gpt 5.2 attempt on Math-656

OUTPUT 1) FORMAL RESTATEMENT Definitions/Conventions. - $S\subset\mathbb Z$ is Sidon if all sums $s_1+s_2$ with $s_1\le s_2$ in $S$ are distinct.

Question. Given $N\ge 1$, a Sidon set $A\subset\{1,\dots,N\}$, and $\varepsilon>0$, does there exist $M\ge N$ and $B\subset\{N+1,\dots,M\}$ such that: 1. $A\cup B$ is Sidon, and 2. $|A\cup B|\ge (1-\varepsilon)\sqrt M$?

2) QUICK LITERATURE/CONTEXT CHECK Web browsing is not available in this session. I only note that the problem text connects this to problems [329] and [707] and cites Guy's collection; I do not use any external results here.

3) ATTACK PLAN Proof track ideas: 1. Try to build $B$ greedily inside $\{N+1,\dots,M\}$ while maintaining Sidon. Since $M$ can depend on $A$ and $\varepsilon$, one can afford to take $M$ extremely large and try to reach density $(1-\varepsilon)\sqrt M$. 2. Reformulate the Sidon constraint for $A\cup B$ in terms of disjointness between certain difference sets, and attempt to show that the forbidden-difference set coming from $A$ does not prevent constructing a near-optimal Sidon set later. 3. Try a probabilistic method: pick $B$ of near-optimal size and estimate probability that it creates any collision with $A$.

Disproof track ideas: 1. Search for "rigid" Sidon sets $A$ with very large difference sets that might force any extension past $N$ to be sparse. 2. Look for local obstructions: forced collisions between $A+A$ and $A+B$ when $B$ sits just above $N$.

Chosen path in this attempt: prove a few unconditional extension lemmas (sparse extension and clean separation tricks), and test greedy extension on a small example.

4) WORK Lemma 1 (One-step extension far to the right). Let $A\subset\{1,\dots,N\}$ be Sidon, and let $x$ be an integer with $x>2N$. Then $A\cup\{x\}$ is Sidon.

Proof. All sums of two elements from $A$ lie in $\{2,3,\dots,2N\}$. All sums involving $x$ are of the form $x+a$ ($a\in A$) or $x+x$, and these are all $>2N$ since $x>2N$ and $a\ge 1$. Thus no sum involving $x$ can equal a sum from $A+A$. Within the new sums involving $x$, we have $x+a=x+a'$ implies $a=a'$, and $x+x=x+a$ implies $x=a$, impossible since $x>N\ge a$. Therefore all 2-sums in $A\cup\{x\}$ are distinct aside from swapping the order, i.e. $A\cup\{x\}$ is Sidon. $\square$

Lemma 2 (Sparse infinite extension by rapidly growing elements). Let $A\subset\{1,\dots,N\}$ be Sidon. Define a sequence $(x_k)_{k\ge 1}$ by choosing $x_1>2N$ and \[x_{k+1}>2x_k\quad\text{for all }k\ge 1.\] Then $A\cup\{x_k:k\ge 1\}$ is Sidon.

Proof. Let $S:=A\cup\{x_k:k\ge 1\}$. We must show: if $u+v=u'+v'$ with $u\le v$ and $u'\le v'$ in $S$, then $\{u,v\}=\{u',v'\}$.

If all four elements lie in $A$, this holds because $A$ is Sidon. If at least one of $v,v'$ is some $x_k$, compare the sizes of the sums. Fix $k$. Any sum involving $x_k$ is at least $x_k+1$ and at most $x_k+x_k=2x_k$. Any sum involving only elements $\le x_{k-1}$ is at most $2x_{k-1}<x_k$ by the growth condition. Therefore, if a sum uses $x_k$, then it cannot equal a sum using only smaller elements.

So in any equality $u+v=u'+v'$, the largest element among $\{u,v,u',v'\}$ must appear on both sides. Let that largest element be $x_k$. Cancel it from both sides: we get $x_k+t=x_k+t'$ for some $t,t'\in S$, hence $t=t'$. This forces the pairs to agree. $\square$

Lemma 3 (A clean sufficient condition for adding a whole block $B$). Let $A\subset\{1,\dots,N\}$ be Sidon, and let $B\subset\{2N+1,\dots,M\}$ be Sidon. Assume additionally that \[(A-A)\cap(B-B)=\{0\}.\] Then $A\cup B$ is Sidon.

