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Math-5462-A3 · revision r1 · on Math-5462

gpt pro 5.4 attempt on Math-5462

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Manuscript

Contents
  1. Erdős Problem #152: isolated sums in the sumset of a Sidon set (ROUND 2)
  2. 1) ROUND-2 OBJECTIVE
  3. 2) Round-1 FOUNDATION USED
  4. 3) NEW INSIGHT / TOOL (ROUND 2)
  5. 4) ATTACK PLAN (ROUND 2)
  6. 5) WORK (ROUND 2)
  7. 6) ADVERSARIAL VERIFICATION
  8. 7) FINAL
  9. 8) COMPLETION ESTIMATE (MANDATORY)
  10. References
  11. References

Erdős Problem #152: isolated sums in the sumset of a Sidon set (ROUND 2)

1) ROUND-2 OBJECTIVE

I pursue path (C). The problem is Erdős Problem #152 [Bloom152], and the relevant sumset background comes from Erdős--Sárközy--Sós and O'Bryant [ESS94,OBryant04]. The Round-1 writeup reduced the problem to understanding whether many runs in $A+A$ force many singleton runs, but left open the possibility that $A+A$ might be built mostly from runs of length $2$. In this round I rule out the extreme version of that obstruction completely: a finite Sidon set with at least $3$ elements can never have zero isolated sums, and every finite Sidon set with at least $4$ elements has at least $2$ isolated sums. This is not the full Erdős statement, but it is a genuine structural theorem beyond Round 1.

2) Round-1 FOUNDATION USED

From Round 1 I use only the following already-established facts.
  • If $A$ is a finite Sidon set of size $n$, then $|A+A|=n(n+1)/2$.
  • If $S\subset \mathbf Z$ is finite and $B(S)$ denotes the number of maximal runs of consecutive integers in $S$, while $I(S)$ denotes the number of singleton runs, then \[ I(S)\ge 2B(S)-|S|. \]
  • The Round-1 obstacle was that a lower bound on $B(A+A)$ alone does not exclude the possibility that every run has length $2$.
I do not re-prove those facts here.

3) NEW INSIGHT / TOOL (ROUND 2)

The new tool is an elementary but decisive observation about consecutive elements.

Key new observation. A Sidon set contains at most one pair of consecutive integers.

This single lemma forces isolated sums at the edges of the sumset and kills the most obvious zero-isolated-sum constructions.

4) ATTACK PLAN (ROUND 2)

The exact Round-1 gap was this: many runs in $A+A$ need not imply any singleton run.

I now attack that gap directly by working with the outermost sums of $A+A$. Write \[ A=\{a_1<a_2<\cdots<a_n\}. \] The sums $2a_1$, $a_1+a_2$, $a_{n-1}+a_n$, and $2a_n$ are the natural candidates for isolated points. The plan is:

  1. prove that $A$ has at most one consecutive pair;
  2. show that if the bottom end of $A$ has no consecutive pair then $2a_1$ is isolated, and similarly at the top end $2a_n$ is isolated;
  3. if there is a consecutive pair at one end, prove that the two largest (or two smallest) extremal sums are isolated instead.
This overcomes the Round-1 obstruction because it proves that the configuration $I(A)=0$ is impossible for all $n\ge 3$, and moreover gives a sharp universal lower bound $I(A)\ge 2$ for $n\ge 4$.

5) WORK (ROUND 2)

Throughout this section, \[ A=\{a_1<a_2<\cdots<a_n\}\subset \mathbf Z \] is a finite Sidon set, and \[ I(A):=\left|\{s\in A+A: s-1\notin A+A\text{ and }s+1\notin A+A\}\right|. \] By translation invariance, nothing depends on where $A$ sits in $\mathbf Z$.

Lemma 5.1. A Sidon set contains at most one pair of consecutive integers.

Proof. Suppose first that $x,x+1,y,y+1\in A$ with $x+1<y$. Then \[ x+(y+1)=(x+1)+y, \] and the two unordered pairs $\{x,y+1\}$ and $\{x+1,y\}$ are distinct, contradicting the Sidon property.

If instead the two consecutive pairs overlap, then $x,x+1,x+2\in A$, and \[ x+(x+2)=2(x+1), \] again contradicting the Sidon property. Therefore there is at most one consecutive pair in $A$. $\square$

Lemma 5.2. If $a_2\ge a_1+2$, then $2a_1$ is isolated in $A+A$.

