Erdős Problem #92

Erdős Problem 92: many equidistant neighbors at every point

1) Statement and the source of a counterexample

The explicit external input is the theorem of OpenAI's Planar Point Sets with Many Unit Distances: there is a fixed $\delta>0$ and sets $P_j$ of $N_j\to\infty$ planar points whose unit-distance graphs have at least $N_j^{1+\delta}$ edges. The geometric part of a reconstruction is also given in the companion attempt for Problem 90; its arithmetic existence input is not reproved here.

2) Converting many edges into a minimum-degree bound

Apply this to the unit-distance graph on $P_j$. The remaining vertices form a planar set $Q_j$ of order $M_j\leq N_j$, with \[ \deg_{Q_j}(x)\geq E_j/N_j\geq N_j^\delta \quad\hbox{for every }x\in Q_j. \] All these neighbors have the same radius $1$ at every vertex, a stronger requirement than allowing a different radius at each point. Since degree is at most $M_j-1$, we also have \[ M_j\geq N_j^\delta+1\longrightarrow\infty. \] Finally, $M_j\leq N_j$ implies \[ f(M_j)\geq g(Q_j)\geq N_j^\delta\geq M_j^\delta. \] Thus the counterexamples occur at unbounded orders, not merely in a bounded core repeatedly extracted from larger sets.

3) Negating both asymptotic assertions

4) An elementary upper-bound consistency check

Outcome: SOLVED, known negative answer reconstructed using the explicit unit-distance theorem. The pruning argument and quantifier checks are proved in full. The batch verification exhausts small graphs to check the fixed-threshold pruning procedure; it is not a formal or numerical verification of the number-field theorem.

5) Sources

• OpenAI, Planar Point Sets with Many Unit Distances, Theorem 1.1 and the discussion immediately following it, https://cdn.openai.com/pdf/74c24085-19b0-4534-9c90-465b8e29ad73/unit-distance-proof.pdf.
• T. F. Bloom, Problem 92, https://www.erdosproblems.com/92, checked 23 September 2026.

6) COMPLETION ESTIMATE

1) FORMAL RESTATEMENT

I attach Erdős Problem #92 (equidistant points). The problem statement on the Erdős Problems site is: ([Erdős Problems][1])

Let (f(n)) be maximal such that there exists a set (A) of (n) points in $\mathbb{R}^2$ in which every $x\in A$ has at least (f(n)) points in (A) equidistant from (x). Is it true that $f(n)\le n^{o(1)}$? Or even $f(n) < n^{O(1/\log\log n)}$?

Definitions with explicit quantifiers

* Let $\mathbb N:={1,2,3,\dots}$. * For a finite set $A\subset\mathbb R^2$ and $x\in A$, define for each $r>0$ \[ N_A(x,r):=\left|{y\in A\setminus{x}:|x-y|=r}\right|. \] * Define \[ g(A):=\min_{x\in A}\ \max_{r>0} N_A(x,r). \] So (g(A)) is the largest integer (t) such that every $x\in A$ has some radius $r=r(x)>0$ with at least (t) other points of (A) on the circle centered at (x) of radius (r). * Then \[ f(n):=\max{g(A): A\subset\mathbb R^2,\ |A|=n}. \]

Meaning of the asymptotic expressions

* $f(n)\le n^{o(1)}$ means: \[ \forall \varepsilon>0\ \exists N(\varepsilon)\ \forall n\ge N(\varepsilon):\ f(n)\le n^\varepsilon. \] * $f(n)<n^{O(1/\log\log n)}$ means: there exists $C>0$ and $N_0$ such that for all $n\ge N_0$, \[ f(n)\le n^{C/\log\log n}. \] (Here $\log$ is the natural logarithm; changing base only changes (C) by a constant factor.)

Edge cases / stress points

* $f(1)=0$ (since there are no other points). * For $n\ge 2$, $f(n)\ge 1$ (take any two points). * The definition allows different radii (r(x)) for different (x). * No general position assumptions: collinearities, many concyclic points, etc. are allowed.

