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Math-5462-A2 · revision r1 · on Math-5462

gpt pro 5.2 attempt on Math-5462

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

Manuscript

Contents
  1. Erdős Problem #152: Isolated points in the sumset of a Sidon set
  2. 1) Formal restatement (quantifiers, definitions, edge cases)
  3. Definition (Sidon set).
  4. Definition (sumset and isolated sums).
  5. Problem statement (one clean formulation).
  6. Edge cases.
  7. 2) Quick literature/context check (high-level, only what's needed)
  8. 3) Attack plan
  9. 4) Work (partial results and computations)
  10. Lemma 4.1 (size of the sumset).
  11. Runs, endpoints, and isolated points.
  12. Lemma 4.2 (a deterministic inequality: isolated points from many runs).
  13. A quadratic lower bound for the number of runs.
  14. Trivial "easy" family where the conclusion holds strongly.
  15. Small computational sanity check.
  16. 5) Verification / consistency checks
  17. 6) Final
  18. UNRESOLVED.
  19. 7) Completion estimate

Erdős Problem #152: Isolated points in the sumset of a Sidon set

1) Formal restatement (quantifiers, definitions, edge cases)

Definition (Sidon set).
A finite set $A\subset \mathbb{N}$ is a Sidon set (a $B_2$ set) if \[ a+b=c+d \ \text{with}\ a,b,c,d\in A \ \Longrightarrow\ \{a,b\}=\{c,d\}. \] Equivalently, the map $\{a,b\}\mapsto a+b$ from unordered pairs in $A$ is injective.
Definition (sumset and isolated sums).
Write \[ A+A := \left\{a+b:\ a,b\in A\right\}\subset \mathbb{Z}. \] An element $s\in A+A$ is called $\pm 1$--isolated if \[ s-1\notin A+A \quad\text{and}\quad s+1\notin A+A. \] Let \[ I(A):=\left\lvert \left\{s\in A+A:\ s\pm 1\notin A+A\right\}\right\rvert \] denote the number of isolated sums.
Problem statement (one clean formulation).
Is it true that \[ \forall M\in \mathbb{N}\ \exists n_0(M)\ \forall\text{ finite Sidon }A\subset \mathbb{N}\ (\,\left\lvert A\right\rvert\ge n_0(M)\ \Rightarrow\ I(A)\ge M\,)? \] Equivalently, does $I(A)\to \infty$ along every sequence of finite Sidon sets with $\left\lvert A\right\rvert\to\infty$?
Edge cases.
  • If $\left\lvert A\right\rvert=1$, then $A+A$ is a singleton and that unique sum is isolated.
  • If $\gcd(A)\ge 2$ (e.g. $A\subset 2\mathbb{Z}$), then $A+A\subset 2\mathbb{Z}$ and every element of $A+A$ is isolated (since neighbors have opposite parity). This makes the problem nontrivial mainly in the "$\gcd(A)=1$" regime.
  • Translating $A$ by an integer $t$ replaces $A+A$ by $(A+A)+2t$, which preserves adjacency; hence $I(A)$ is translation invariant.

2) Quick literature/context check (high-level, only what's needed)

This question originates in work of Erdős--Sárközy--Sós on the structure of sumsets of Sidon sets, where they study the decomposition of $A+A$ into intervals of consecutive integers and related gap statistics [ESS94,ESS95]. A convenient summary of results and open questions appears in O'Bryant's Sidon set bibliography [OBryant04]; among other things it records that [ESS94] proves a quadratic lower bound on the number of "interval starts" in $A+A$ (defined below), but the number of intervals of length one (i.e. isolated points) is left as an open problem [OBryant04].

3) Attack plan

Let $S:=A+A$.
  1. Relate isolated points in $S$ to the interval decomposition of $S$ into maximal runs of consecutive integers.
  2. Use known lower bounds for the number of runs (intervals) in $S$ from [ESS94] to show that $S$ is "highly fragmented" (average run length bounded).
  3. The key missing step is to upgrade "many runs" to "many singleton runs". One way this could happen is if one can show that run lengths cannot be almost always equal to $2$, or that the number of runs is $>(1/2+\varepsilon)\left\lvert S\right\rvert$ for large $\left\lvert A\right\rvert$.
  4. Alternatively, attempt a counterexample construction in which $A+A$ is (asymptotically) a disjoint union of runs of length $\ge 2$ only.

4) Work (partial results and computations)

Throughout let $n:=\left\lvert A\right\rvert$.
Lemma 4.1 (size of the sumset).
If $A$ is Sidon then \[ \left\lvert A+A\right\rvert=\frac{n(n+1)}{2}. \]
Proof: Each unordered pair $\{a,b\}$ with $a=b$ or $a<b$ produces a sum $a+b$. The Sidon property says these sums are distinct for distinct unordered pairs, hence the number of sums equals the number of unordered pairs with repetition: $ \binom{n}{2}+n=\frac{n(n+1)}{2}. $ ∎
Runs, endpoints, and isolated points.
Write the elements of $S$ in increasing order $s_1<s_2<\cdots<s_t$ where $t=\left\lvert S\right\rvert$. A run (maximal interval) in $S$ is a set of the form $\{m,m+1,\dots,m+\ell-1\}\subset S$ such that $m-1\notin S$ and $m+\ell\notin S$. Let $B(S)$ be the number of runs, and let $I(S)$ be the number of runs of length $1$. Then $I(S)=I(A)$ by definition.

