gpt pro 5.2 attempt on Math-643

1) FORMAL RESTATEMENT

Let $A\subseteq\mathbb{N}$. For an integer $n\ge 2$, define the number of unordered representations of $n$ as a sum of two elements of $A$ by \[ r_A(n):=|\{\{a,b\}: a\in A,\ b\in A,\ a\le b,\ a+b=n\}|. \] (So $a=b$ is allowed; it contributes one representation.) Define \[ B:=\{n\in\mathbb{N}: r_A(n)=1\}. \] The questions are:

1. $Q1$ Is it true that for every $\varepsilon>0$ there exists $N_0(\varepsilon)$ such that for all $N\ge N_0(\varepsilon)$, \[ |\{1,\dots,N\}\setminus B| \gg_{\varepsilon} N^{1/2-\varepsilon}? \]
2. $Q2$ Is it possible that $|\{1,\dots,N\}\setminus B|=o(N^{1/2})$ as $N\to\infty$ for some choice of $A$?

Ambiguity note: some authors count ordered representations $(a,b)$ instead of unordered, or forbid $a=b$. The qualitative questions change only by constant factors in typical regimes, but for small $n$ the distinction matters; I proceed with the (common) convention above.

2) QUICK LITERATURE/CONTEXT CHECK

From the problem statement: Erdős claimed a construction with $|\{1,\dots,N\}\setminus B|\ll_\varepsilon N^{1/2+\varepsilon}$ for all large $N$ and also infinitely many $N$ with lower bound $\gg_\varepsilon N^{1/3-\varepsilon}$. Erdős--Freud proved a finite analogue giving an upper bound $<2^{3/2}N^{1/2}$ for some $A\subseteq\{1,\dots,N\}$.

I do not use these claims in proofs; I only give elementary constraints and small-$N$ computations.

3) ATTACK PLAN

Proof track (lower bounds on the complement):

• Relate the number of uniquely represented sums to the additive energy of $A\cap[1,N]$; collisions among sums should force many $n$ with $r_A(n)\ne 1$.
• Use extremal counting: the number of available sums is $N$, but the number of pairs in $A\cap[1,N]$ is quadratic; this tension should force either many missing sums (no representations) or many collisions (multiple representations).

Disproof track (constructing small complement):

• Try to build $A$ so that most $n\le N$ have exactly one representation, balancing coverage (most $n$ represented) with Sidon-type uniqueness (few collisions).
• Computationally search for small $N$ optimal sets and look for a pattern that might extend to an infinite construction.

4) WORK

Lemma 14.1 (local dependence on initial segment). Fix $N\ge 1$. For any $n\le N$, whether $n\in B$ (i.e. $r_A(n)=1$) depends only on the finite set $A\cap\{1,\dots,N\}$.

Proof. If $n\le N$ and $a,b\in A$ satisfy $a+b=n$, then necessarily $a\le n\le N$ and $b\le n\le N$. Thus every representation of $n$ uses only elements of $A\cap[1,N]$. Conversely, any representation using elements of $A\cap[1,N]$ is a representation using elements of $A$. Therefore $r_A(n)=r_{A\cap[1,N]}(n)$ for all $n\le N$. $\square$

Lemma 14.2 (small-number obstruction via the minimum of $A$). Let $m:=\min A$ (assuming $A\ne\emptyset$). Then for every integer $n$ with $1\le n<2m$, one has $r_A(n)=0$, hence $n\notin B$. In particular, \[ |\{1,\dots,N\}\setminus B|\ge \min\{N,2m-1\}. \]

Proof. If $a,b\in A$, then $a\ge m$ and $b\ge m$, so $a+b\ge 2m$. Thus no integer $n<2m$ can be represented as $a+b$ with $a,b\in A$, i.e. $r_A(n)=0$. Counting gives the displayed lower bound for the complement. $\square$

Lemma 14.3 (pair-counting upper bound in the collision-free regime). Fix $N\ge 1$ and set $A_N:=A\cap\{1,\dots,N\}$ with $|A_N|=t$. If every sum $a+b$ with $a,b\in A_N$ and $a\le b$ is distinct (i.e. $r_{A_N}(n)\le 1$ for all $n$), then \[ \binom{t+1}{2} \le 2N-1, \] hence \[ t \le \frac{\sqrt{16N-7}-1}{2} < 2\sqrt{N}. \]