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2) QUICK LITERATURE/CONTEXT CHECK

The Erdős Problems page states the following known bounds and facts: ([Erdős Problems][1])

* Trivial upper bound: $f(n)\ll n^{1/2}$. * Pach–Sharir $1992$ implies $f(n)\ll n^{2/5}$. * A circle–point incidence bound of Janzer–Janzer–Methuku–Tardos $2024$ implies $f(n)\ll n^{4/11}$. * Lattice-point constructions give a lower bound $f(n)>n^{c/\log\log n}$ for some $c>0$. * Small exact values (from Fishburn/Erdős–Fishburn): $6$ is the smallest (n) with $f(n)=3$, and $8$ is the smallest (n) with $f(n)=4$.

I will not assume any deeper “solved status” beyond this; the page is marked open and was last edited Dec 28, 2025. ([Erdős Problems][1])

For the main upper bound $f(n)\ll n^{4/11}$, I will use the explicit circle–point incidence theorem stated in JJMT24:

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3) ATTACK PLAN

Proof-track ideas (upper bounds)

1. Incidence-geometry reduction (circles centered at the points): Choose, for each $x\in A$, a circle centered at (x) that contains $\ge t$ points. Then total point–circle incidences ($\ge$ nt). Apply best known incidence bounds for points vs circles to upper bound incidences, hence bound (t). 2. Crude combinatorial incidence via circle–circle intersections: Use that two distinct circles intersect in at most two points to get $I=O(n^{3/2})\Rightarrow f(n)=O(n^{1/2})$. This is a clean baseline proof. 3. Try to exploit “centers belong to the point set”: Hope for an incidence improvement beyond general circle incidences. I don’t see a rigorous path to beat $4/11$ today without new ideas.

Disproof/construction ideas (lower bounds)

1. Lattice-grid construction with a “popular radius”: Use an $m\times m$ grid. If we can pick one integer (k) with many representations $k=a^2+b^2$ and with $\sqrt{k}\ll m$, then every grid point has many neighbors at distance $\sqrt{k}$. 2. Finite-geometry / algebraic constructions: Try to build configurations where each point is center of many circles rich in points. 3. Recursive “cluster + symmetry” constructions: Hard to keep the minimum over points large; adding clusters tends to create bad points.

Best path now: do $1$ for an upper bound and $1$ for a lower bound, both completely rigorously. This does not resolve the open question, but it gives the cleanest proved “progress envelope” around the conjecture.

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4) WORK

Lemma 0: Small-(n) sanity checks

* $f(1)=0$. * $f(2)=1$. * $f(3)=2$, because an equilateral triangle has each point at equal distance from the other two, and $f(3)\le 2$ since each point has only 2 other points. These don’t touch the asymptotic regime but confirm the definition behaves as intended.

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Part A. Upper bound $f(n)\ll n^{4/11}$ from circle–point incidences

Setup: circles associated to points

Let $A\subset\mathbb R^2$ with $|A|=n$. Suppose $g(A)\ge t$. By definition of (g(A)),

\[ \forall x\in A\ \exists r_x>0:\ N_A(x,r_x)\ge t. \]

For each $x\in A$, define the circle \[ C_x := {z\in\mathbb R^2:\ |z-x|=r_x}. \]

Let $\mathcal C:={C_x:\ x\in A}$. Then $|\mathcal C|=n$ because circles with different centers are different.

Define the incidence set \[ \mathrm{Inc}(A,\mathcal C):={(p,C)\in A\times\mathcal C:\ p\in C}, \] and let $I(A,\mathcal C):=|\mathrm{Inc}(A,\mathcal C)|$.

Claim 1. (I$A,\mathcal C$$\ge$ nt).

Proof. For each $x\in A$, the circle $C_x$ contains at least (t) points of $A\setminus{x}$ by construction, i.e. $|A\cap C_x|\ge t$. Summing over all $x\in A$, \[ I(A,\mathcal C)=\sum_{x\in A} |A\cap C_x|\ \ge\ \sum_{x\in A} t\ = nt. \] ∎

So bounding $I(A,\mathcal C)$ from above bounds (t).