Proof. Clearly $2a_1-1<2a_1$ is smaller than the least element of $A+A$, so only $2a_1+1$ needs attention. If $2a_1+1=u+v$ with $u,v\in A$ and $u\le v$, then $u,v\ge a_1$. Since \[ u+v=2a_1+1, \] we cannot have $u\ge a_1+1$, because then $u+v\ge 2a_1+2$. Hence $u=a_1$, so $v=a_1+1$, contradicting $a_2\ge a_1+2$. Thus $2a_1+1\notin A+A$, and $2a_1$ is isolated. $\square$

Lemma 5.3. If $a_n\ge a_{n-1}+2$, then $2a_n$ is isolated in $A+A$.

Proof. Now $2a_n+1$ is larger than the maximum of $A+A$, so only $2a_n-1$ matters. If $2a_n-1=u+v$ with $u,v\in A$ and $u\le v$, then $u,v\le a_n$. Since \[ u+v=2a_n-1, \] we cannot have $v\le a_n-1$, because then $u+v\le 2a_n-2$. Hence $v=a_n$, so $u=a_n-1$, contradicting $a_n\ge a_{n-1}+2$. Thus $2a_n-1\notin A+A$, and $2a_n$ is isolated. $\square$

Lemma 5.4. Assume $n\ge 4$ and $a_2=a_1+1$. Then both $a_{n-1}+a_n$ and $2a_n$ are isolated in $A+A$.

Proof. By Lemma 5.1, the consecutive pair $\{a_1,a_2\}$ is the unique consecutive pair in $A$. In particular, \[ a_n\ge a_{n-1}+2. \] So $2a_n$ is isolated by Lemma 5.3.

It remains to show that $a_{n-1}+a_n$ is isolated.

First, no larger sum lies at distance $1$ above it. Indeed, the next larger sum in $A+A$ is at least $2a_n$, and \[ 2a_n-(a_{n-1}+a_n)=a_n-a_{n-1}\ge 2. \] Hence \[ a_{n-1}+a_n+1\notin A+A. \]

Second, suppose for contradiction that \[ u+v=a_{n-1}+a_n-1 \] with $u,v\in A$ and $u\le v$. If $v=a_n$, then $u=a_{n-1}-1$. But then $u,u+1=a_{n-1}\in A$, giving a second consecutive pair in $A$, impossible by Lemma 5.1. If instead $v\le a_{n-1}$, then \[ u+v\le 2a_{n-1}<a_{n-1}+a_n-1, \] since $a_n\ge a_{n-1}+2$. Both cases are impossible, so \[ a_{n-1}+a_n-1\notin A+A. \] Therefore $a_{n-1}+a_n$ is isolated. $\square$

Corollary 5.5. Assume $n\ge 4$ and $a_n=a_{n-1}+1$. Then both $2a_1$ and $a_1+a_2$ are isolated in $A+A$.

Proof. Apply Lemma 5.4 to the reflected Sidon set $-A=\{-a_n<\cdots<-a_1\}$ and then negate the resulting isolated sums. $\square$

Theorem 5.6. Let $A$ be a finite Sidon set.

  1. $I(A)=0$ if and only if $A=\{x,x+1\}$ for some integer $x$.
  2. If $|A|\ge 3$, then $I(A)\ge 1$.
  3. If $|A|\ge 4$, then $I(A)\ge 2$.

Proof. For $|A|=2$, write $A=\{x,y\}$ with $x<y$. Then \[ A+A=\{2x,x+y,2y\}. \] If $y=x+1$, this is $\{2x,2x+1,2x+2\}$ and indeed $I(A)=0$. If $y\ge x+2$, then $2x$ and $2y$ are isolated, so $I(A)=2$. This proves part $1$ for $|A|=2$.

Now assume $|A|\ge 3$. By Lemma 5.1, at least one of the end gaps $a_2-a_1$ and $a_n-a_{n-1}$ is at least $2$, so Lemma 5.2 or Lemma 5.3 gives at least one isolated sum. Hence $I(A)\ge 1$, proving part $2$.

Finally assume $|A|\ge 4$. There are three cases.