Proof. Sums in $A+A$ lie in $\{2,\dots,2N\}$. Since $\min(B)\ge 2N+1$, any sum in $A+B$ is at least $1+(2N+1)=2N+2$, and any sum in $B+B$ is at least $2(2N+1)=4N+2$. Hence there are no collisions between $A+A$, $A+B$, and $B+B$ by range separation.

It remains to show that sums in $A+B$ are unique. Suppose $a+b=a'+b'$ with $a,a'\in A$ and $b,b'\in B$. Then $a-a'=b'-b$ lies in both $A-A$ and $B-B$, hence equals $0$ by hypothesis. Thus $a=a'$ and $b=b'$. So cross sums are unique, and $A\cup B$ is Sidon. $\square$

FAST REALITY CHECK (local computation; one example). Take $N=20$ and the Sidon set $A=\{1,2,4,8,13\}\subset[1,20]$. I greedily extended $A$ by scanning $x=N+1,N+2,\dots,M$ and adding $x$ whenever Sidon-ness is preserved. Results: \[ \begin{array}{c|c|c} M & |A\cup B| & |A\cup B|/\sqrt M \\\hline 40 & 7 & 1.11\\ 100 & 11 & 1.10\\ 200 & 14 & 0.99\\ 400 & 18 & 0.90 \end{array} \] This shows that for this particular $A$, it is possible (at least for moderate $M$) to extend using only elements $>N$ and stay within a constant factor of $\sqrt M$. This is not evidence for the $(1-\varepsilon)$ asymptotic claim for arbitrary $A$.

5) VERIFICATION - Lemma 1: uses strict range separation ($x>2N$) to prevent collisions with $A+A$. - Lemma 2: the "largest element cancels" argument is valid because the growth condition forces sum ranges of different scales to be disjoint. - Lemma 3: checks both range separation and the cross-sum uniqueness reduced to disjointness of difference sets. - Computation: greedy procedure checks sums explicitly; the ratio $>1$ for small $M$ is possible because $f(M)$ can exceed $\sqrt M$ at small $M$.

6) FINAL UNRESOLVED

(i) Strongest fully proved partial result: Any Sidon $A\subset[1,N]$ can be extended to an infinite Sidon set (Lemma 2), and if one can find a Sidon block $B\subset[2N+1,M]$ with $(A-A)\cap(B-B)=\{0\}$, then $A\cup B$ is Sidon (Lemma 3).

(ii) First gap: Produce, for arbitrary given $A\subset[1,N]$ and $\varepsilon>0$, a large $B\subset[N+1,M]$ with $|A\cup B|\ge (1-\varepsilon)\sqrt M$. The obstacle is constructing $B$ so that it both is Sidon and avoids creating cross-sum collisions with $A$ at near-optimal density.

(iii) Top 3 next moves (concrete): 1. Quantify the forbidden-difference set $F:=A-A$ and attempt a probabilistic construction of a large Sidon $B\subset[2N+1,M]$ conditioned on $(B-B)\cap F=\{0\}$; estimate how large $|B|$ can be in terms of $M$ and $|F|$. 2. Explore whether a greedy extension algorithm can be analyzed: show that after extending to size $k$, the number of forbidden next elements is $O(k^2)$, which might still permit $k\approx (1-\varepsilon)\sqrt M$ if one has enough slack. 3. Search for explicit "worst-case" Sidon sets $A$ whose difference sets maximize obstructions, and test greedy/probabilistic extensions computationally for those.

(iv) What a minimal counterexample would likely look like (structure/parameters): A family of Sidon sets $A_N\subset[1,N]$ with $N\to\infty$ such that any Sidon extension using only elements $>N$ must have size $o(\sqrt M)$ in $[1,M]$, i.e. the initial differences in $A_N$ force pervasive cross-sum collisions that cannot be repaired by taking $M$ large.