External theorem: point–circle incidence bound (JJMT24)

I will use the following corollary (specialized to circles) from Janzer–Janzer–Methuku–Tardos $2024$:

If $\mathcal C$ is a collection of (n) circles in the plane and (P) is a set of (m) points, then \[ I(\mathcal C,P)=O\left(m^{2/3}n^{2/3}+m^{6/11}n^{9/11}+m+n\right). \]

Interpreting the big-(O): there exists an absolute constant $K>0$ such that for all finite (P) and circle collections $\mathcal C$, \[ I(\mathcal C,P)\le K\left(m^{2/3}n^{2/3}+m^{6/11}n^{9/11}+m+n\right), \] where $m=|P|$, $n=|\mathcal C|$.

Apply this with $P=A$ and $\mathcal C$ as above, so $m=n$. Then \[ I(A,\mathcal C)\le K\left(n^{2/3}n^{2/3}+n^{6/11}n^{9/11}+n+n\right) =K\left(n^{4/3}+n^{15/11}+2n\right). \]

For $n\ge 2$, $n^{4/3}\le n^{15/11}$ because $15/11-4/3=1/33>0$, and $2n\le 2n^{15/11}$ for $n\ge 2$. Hence there is a constant $K'>0$ with \[ I(A,\mathcal C)\le K' n^{15/11}\qquad(n\ge 2). \]

Combine with Claim 1 ((nt$\le$ I)) to get \[ nt \le K' n^{15/11}\quad\Rightarrow\quad t\le K' n^{4/11}. \]

Since this holds for every (A) with $g(A)\ge t$, we conclude:

Theorem 1 (proved upper bound)

There exists an absolute constant $C>0$ such that for all $n\ge 2$, \[ f(n)\le C, n^{4/11}. \]

This matches the “$n^{4/11}$” upper bound reported on the Erdős Problems page. ([Erdős Problems][1])

∎

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Part B. A clean lattice lower bound $f(n)\ge n^{c/\log\log n}$ along an infinite sequence

I’ll prove a self-contained version of the lower bound claim from the problem page. ([Erdős Problems][1])

Number-theory input: the sum-of-two-squares representation function

Let \[ r_2(k):=\left|{(a,b)\in\mathbb Z^2:\ a^2+b^2=k}\right|. \] A standard formula (e.g., MathWorld) states that if (k) has prime factorization \[ k = 2^a \prod_i p_i^{\alpha_i}\prod_j q_j^{\beta_j}, \] where $p_i\equiv 1\pmod 4$ and $q_j\equiv 3\pmod 4$, then $r_2(k)=0$ if any $\beta_j$ is odd, and otherwise \[ r_2(k)=4\prod_i (\alpha_i+1). \] ([mathworld.wolfram.com][2])

In particular, if $k=\prod_{i=1}^t p_i$ is a product of (t) distinct primes $p_i\equiv 1\pmod 4$, then $\alpha_i=1$ for all (i) and \[ r_2(k)=4\cdot 2^t. \] ([mathworld.wolfram.com][2])

Geometric construction: an $m\times m$ integer grid

Fix an integer $m\ge 3$, and let \[ A:={1,2,\dots,m}^2\subset\mathbb R^2,\qquad n:=|A|=m^2. \]

Let (k) be a positive integer such that \[ \sqrt{k}\le \frac{m-1}{2} \quad\text{and}\quad k\ \text{is not a perfect square}. \] (The “not a square” condition ensures there are no solutions with a coordinate $0$, which simplifies a counting step.)

Claim 2. For every $x\in A$, there are at least $r_2(k)/4$ points $y\in A\setminus{x}$ with $|x-y|=\sqrt{k}$.

Proof.

Let $x=(x_1,x_2)\in A$. Consider the set \[ S:={(a,b)\in\mathbb Z^2:\ a>0,\ b>0,\ a^2+b^2=k}. \]

Step 1: $|S|=r_2(k)/4$. Because (k) is not a square, every integer solution ((a,b)) to $a^2+b^2=k$ has $a\ne 0$ and $b\ne 0$. For any solution ((a,b)), exactly one of the four sign choices $(\pm a,\pm b)$ lies in the first quadrant with $a>0,b>0$. Thus the mapping \[ (a,b)\mapsto (|a|,|b|) \] partitions the $r_2(k)$ solutions into 4-element sign orbits, and exactly one element of each orbit lies in (S). Therefore $|S|=r_2(k)/4$.