Proof. There are exactly $\binom{t+1}{2}$ unordered pairs $\{a,b\}$ with $a,b\in A_N$ and $a\le b$. Each such pair produces a sum $s=a+b$ in the range $2\le s\le 2N$. If all these sums are distinct, then the number of pairs is at most the number of possible sums in $[2,2N]$, which is $2N-1$. This gives $\binom{t+1}{2}\le 2N-1$. Solving the quadratic inequality $t(t+1)\le 4N-2$ yields the stated bound. $\square$

FAST REALITY CHECK / COMPUTATION. For each $N\le 20$, I brute-forced over all subsets $A\subseteq\{1,\dots,N\}$ and computed \[ \mathrm{bad}(A,N):=|\{1,\dots,N\}\setminus B| \] with $B$ defined via unordered representations $a\le b$. The minimum value $\min_A \mathrm{bad}(A,N)$ for $N\le 20$ is: \[ \begin{array}{c|cccccccccccccccccccc} N & 1&2&3&4&5&6&7&8&9&10&11&12&13&14&15&16&17&18&19&20\\\hline \min \mathrm{bad}(A,N) & 1&1&1&1&1&1&1&2&2&2&2&2&2&2&2&2&2&3&3&3 \end{array} \] Example optimizers found by brute force include:

• $N=16$: $A=\{1,2,4,6,7,14\}$ gives complement $\{1,8\}$ (size $2$).
• $N=20$: $A=\{1,3,4,8,9,11\}$ achieves complement size $3$.

These are not consistent initial segments of a single infinite $A$; they are separately optimized for each $N$.

5) VERIFICATION

• Lemma 14.1 uses only positivity of elements of $A$.
• Lemma 14.2: verified the inequality $a+b\ge 2m$ is sharp when $a=b=m$ (allowed in our convention).
• Lemma 14.3: checked that all sums lie in $[2,2N]$ and the count of possible sums is $2N-1$.
• Computation: brute force over $2^N$ subsets is exact for $N\le 20$.

6) FINAL

UNRESOLVED

(i) Strongest proved partial result: Lemma 14.2 gives a necessary lower bound on the complement in terms of $\min A$, forcing $1\in A$ if one hopes for $o(N)$ complement. Lemma 14.3 gives a sharp constraint in the extreme "no collisions" regime. Computationally, optimizing $A\subseteq[1,N]$ separately for each $N\le 20$ yields very small complements (as low as $2$ or $3$).

(ii) First gap (crisp): Either prove that for every infinite $A\subseteq\mathbb{N}$, \[ |\{1,\dots,N\}\setminus B| \ge c_\varepsilon N^{1/2-\varepsilon} \quad\text{for all large }N, \] for some $c_\varepsilon>0$, or construct a single infinite $A$ with $|\{1,\dots,N\}\setminus B|=o(N^{1/2})$.

(iii) Top 3 next moves:

1. Search for a consistent infinite construction: attempt to extend optimized small-$N$ sets by a backtracking algorithm imposing that $A\cap[1,N]$ is the restriction of $A\cap[1,N+1]$.
2. Develop an energy-based inequality: express counts of $r_A(n)=0,1,\ge 2$ in terms of $|A\cap[1,N]|$ and additive energy, then lower bound the number of "bad" $n$.
3. Compute exact minima for larger $N$ via ILP/SAT (variables indicating membership of each $a\le N$) to get better evidence about growth of $\min_A \mathrm{bad}(A,N)$.

(iv) Minimal counterexample structure: a counterexample to Q1 (i.e. showing $|[1,N]\setminus B|=o(\sqrt{N})$ is possible) would require an infinite $A$ with $1\in A$ and such that for most $n\le N$ there is exactly one pair $a\le b$ in $A\cap[1,N]$ summing to $n$. This simultaneously demands (a) near-complete coverage of $[1,N]$ by pair-sums and (b) strong Sidon-type control on collisions among sums in the range $\le N$.