gpt 5.2 attempt on Math-5386

OUTPUT 1) FORMAL RESTATEMENT Definitions. - The Schnirelmann density of $A\subseteq\mathbb N$ is \[d_s(A):=\inf_{N\ge 1} \frac{|A\cap\{1,\dots,N\}|}{N}.\] So $d_s(A)=\alpha$ means $|A\cap[1,N]|\ge \alpha N$ for every $N$ and the inequality is essentially sharp for some $N$. - For $b\in\mathbb N$, write $A+b:=\{a+b:a\in A\}$. - "$B$ is an additive basis" (as used in additive number theory) typically means: there exists $k\ge 1$ such that every sufficiently large integer is a sum of $k$ (not necessarily distinct) elements of $B$.

Question. Does there exist a set $B\subset\mathbb N$ that is not an additive basis, and a function $f:(0,1)\to (0,\infty)$, such that for every $\alpha\in(0,1)$, every $A\subseteq\mathbb N$ with $d_s(A)=\alpha$, and every $N\ge 1$, there exists $b\in B$ with \[ |(A\cup(A+b))\cap[1,N]| \ge (\alpha+f(\alpha))N? \]

2) QUICK LITERATURE/CONTEXT CHECK Web browsing is not available in this session. I only record what is stated in the problem text: - Erdős proved the inequality with gain $\alpha(1-\alpha)/(2k)$ when $B$ is an additive basis of order $k$. - There are connections to "essential components" and to work of Linnik. I have not verified any of these external claims.

3) ATTACK PLAN Proof track: 1. Derive necessary conditions on $B$ and on $f(\alpha)$ by choosing special sets $A$ with controlled translation behavior. 2. Try to leverage known structure theorems for Schnirelmann density and sumsets to build (or rule out) such a $B$.

Disproof track: 1. Find an obstruction: a family of sets $A$ (with fixed Schnirelmann density) such that for all $b$ in a candidate $B$, the union $A\cup(A+b)$ cannot gain a positive proportion. 2. In particular, test periodic sets $A$ and use gcd/parity obstructions for $B$.

Chosen path: obtain elementary necessary conditions on $f$ and $B$ (fully proved).

4) WORK Lemma 1 (Trivial upper bound on $f$). For any $\alpha\in(0,1)$, any $A\subseteq\mathbb N$, any $b\in\mathbb N$, and any $N\ge 1$, one has \[|(A\cup(A+b))\cap[1,N]|\le N.\] Consequently, any function $f$ appearing in the problem must satisfy \[f(\alpha)\le 1-\alpha\qquad(0<\alpha<1).\]

Proof. The set $(A\cup(A+b))\cap[1,N]$ is a subset of $[1,N]$, so its cardinality is at most $N$. If the problem's inequality holds, then for each $\alpha$ there exist choices with $(\alpha+f(\alpha))N\le N$, i.e. $f(\alpha)\le 1-\alpha$. $\square$

Lemma 2 (If the property holds, then $\gcd(B)=1$). Assume $B\subset\mathbb N$ satisfies the stated property for some function $f$ with $f(\alpha)>0$ for all $\alpha\in(0,1)$. Then $\gcd(B)=1$.

Proof. Suppose for contradiction that $g:=\gcd(B)\ge 2$. Then every $b\in B$ is a multiple of $g$. Define \[A:=\{n\in\mathbb N: n\not\equiv 0\pmod g\}.\] Claim: $d_s(A)=1-\frac{1}{g}$. Indeed, for any $N\ge 1$ we have \[|A\cap[1,N]| = N-\left\lfloor\frac{N}{g}\right\rfloor \ge N-\frac{N}{g} = \left(1-\frac{1}{g}\right)N,\] so $d_s(A)\ge 1-1/g$. For $N=g$ we have $|A\cap[1,g]|=g-1$, hence \[\frac{|A\cap[1,g]|}{g}=1-\frac{1}{g},\] so $d_s(A)\le 1-1/g$. This proves the claim.

Now fix any $b\in B$. Since $b\equiv 0\pmod g$, translation by $b$ preserves the residue class $0\pmod g$. Therefore $n\equiv 0\pmod g$ iff $n-b\equiv 0\pmod g$, which implies $A+b=A$. Hence for every $N$, \[|(A\cup(A+b))\cap[1,N]| = |A\cap[1,N]|.\] In particular, for $N$ a multiple of $g$ we have $|A\cap[1,N]|=(1-1/g)N$. Let $\alpha:=d_s(A)=1-1/g\in(0,1)$. The stated property would require that for this $A$ and this $N$ there exists $b\in B$ such that \[|(A\cup(A+b))\cap[1,N]|\ge (\alpha+f(\alpha))N.\] But the left-hand side equals $\alpha N$, so this inequality forces $f(\alpha)\le 0$, contradicting $f(\alpha)>0$. Thus $g=1$. $\square$

5) VERIFICATION - Lemma 1 is a pure set-theoretic bound and is exact. - Lemma 2 uses an explicit periodic test set $A$ and computes its Schnirelmann density exactly. The contradiction uses $N$ a multiple of $g$ so that $|A\cap[1,N]|=(1-1/g)N$.

6) FINAL UNRESOLVED

(i) Strongest fully proved partial result: Any admissible gain function must satisfy $f(\alpha)\le 1-\alpha$, and any $B$ with the stated property must have $\gcd(B)=1$ (Lemmas 1--2).

(ii) First gap: Either (a) construct a specific $B$ with $\gcd(B)=1$ that is not an additive basis but still forces a uniform Schnirelmann-density gain, or (b) prove that no such $B$ can exist even when $\gcd(B)=1$.

(iii) Top 3 next moves (concrete): 1. Try to strengthen Lemma 2: for any fixed finite set of residues $R\subset\mathbb Z/m\mathbb Z$, analyze sets $A$ that are unions of residue classes and quantify the best possible gain under translation by $b\in B$. 2. Search for further necessary conditions on $B$ by choosing $A$ as complements of sparse periodic sets (to test many $\alpha$). 3. Attempt constructive candidates for $B$ based on known "essential components" that are not additive bases, and check whether they can enforce a uniform gain $f(\alpha)>0$.

(iv) What a minimal counterexample would likely look like (if existence fails): A set $B$ with $\gcd(B)=1$ but so arithmetically constrained that one can build, for some $\alpha\in(0,1)$, a periodic (or almost periodic) set $A$ of Schnirelmann density $\alpha$ whose translates by all $b\in B$ overlap $A$ too heavily to yield any positive proportional gain.