Step 2: For each $(a,b)\in S$, build a point $y\in A$ at distance $\sqrt{k}$ from (x). Fix $(a,b)\in S$. Because $a\le \sqrt{k}\le (m-1)/2$, we have $a\le m-x_1$ or else $x_1>m-a$. In the first case, set $s_1=+1$, so $x_1+s_1 a=x_1+a\le m$. In the second case $x_1>m-a$, and since $a\le (m-1)/2$ we have $m-a\ge a+1$, hence $x_1\ge a+1$, so setting $s_1=-1$ gives $x_1+s_1 a=x_1-a\ge 1$. In either case, we can choose $s_1\in{\pm1}$ such that $x_1+s_1 a\in{1,\dots,m}$.

Similarly, choose $s_2\in{\pm1}$ such that $x_2+s_2 b\in{1,\dots,m}$ (using $b\le (m-1)/2$).

Now define \[ y:=\left(x_1+s_1 a,\ x_2+s_2 b\right)\in A. \] Then \[ |x-y|^2 = (s_1 a)^2 + (s_2 b)^2 = a^2+b^2 = k, \] so $|x-y|=\sqrt{k}$. Also $y\neq x$ because $a,b>0$.

Step 3: Different $(a,b)\in S$ yield different points (y). Suppose $(a,b),(a',b')\in S$ yield the same (y) for the same fixed (x). Then \[ (s_1 a,\ s_2 b)=(s_1' a',\ s_2' b') \] for some $s_1,s_2,s_1',s_2'\in{\pm1}$. Taking absolute values gives $(a,b)=(a',b')$ since all coordinates are positive in (S). Hence the mapping $S\to A$ is injective, and (x) has at least $|S|=r_2(k)/4$ distinct neighbors in (A) at distance $\sqrt{k}$.

This proves Claim 2. ∎

Therefore $g(A)\ge r_2(k)/4$, and hence \[ f(n)=f(m^2)\ \ge\ r_2(k)/4. \]

Choosing (k) with large $r_2(k)$

We want many primes $\equiv 1\pmod 4$. The prime number theorem for arithmetic progressions gives \[ \pi_{4,1}(x)\sim \frac{\mathrm{Li}(x)}{\varphi(4)}=\frac{\mathrm{Li}(x)}{2}, \] so in particular $\pi_{4,1}(x)$ grows on the order of $x/\log x$. ([Wikipedia][3])

Let \[ X := (\log m)^2, \qquad t := \left\lfloor \frac{\log m}{\log\log m}\right\rfloor - 1. \] For all sufficiently large (m), the asymptotic $\pi_{4,1}(X)\sim X/(2\log X)$ implies $\pi_{4,1}(X)\ge t$, because \[ \frac{X}{\log X} \asymp \frac{\log^2 m}{\log\log m}\ \gg\ \frac{\log m}{\log\log m}\asymp t. \] (Here $\asymp$ and $\gg$ are in the standard asymptotic sense, and the inequality eventually holds because $\log^2 m$ dominates $\log m$.)

So for all sufficiently large (m), we can select (t) distinct primes $p_1,\dots,p_t\equiv 1\pmod 4$ with $p_i\le X$. Define \[ k:=\prod_{i=1}^t p_i. \] Then (k) is not a square and all its prime factors are $1\pmod4$, so (by the formula above) \[ r_2(k)=4\cdot 2^t. \tag{*} \] ([mathworld.wolfram.com][2])

Also, \[ k \le X^t = \exp(t\log X). \] Since $\log X = 2\log\log m$, we have \[ t\log X \le \left(\frac{\log m}{\log\log m}\right)\cdot 2\log\log m = 2\log m \] up to the “$-1$” slack in (t). More precisely, because $t \le \frac{\log m}{\log\log m}-1$, \[ t\log X \le 2\log m - 2\log\log m, \] hence \[ k \le \exp(2\log m - 2\log\log m)=\frac{m^2}{(\log m)^2}. \] For sufficiently large (m), $\frac{m^2}{(\log m)^2}\le \frac{(m-1)^2}{4}$, so $\sqrt{k}\le (m-1)/2$, as required to apply Claim 2.

