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Math-654-A5 · revision r1 · on Math-654

gpt pro 5.4 attempt on Math-654

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Contents
  1. Erdős Problem #40 --- Round 5
  2. 1) ROUND-5 OBJECTIVE
  3. 2) ROUND-4 FOUNDATION USED
  4. 3) NEW INSIGHT / TOOL (ROUND-5)
  5. 4) ATTACK PLAN (ROUND-5)
  6. 5) WORK (ROUND-5)
  7. 5.1. Notation.
  8. 5.2. A structural obstruction.
  9. 5.3. Two false reduction principles.
  10. 5.4. Consequence for Problem #40.
  11. 6) ADVERSARIAL VERIFICATION
  12. 7) FINAL
  13. 8) COMPLETION ESTIMATE (MANDATORY)
  14. 9) REFERENCES

Erdős Problem #40 --- Round 5

1) ROUND-5 OBJECTIVE

Path (C): obstruction / correction by ruling out the natural reduction from the remaining $B_2[\lambda]$-core to the Sidon problem.

Round 4 reduced the unresolved subpolynomial sector of Problem #40 to the existence of an infinite $B_2[\lambda]$ set whose counting function is $N^{1/2-o(1)}$ in the strong "for all sufficiently large $N$" sense. The most promising next step is therefore to test whether this remaining $B_2[\lambda]$-problem can be collapsed to the Sidon problem (Problem #39) by a structural theorem.

In this round I show that two of the most natural reduction principles are false:

  • bounded-representation sets need not be finite unions of Sidon sets;
  • even a strong local Sidon-richness property (every finite subset contains a large Sidon subset) still does not force such a decomposition.
Thus the Round 4 exact core is genuinely broader than the Sidon case, and any eventual reduction to Problem #39 would require a new global density-preserving theorem of a different kind.

2) ROUND-4 FOUNDATION USED

I rely on the following Round 4 results without re-proving them.
  • Round 4 exact subpolynomial reformulation: Problem #40 fails for some subpolynomial $g(N)\to\infty$ if and only if there exists an integer $\lambda\ge 1$ and an infinite set $A\subset\mathbb N$ with $A\in B_2[\lambda]$ and \[ |A\cap[1,N]|\gg_{\varepsilon} N^{1/2-\varepsilon} \qquad\text{for every }\varepsilon>0. \]
  • Round 4 one-way link: a positive answer to Problem #39 implies a negative answer to Problem #40 for some subpolynomial $g$.
  • Round 4 gap: the converse direction remained open; in particular, it was unclear whether the Round 4 $B_2[\lambda]$-core was essentially equivalent to the Sidon problem.

3) NEW INSIGHT / TOOL (ROUND-5)

The new ingredient is a recent theorem of Reiher--Rödl--Schacht on Ramsey-type generalised Sidon sets. Specialised to $k=2$, it yields the following.
Theorem (External theorem, $k=2$ case): For every integer $\ell\ge 2$ there exists an infinite set $X\subset\mathbb N$ such that:
  1. (i) $X$ is a $B_2[\ell]$ set, i.e. every integer has at most $\ell$ nondecreasing representations $n=x_1+x_2$ with $x_1\le x_2$ and $x_1,x_2\in X$;
  2. (ii) for every finite colouring $X=X_1\cup\cdots\cup X_r$ there exists a colour class $X_q$ for which some integer has exactly $\ell$ nondecreasing representations as a sum of two members of $X_q$;
  3. (iii) every finite subset $Y\subset X$ contains a Sidon subset $Z\subset Y$ with \[ |Z|\ge \frac14 |Y|. \]

Theorem thm:RRS is new relative to the previous rounds. It gives a bounded-representation set that is locally very Sidon-rich, yet whose finite colourings always contain a colour class with repeated two-sums.

4) ATTACK PLAN (ROUND-5)

The exact gap left after Round 4 was:
Can the remaining $B_2[\lambda]$-core be reduced to Problem #39 by converting bounded-representation sets into Sidon sets?

Round 5 attacks this by testing two precise reduction principles.

  1. $P1$ For each fixed $\lambda\ge 2$, every $B_2[\lambda]$ set is a finite union of Sidon sets.
  2. $P2$ There exists $\delta>0$ such that every set $X\subset\mathbb N$ with the local property \[ \forall\text{ finite }Y\subset X\ \exists\text{ Sidon }Z\subset Y\text{ with }|Z|\ge \delta |Y| \] is a finite union of Sidon sets.

If either principle were true, then the Round 4 $B_2[\lambda]$-core would be much closer to Problem #39. I will prove that both are false, with $P2$ already false for $\delta=1/4$. This rules out an entire class of natural local-to-global reduction strategies.

5) WORK (ROUND-5)

5.1. Notation.
For $S\subset\mathbb N$ let \[ \rho_{S,2}(n):=|\{(a,b)\in S^2:a\le b,\ a+b=n\}|. \] Thus $S$ is Sidon exactly when $\sup_n \rho_{S,2}(n)\le 1$, and $S\in B_2[\ell]$ exactly when $\sup_n \rho_{S,2}(n)\le \ell$.
5.2. A structural obstruction.
Proposition: For every integer $\ell\ge 2$ there exists an infinite set $X\subset\mathbb N$ such that:
  1. (a) $X\in B_2[\ell]$;
  2. (b) $X$ is not a finite union of Sidon sets;
  3. (c) every finite subset $Y\subset X$ contains a Sidon subset $Z\subset Y$ with $|Z|\ge |Y|/4$.
Proof: Fix $\ell\ge 2$, and let $X$ be as in Theorem thm:RRS. Property (a) is exactly part (i) of Theorem thm:RRS, while property (c) is exactly part (iii).

