codex 5.2 extra high attempt on Math-643

OUTPUT 1) FORMAL RESTATEMENT Let $A\subseteq\mathbb N$. For each integer $n$, consider the number of representations \[n=a+b,\qquad a,b\in A,\ a\le b.\] Let $B$ be the set of $n$ for which there is exactly one such representation. Questions: (a) Is it true that for all $\epsilon>0$ and large $N$, $|\{1,\dots,N\}\setminus B|\gg_\epsilon N^{1/2-\epsilon}$? (b) Can it happen that $|\{1,\dots,N\}\setminus B|=o(N^{1/2})$?

2) QUICK LITERATURE/CONTEXT CHECK Browsing is not available. I only record what the problem statement itself claims. I have not verified those results here.

3) ATTACK PLAN Proof track: 1. Relate unique representation counts to additive energy of $A$. 2. Use sumset growth to force many collisions in $A+A$.

Disproof track: 1. Construct very thin $A$ (Sidon-type) to maximize uniqueness. 2. Construct dense $A$ to force many multiple representations and large complement.

Chosen path: prove elementary bounds and work out a concrete example.

4) WORK Lemma 1 (Trivial bound via number of pairs). Let $A_N=A\cap\{1,\dots,N\}$ and $m=|A_N|$. Then \[|B\cap\{1,\dots,N\}|\le \binom{m+1}{2}. \]

Proof. Each unordered pair $(a,b)$ with $a\le b$ gives at most one integer $n=a+b$ with a unique representation. There are $\binom{m+1}{2}$ such pairs. $\square$

Lemma 2 (Unique sums for an interval). Let $A=\{1,2,\dots,m\}$ with $m\ge 3$. Then the integers with a unique representation as $a+b$ with $a,b\in A$ are exactly $2,3,2m-1,2m$.

Proof. For $s\le m+1$, the number of pairs $(a,b)$ with $a\le b$ and $a+b=s$ equals $\lfloor s/2\rfloor$: indeed $a$ can be any integer from $1$ to $\lfloor s/2\rfloor$, and then $b=s-a$ automatically satisfies $a\le b$ and $b\le m$. Thus $s$ has a unique representation iff $\lfloor s/2\rfloor=1$, i.e., $s=2$ or $3$. For $s>m+1$, the number of representations of $s$ equals that of $2m+2-s$ by the map $(a,b)\mapsto(m+1-b, m+1-a)$, so unique sums are also $2m-1$ and $2m$. $\square$

Corollary 3 (Large complement example). For $A=\{1,\dots,m\}$ and $N\ge 2m$, we have $|\{1,\dots,N\}\setminus B|\ge N-4$.

Proof. By Lemma 2, $B\cap\{1,\dots,2m\}$ has size $4$. All other integers up to $N$ lie outside $B$. $\square$

5) VERIFICATION - Lemma 2 uses an explicit count of representations in an interval, which can be checked directly. - Corollary 3 applies only to the specific choice $A=\{1,\dots,m\}$.

6) FINAL UNRESOLVED

(i) Strongest fully proved partial result: For any $A$, $|B\cap[1,N]|\le \binom{|A\cap[1,N]|+1}{2}$, and for $A=\{1,\dots,m\}$ we have $|B\cap[1,2m]|=4$.

(ii) First gap: No general lower bound on $|\{1,\dots,N\}\setminus B|$ of order $N^{1/2}$ is proved.

(iii) Top 3 next moves: 1. Relate $|\{1,\dots,N\}\setminus B|$ to additive energy and use sumset estimates. 2. Study Sidon-type constructions where $A+A$ has mostly unique sums. 3. Optimize between dense and sparse $A$ to minimize the complement size.

(iv) Minimal counterexample structure (if the lower bound were false): A family of sets $A$ whose sumset $A+A$ has unusually many unique representations, forcing $|\{1,\dots,N\}\setminus B|=o(N^{1/2})$.