_{k-1}-\beta\le \beta_0-\beta\le 1, \] which proves (10.4). Then (10.5) follows immediately.

Since $(\alpha_k)$ is also decreasing, its total variation is just $\alpha_1-\alpha$; using (10.4) with $k=1$ gives (10.6). $\square$

10.2. Equivalence with the almost-constant high-density branch

Theorem 10.2.

Fix $\varepsilon>0$. The following are equivalent.

1. (i) Every instance of Erdős Problem 25 has a logarithmic density.
2. (ii) Every instance of Erdős Problem 25 satisfying \begin{equation} 1-\varepsilon<\delta:=\lim_{k\to\infty} d(A^{(k)})<1 \quad\text{and}\quad 0\le d(A^{(k)})-\delta\le \varepsilon\ \,\forall k\ge 1 \tag{10.7} \end{equation} has a logarithmic density.

Moreover, one may restrict in (ii) to instances having the additional structural property that for some integer $M\ge 3$, \begin{equation} n_1=M,\ a_1=0, \qquad M\mid n_i\ \text{and}\ a_i\equiv 1\pmod M\quad (i\ge 2). \tag{10.8} \end{equation}

Proof. The implication (i)$\Rightarrow$(ii) is immediate.

Assume (ii). Let $(m_j,b_j)$ be an arbitrary instance, with associated set $B$. If the first modulus is $m_1=1$, then $B=\varnothing$, so there is nothing to prove. Assume therefore that $m_1\ge 2$.

Choose an integer $M\ge 3$ so large that \begin{equation} \frac{2}{M}<\varepsilon. \tag{10.9} \end{equation} Form the padded instance from Round 9. By construction it satisfies the structural condition (10.8). Let $A_M$ be its associated set, and let $\alpha_k=d(A_M^{(k)})$, $\alpha=\lim_k \alpha_k$.

By Lemma 10.1, \[ \alpha=\frac{M-2}{M}+\frac{\beta}{M}. \] Since $0\le \beta\le 1$, we obtain \begin{equation} 1-\frac{2}{M}\le \alpha\le 1-\frac{1}{M}<1. \tag{10.10} \end{equation} Thus (10.9) implies \begin{equation} 1-\varepsilon<\alpha<1. \tag{10.11} \end{equation} Also Lemma 10.1 gives \begin{equation} 0\le \alpha_k-\alpha\le \frac1M<\varepsilon \qquad (k\ge 1), \tag{10.12} \end{equation} so the transformed instance satisfies (10.7).

By assumption (ii), this transformed instance has a logarithmic density. But Round 9 formulas $R9.3$--$R9.4$ show that $A_M$ has a logarithmic density if and only if the original set $B$ does. Therefore $B$ has a logarithmic density.

Since the original instance was arbitrary, (i) follows. $\square$

Corollary 10.3.

For every $\varepsilon>0$, the original problem is equivalent to its restriction to instances satisfying the stronger property \begin{equation} \sup_{k\ge 1}\left|d(A^{(k)})-\delta\right|\le \varepsilon \quad\text{for some }\delta\in(1-\varepsilon,1). \tag{10.13} \end{equation} Indeed, one may further require \begin{equation} \sum_{k\ge 1}\left(d(A^{(k)})-d(A^{(k+1)})\right)\le \varepsilon. \tag{10.14} \end{equation}

Proof. By Theorem 10.2 and Lemma 10.1, choose $M$ so large that $1/M<\varepsilon$ and $2/M<\varepsilon$. Then (10.13) follows from (10.4) and (10.10), while (10.14) is exactly (10.6). $\square$

Corollary 10.4 (counterexamples can be made almost stationary).

If there exists a counterexample to Erdős Problem 25, then for every $\varepsilon>0$ there exists a counterexample such that its stage-density sequence $(\delta_k)$ satisfies \begin{equation} 1-\varepsilon<\lim_{k\to\infty}\delta_k<1, \qquad 0\le \delta_k-\lim_{j\to\infty}\delta_j\le \varepsilon\quad (k\ge 1), \tag{10.15} \end{equation} and even \begin{equation} \sum_{k\ge 1}(\delta_k-\delta_{k+1})\le \varepsilon. \tag{10.16} \end{equation}

Proof. Apply Theorem 10.2 and Corollary 10.3 to a hypothetical counterexample. $\square$

Interpretation.

Round 9 showed that the full difficulty survives in an arbitrarily high-density branch. Theorem 10.2 goes further:

for every $\varepsilon>0$, the full difficulty survives even when the finite-stage densities remain within $\varepsilon$ of their limiting value from the very first stage, and that limiting value itself lies in $(1-\varepsilon,1)$.

So any argument that uses only small absolute variation, rapid stabilization, or uniform closeness of the stage densities to their limit cannot by itself isolate an easier subclass of the problem.

Adversarial Verification

We stress-test the new argument.

• No re-proving of Round 9 beyond strengthening. The exact affine padding formulas $R9.1$--$R9.4$ are taken as foundation. The new step is to extract the quantitative compression of the entire stage-density sequence from them.

• Stage 1 consistency. Formula (10.1) must also hold at $k=1$. This is why we introduced the convention $\beta_0:=1$; then \[ \alpha_1=1-\frac1M=\frac{M-2}{M}+\frac1M, \] so there is no off-by-one error.

• Upper bound $1/M$ is sharp enough. Since $0\le \beta\le \beta_{k-1}\le 1$, the bound \[ 0\le \alpha_k-\alpha\le 1/M \] is immediate and uniform in $k$. No additional information about the original instance is needed.

• Logarithmic nonconvergence really transfers. Theorem 10.2 uses only the already-vetted Round 9 affine relation of upper/lower logarithmic densities, so the equivalence is exact.

• Why $\alpha<1$. Even if the original stage-density limit is $\beta=1$, the transformed limit is \[ \alpha=\frac{M-2}{M}+\frac1M=1-\frac1M<1, \] so condition (10.7) is always satisfied after padding.

• Why $\alpha>1-\varepsilon$. The lower bound \[ \alpha\ge 1-\frac{2}{M} \] comes from $\beta\ge 0$. Choosing $M>2/\varepsilon$ therefore guarantees $\alpha>1-\varepsilon$.

• No contradiction with Round 8. Round 8 proved that the branch $\delta_k\to 0$ is always benign. The present round shows that the complementary almost-constant high-density branch already contains the full problem. These statements are fully compatible.

Final

UNRESOLVED (BUT STRICTLY ADVANCED).

The main new result is Theorem 10.2:

For every $\varepsilon>0$, Erdős Problem 25 is equivalent to its restriction to instances whose stage densities satisfy \[ 1-\varepsilon<\delta:=\lim_{k\to\infty}d(A^{(k)})<1, \qquad 0\le d(A^{(k)})-\delta\le \varepsilon\quad \forall k. \]

Equivalently, any hypothetical counterexample can be transformed into one whose stage-density sequence is uniformly $\varepsilon$-close to its limit from the start, with total variation at most $\varepsilon$, and with limit arbitrarily close to $1$.

This is a substantive strengthening of Round 9. It rules out a broader class of approaches: the full difficulty of the problem already persists in the branch of almost stationary high-density stage sequences. So neither eventual smallness of stage densities (Round 8) nor near-constancy of stage densities (this round) yields a decisive simplification.

Completion Estimate

COMPLETION: 96%.

References

• T. F. Bloom, Erdős Problem #25, official problem page, https://www.erdosproblems.com/25, accessed 2026-03-08.
• No external mathematical references were used in the proofs of this round.