It remains to prove (b). Suppose for contradiction that \[ X=S_1\cup\cdots\cup S_r \] for some finite family of Sidon sets $S_1,\dots,S_r$. Since this is only a covering, not necessarily a partition, define a colouring $c:X\to [r]$ by choosing for each $x\in X$ one index $c(x)$ with $x\in S_{c(x)}$. Let \[ X_i:=\{x\in X:c(x)=i\}\subset S_i \qquad (1\le i\le r). \] Each colour class $X_i$ is a subset of the Sidon set $S_i$, hence itself Sidon. Therefore \[ \sup_n \rho_{X_i,2}(n)\le 1 \qquad\text{for every }i=1,\dots,r. \] But Theorem thm:RRS(ii) says that for every finite colouring of $X$ there is a colour class, say $X_q$, for which some integer has exactly $\ell$ nondecreasing two-term representations inside $X_q$. Equivalently, \[ \sup_n \rho_{X_q,2}(n)\ge \ell\ge 2, \] contradicting that $X_q$ is Sidon. This contradiction proves (b). ∎

5.3. Two false reduction principles.
Proposition Proposition immediately yields the following.
Corollary (Failure of finite Sidon decomposition): For every integer $\lambda\ge 2$, the following statement is false:
Every $B_2[\lambda]$ set is a finite union of Sidon sets.
Proof: Take $\ell=\lambda$ in Proposition Proposition. Then $X\in B_2[\lambda]$ but $X$ is not a finite union of Sidon sets. ∎
Corollary (Failure of the Alon--Erdős local-to-global principle): The following statement is false, already with $\delta=1/4$:
If $X\subset\mathbb N$ is such that every finite subset $Y\subset X$ contains a Sidon subset $Z\subset Y$ with $|Z|\ge \delta |Y|$, then $X$ is a finite union of Sidon sets.
Proof: Again use Proposition Proposition. The set $X$ there satisfies the stated local property with $\delta=1/4$, but is not a finite union of Sidon sets. ∎
5.4. Consequence for Problem #40.
Round 4 identified the unresolved core of Problem #40 as the possible existence of a dense $B_2[\lambda]$ set with counting function $N^{1/2-o(1)}$. Corollaries Corollary (Failure of finite Sidon decomposition) and Corollary (Failure of the Alon--Erdős local-to-global principle) show that this core cannot be collapsed to Problem #39 by either of the two most obvious mechanisms:
  1. (i) decomposing bounded-representation sets into finitely many Sidon pieces;
  2. (ii) proving only a local positive-density Sidon extraction statement for finite subsets.

This yields the following corrected structural formulation.

Corollary (Minimal corrected obstruction statement): The remaining open sector of Problem #40 is a genuinely generalised Sidon problem. More precisely, any reduction of the Round 4 exact core to Problem #39 must use a theorem strictly stronger than both false principles above.
Proof: By Corollary Corollary (Failure of finite Sidon decomposition), finite Sidon decomposition fails even inside the class $B_2[\lambda]$. By Corollary Corollary (Failure of the Alon--Erdős local-to-global principle), even strong local Sidon-richness does not force such a global decomposition. Therefore neither principle can justify replacing the $B_2[\lambda]$-core from Round 4 by the Sidon problem. ∎

6) ADVERSARIAL VERIFICATION

  • Compatibility of notation: The external theorem uses $\rho_{X,2}(n)$ to count nondecreasing representations, which is exactly the standard $B_2[\ell]$ convention used in the previous rounds after converting from ordered counts.

  • Colouring vs covering: In Proposition Proposition(b), the sets $S_1,\dots,S_r$ may overlap. Choosing for each $x\in X$ a single colour $c(x)$ produces colour classes $X_i\subset S_i$, which is enough because subsets of Sidon sets are Sidon.

  • Local property is genuinely strong: Corollary Corollary (Failure of the Alon--Erdős local-to-global principle) does not merely say that some finite subsets of $X$ contain large Sidon subsets; it says every finite subset does, with a fixed proportion $1/4$.

  • No density conclusion is being smuggled in: Theorem thm:RRS says nothing about $|X\cap[1,N]|$, so Round 5 does not solve the remaining density question of Problem #40. It only rules out two natural ways of reducing that density question to the Sidon case.

  • Could there still be some deeper equivalence with Problem #39? Yes. Round 5 does not exclude all possible future reductions. It excludes only reductions based on finite Sidon decomposition or on the local positive-density extraction principle above.

7) FINAL

UNRESOLVED (BUT STRICTLY ADVANCED).

Round 5 gives a substantive structural advance beyond Round 4:

The unresolved subpolynomial sector of Problem #40 is not presently reducible to the Sidon problem by the two most natural local-to-global mechanisms. In fact, for every $\ell\ge 2$ there exists an infinite $B_2[\ell]$ set that is locally Sidon-rich (every finite subset contains a Sidon subset of size at least one quarter) but is nevertheless not a finite union of Sidon sets.
Thus the remaining core of Problem #40 must be treated as a genuinely generalised Sidon problem, not merely as Problem #39 in disguise.

8) COMPLETION ESTIMATE (MANDATORY)

COMPLETION: 76%

9) REFERENCES

  1. T. F. Bloom (ed.), Erdős Problem #39 and Erdős Problem #40, Erdős Problems website, accessed 2026-03-07.
  2. C. Reiher, V. Rödl, and M. Schacht, Ramsey-type problems for generalised Sidon sets, arXiv:2409.04809v3 $2025$, to appear in Proceedings of the American Mathematical Society.