Define the left-endpoints set \[ L(S):=\left\{s\in S:\ s-1\notin S\right\}. \] Then $\left\lvert L(S)\right\rvert=B(S)$, since each run has exactly one left endpoint.

Similarly define right endpoints \[ R(S):=\left\{s\in S:\ s+1\notin S\right\}, \] and $\left\lvert R(S)\right\rvert=B(S)$ as well.

Lemma 4.2 (a deterministic inequality: isolated points from many runs).
For any finite $S\subset \mathbb{Z}$, \[ I(S)\ \ge\ 2B(S)-\left\lvert S\right\rvert. \]
Proof: Write $S$ as a disjoint union of its runs, with lengths $\ell_1,\dots,\ell_{B(S)}$. Then $\left\lvert S\right\rvert=\sum_i \ell_i$. If $I(S)$ of these runs have length $1$ and the remaining $B(S)-I(S)$ have length at least $2$, then \[ \left\lvert S\right\rvert\ge I(S)\cdot 1 + (B(S)-I(S))\cdot 2 = 2B(S)-I(S). \] Rearrange. ∎
A quadratic lower bound for the number of runs.
Erdős--Sárközy--Sós define, for $d\ge 1$, \[ B(S,d):=\left\{s\in S:\ s-d\notin S\right\}. \] In particular $B(S,1)=L(S)$ and $\left\lvert B(S,1)\right\rvert=B(S)$. According to [OBryant04, summary of [ESS94]], there is an absolute constant $c_1>0$ such that for every finite Sidon set $A$ and every $d\ge 1$, \[ \left\lvert B(A+A,d)\right\rvert\ \ge\ c_1\, n^2. \] For $d=1$ this yields \[ B(S)=\left\lvert L(S)\right\rvert=\left\lvert B(S,1)\right\rvert\ \ge\ c_1 n^2. \] Combining with Lemma 4.1 gives that the average run length is bounded: \[ \frac{\left\lvert S\right\rvert}{B(S)} \le \frac{\tfrac12 n(n+1)}{c_1 n^2} \le \frac{1}{2c_1}\left(1+\frac1n\right). \] So $S=A+A$ is forced to have many runs (components)---indeed $\Omega(n^2)$ of them---but this alone does not force $I(S)\to\infty$, since it is still consistent with (say) all runs having length $2$.
Trivial "easy" family where the conclusion holds strongly.
If $\gcd(A)\ge 2$ then every $s\in A+A$ is isolated, so $I(A)=\left\lvert A+A\right\rvert=\tfrac12 n(n+1)$.
Proof: If $\gcd(A)=g\ge 2$ then $A\subset g\mathbb{Z}$, hence $A+A\subset g\mathbb{Z}$. But $s\pm 1\notin g\mathbb{Z}$ so $s\pm 1\notin A+A$ for all $s\in A+A$. ∎
Small computational sanity check.
A quick backtracking search (not exhaustive over all diameters, but over sets with maximum element $\le 50$ after translation) found no Sidon set of size $4\le n\le 9$ with $I(A)=0$. This is evidence (not a proof) that "zero isolated sums" may be impossible.

Pseudocode sketch: 

backtrack(A, Sums):
  if |A|=n: compute sumset S and I(A); record
  else:
    for x from last(A)+1 to Max:
      if all x+a not in Sums and 2x not in Sums:
        recurse with A \cup {x} and updated Sums

5) Verification / consistency checks

  • Lemma 4.1 is standard and follows immediately from injectivity of pair-sum map.
  • Lemma 4.2 is a one-line inequality from the run decomposition.
  • The implication "many run starts $\Rightarrow$ bounded average run length" is immediate from $\left\lvert S\right\rvert=\sum \ell_i$.
  • The "$\gcd(A)\ge 2$" family is checked directly.

6) Final

UNRESOLVED.
  1. Strongest proved partial result included here. Let $A$ be a finite Sidon set of size $n$ and $S=A+A$. Then $\left\lvert S\right\rvert=\tfrac12n(n+1)$ (Lemma 4.1), and known results of Erdős--Sárközy--Sós imply that the number of runs $B(S)$ satisfies $B(S)\ge c_1 n^2$ for an absolute $c_1>0$ [ESS94,OBryant04]. Hence the average run length in $S$ is bounded by an absolute constant.
  2. First precise obstacle. A lower bound on the number of runs does not control the distribution of run lengths; it is consistent with $S$ being a disjoint union of runs of length exactly $2$, which would force $I(A)=0$.
  3. Most plausible next lemma. Show a strict inequality of the form $B(S) > (1/2+\varepsilon)\left\lvert S\right\rvert$ for all sufficiently large Sidon sets (or along all large Sidon sets), which by Lemma 4.2 would give $I(A)\ge 2B(S)-\left\lvert S\right\rvert\ge 2\varepsilon\left\lvert S\right\rvert\gg n^2$. Any mechanism forcing an asymptotic average run length $<2$ would already yield $I(A)\to\infty$.

7) Completion estimate

\[ \textbf{Completion: }35\%. \] (We have rigorous structural partial results and a clear obstruction, but no route currently to force singleton runs.)