  • If $a_2\ge a_1+2$ and $a_n\ge a_{n-1}+2$, then Lemmas 5.2 and 5.3 show that $2a_1$ and $2a_n$ are both isolated.
  • If $a_2=a_1+1$, then Lemma 5.4 gives two isolated sums, namely $a_{n-1}+a_n$ and $2a_n$.
  • If $a_n=a_{n-1}+1$, then Corollary 5.5 gives two isolated sums, namely $2a_1$ and $a_1+a_2$.
These cases are exhaustive because Lemma 5.1 forbids simultaneous consecutive pairs at both ends. Thus $I(A)\ge 2$ for every Sidon set with at least $4$ elements. $\square$

Sharpness. The bound in part $3$ is best possible. For example, \[ A=\{0,1,3,7\} \] is Sidon and \[ A+A=\{0,1,2,3,4,6,7,8,10,14\}, \] whose isolated sums are exactly $10$ and $14$. Likewise, \[ A=\{0,1,6,9,13\} \] is Sidon and has exactly two isolated sums, namely $22$ and $26$.

6) ADVERSARIAL VERIFICATION

I now try to break the new argument.
  • Boundary case $|A|=2$. This is the unique genuine exception. The theorem explicitly isolates it and classifies it exactly.
  • Could there be two consecutive pairs, one of them internal and one at an end? No. Lemma 5.1 rules out all pairs of distinct consecutive pairs, whether disjoint or overlapping.
  • Does the proof of Lemma 5.4 accidentally use $n\ge 5$? No. The only place where size matters is to ensure that after a bottom consecutive pair there is still a top end to talk about; $n\ge 4$ is enough.
  • Could $a_{n-1}+a_n-1$ be represented using two elements below $a_{n-1}$? No, because then it would be at most $2a_{n-1}$, which is strictly smaller than $a_{n-1}+a_n-1$ when $a_n-a_{n-1}\ge 2$.
  • Does translation matter? No. Replacing $A$ by $A+t$ adds $2t$ to every sum and preserves isolation.
No hidden quantifier issue remains: the theorem is uniform over all finite Sidon sets.

7) FINAL

UNRESOLVED (BUT STRICTLY ADVANCED).

Round 2 proves a new theorem not present in Round 1:

If $A$ is a finite Sidon set with at least $4$ elements, then $A+A$ contains at least $2$ isolated sums. Equivalently, the Erdős property in Problem #152 is already true with $M=2$ once $|A|\ge 4$.
In particular, the Round-1 candidate obstruction "perhaps $A+A$ can have no isolated points" is now completely ruled out for all $|A|\ge 3$, and the stronger candidate "perhaps every run has length $2$" is impossible as well.

The full statement that the number of isolated sums must tend to infinity with $|A|$ remains open.

8) COMPLETION ESTIMATE (MANDATORY)

COMPLETION: 48%

References

References

  • [Bloom152] T. F. Bloom, Erdős Problem #152, https://www.erdosproblems.com/152, accessed 2026-03-07.
  • [Bloom200] T. F. Bloom, Erdős Problem #200, https://www.erdosproblems.com/200, accessed 2026-03-07.
  • [Bloom201] T. F. Bloom, Erdős Problem #201, https://www.erdosproblems.com/201, accessed 2026-03-07.
  • [ESS94] P. Erdős, A. Sárközy, and T. Sós, On sum sets of Sidon sets. I, Journal of Number Theory 47 $1994$, 329--347.
  • [GT08] B. Green and T. Tao, The primes contain arbitrarily long arithmetic progressions, Annals of Mathematics 167 $2008$, 481--547.
  • [KSS75] J. Komlós, M. Sulyok, and E. Szemerédi, Linear problems in combinatorial number theory, Acta Mathematica Academiae Scientiarum Hungaricae 26 $1975$, 113--121.
  • [MV07] H. Montgomery and R. Vaughan, Multiplicative Number Theory I. Classical Theory, Cambridge Studies in Advanced Mathematics 97, Cambridge University Press, 2007.
  • [OBryant04] K. O'Bryant, A complete annotated bibliography of work related to Sidon sequences, Electronic Journal of Combinatorics 11 $2004$, Dynamic Survey 11.
  • [OBryant10] K. O'Bryant, Thick subsets that do not contain arithmetic progressions, arXiv:0912.1494.
  • [Semchankau20] A. Semchankau, Maximal subsets free of arithmetic progressions in arbitrary sets, arXiv:2010.04490.