gpt pro 5.2 attempt on Math-6015

Erdos Problem 956.

1) FORMAL RESTATEMENT. For compact sets $C,D\subseteq\mathbb{R}^2$, define the Euclidean distance between them by \[ \delta(C,D):=\inf\{\|c-d\|: c\in C,\ d\in D\}. \] For $n\in\mathbb{N}$, define $h(n)$ as the maximum possible number of pairs $\{x_1,x_2\}$ with $x_1\ne x_2$ in a set $X\subseteq\mathbb{R}^2$ of size $n$ such that there exists a compact convex set $C\subseteq\mathbb{R}^2$ with:

• The translates $\{C+x: x\in X\}$ are pairwise disjoint, and

• For each counted pair $\{x_1,x_2\}$ we have $\delta(C+x_1, C+x_2)=1$.

Question: determine $h(n)$. In particular, is there a constant $c>0$ such that for all sufficiently large $n$, \[ h(n) > n^{1+c}? \]

2) QUICK LITERATURE/CONTEXT CHECK. The problem file states a known upper bound $h(n)\ll n^{4/3}$ due to Erdős and Pach, and asks for superlinear lower bounds.

3) ATTACK PLAN. (A) Rewrite distances between translates in terms of translation vectors and the Minkowski difference $C-C$. (B) Compare $h(n)$ to the classical unit-distance extremal function $f(n)$ for point sets. (C) Do small $n$ sanity checks (what happens for $n=2,3$?).

4) WORK.

FAST REALITY CHECK (tiny $n$). For $n=2$, choose any compact convex $C$ and two translation vectors so that the two disjoint translates are exactly distance $1$ apart; hence $h(2)=1$. For $n=3$, taking $C$ to be a disk of radius $r>0$ and placing centers at the vertices of an equilateral triangle of side length $2r+1$ gives three disjoint translates with pairwise distance $1$, so $h(3)=3$.

Lemma 956.1 (Distance between translates via Minkowski difference). Let $C\subseteq\mathbb{R}^2$ be nonempty and compact. Define the (compact) set \[ K:=C-C:=\{c_1-c_2: c_1,c_2\in C\}. \] Then for all $x,y\in\mathbb{R}^2$, \[ \delta(C+x, C+y) = \operatorname{dist}(y-x, K), \] where $\operatorname{dist}(v,K):=\inf\{\|v-k\|:k\in K\}$.

Proof. By definition, \[ \delta(C+x,C+y)=\inf_{c_1,c_2\in C}\| (c_1+x)-(c_2+y)\| =\inf_{c_1,c_2\in C}\|(c_1-c_2)-(y-x)\|. \] The set of differences $c_1-c_2$ as $c_1,c_2$ range over $C$ is exactly $K$, so the right-hand side equals $\inf_{k\in K}\|k-(y-x)\|=\operatorname{dist}(y-x,K)$. $\square$

Lemma 956.2 (Trivial reduction: $h(n)\ge f(n)$). Let $f(n)$ denote the maximum number of unit-distance pairs among $n$ points in the plane (the classical unit distance problem). Then $h(n)\ge f(n)$ for all $n$.

Proof. Let $P\subset\mathbb{R}^2$ be a set of $n$ points determining $f(n)$ unit-distance pairs. Let $d_{\min}:=\min\{\|p-q\|:p,q\in P,\ p\ne q\}>0$. Choose $r>0$ so small that $(1+2r)d_{\min}>2r$. Let $C$ be the closed disk of radius $r$ centered at the origin. Define $X:=(1+2r)P:=\{(1+2r)p:p\in P\}$.

Disjointness: for any distinct $p,q\in P$ we have \[\|(1+2r)p-(1+2r)q\|=(1+2r)\|p-q\|\ge (1+2r)d_{\min}>2r,\] so the disks $C+(1+2r)p$ and $C+(1+2r)q$ are disjoint.

Unit-distance pairs transfer: if $\|p-q\|=1$, then the distance between the two disks equals \[\delta\left(C+(1+2r)p,\ C+(1+2r)q\right)=\|(1+2r)(p-q)\|-2r=(1+2r)\cdot 1-2r=1.\] Thus every unit-distance pair in $P$ yields a unit-distance pair of disjoint translates. Therefore $h(n)\ge f(n)$. $\square$

5) VERIFICATION. • Lemma 956.1 is an exact identity obtained by rewriting the infimum. • Lemma 956.2 depends only on choosing $r$ small enough compared to the minimum separation in $P$.

6) FINAL. UNRESOLVED. (i) Strongest proved partial result: $h(n)\ge f(n)$ for all $n$, and for disks $C$ the translate-distance condition reduces to a fixed-distance condition among centers. (ii) First gap: produce an explicit construction of disjoint translates with more than $n^{1+c}$ pairs at distance $1$ for some fixed $c>0$ (without appealing to external results about $f(n)$ beyond the definition). (iii) Top 3 next moves: $1$ Choose special convex bodies $C$ (very elongated, “almost flat”) and try to construct configurations where many translation differences lie on the distance-$1$ offset of $C-C$. $2$ Attempt to adapt lattice-based constructions: place translation vectors on a grid and choose $C$ so that $\operatorname{dist}(\cdot,C-C)=1$ holds for many displacement vectors. $3$ Explore whether the disjointness constraint can be relaxed via packing arguments to permit denser graphs than in the classical unit distance problem. (iv) Minimal counterexample structure: if the conjectured superlinear lower bound fails, then for every convex $C$ and every $X$ of size $n$, the number of differences $y-x$ at distance exactly $1$ from $C-C$ is $O(n^{1+o(1)})$.