Putting everything together: \[ f(m^2)\ \ge\ r_2(k)/4 \stackrel{(*)}= 2^t. \]

Finally, \[ 2^t \ge 2^{\frac12\cdot\frac{\log m}{\log\log m}} = \exp!\left(\frac{\log 2}{2}\cdot\frac{\log m}{\log\log m}\right) \] for all sufficiently large (m) (since (t) is asymptotic to $\frac{\log m}{\log\log m}$). Translating from (m) to $n=m^2$ changes $\log\log$ only by an additive constant $(\log\log(m^2)=\log(2\log m)=\log\log m + O(1))$, so there exists an absolute constant $c>0$ and an infinite sequence of (n) (namely $n=m^2$ for large (m)) such that \[ f(n)\ \ge\ n^{c/\log\log n}. \]

This is a fully rigorous lattice-style lower bound, consistent with the statement on the problem page. ([Erdős Problems][1])

∎

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5) VERIFICATION

Check of quantifiers and definitions

* In the incidence upper bound, I used: “$g(A)\ge t$” $\Rightarrow$ for each $x\in A$ choose a radius $r_x$ with at least (t) points on that circle. This is exactly the definition of (g(A)). * The incidence count inequality (I$\ge$ nt) is a direct sum over circles $C_x$; no multiplicity issues.

Possible failure modes in the incidence argument

* Are circles distinct? Yes: two circles with different centers cannot be identical as sets in $\mathbb R^2$. * Does JJMT24 apply to arbitrary circles? Corollary 1.12 explicitly states the bound for “a collection of circles” in the plane with no extra hypotheses. * Dominant term when $m=n$: I checked exponents $4/3<15/11$, so $n^{15/11}$ dominates for large (n).

Possible failure modes in the lattice lower bound

* Boundary effects: handled by forcing $\sqrt{k}\le (m-1)/2$, guaranteeing that for every coordinate, at least one sign choice keeps you inside ${1,\dots,m}$. * Injection from representations to points: proved: if two signed displacement vectors coincide, their absolute values coincide; because we restrict to $a,b>0$, this forces the representation pair to coincide. * Existence of enough primes $1\bmod 4$ below $(\log m)^2$: follows from the prime number theorem for arithmetic progressions. ([Wikipedia][3])

Everything above checks out.

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6) FINAL — UNRESOLVED

(i) Strongest fully proved partial results obtained here

* Upper bound (proved): There is an absolute (C) such that for all $n\ge 2$, \[ f(n)\le C,n^{4/11}, \] via the JJMT24 point–circle incidence theorem. ([Erdős Problems][1])

* Lower bound along an infinite sequence (proved): There exists an absolute $c>0$ and infinitely many (n) (e.g. large squares $n=m^2$) such that \[ f(n)\ge n^{c/\log\log n}, \] via an explicit $m\times m$ lattice construction and a choice of (k) with many sum-of-two-squares representations. ([mathworld.wolfram.com][2])

(ii) Exact first gap

I cannot prove either:

* $f(n)\le n^{o(1)}$ (i.e. $f(n)\le n^\varepsilon$ for every fixed $\varepsilon>0$ and all large (n)), nor * an explicit counterexample showing $f(n)\ge n^\varepsilon$ for some fixed $\varepsilon>0$ infinitely often (which would refute $n^{o(1)}$).

(iii) Top 3 next moves

1. Exploit the extra structure “each circle’s center is itself a point of (A)” to seek an incidence bound better than the general $m^{6/11}n^{9/11}$ term (which is what forces exponent $4/11$). Concretely: prove a specialized bound on incidences between a point set and circles centered at those points. 2. Try an energy/double-counting approach on bisectors: incidences $y\in C_x$ correspond to constraints $|x-y|=|x-y'|$. Large (f(n)) forces many equal-distance relations; rephrase as many point incidences on perpendicular bisectors and apply line-incidence technology. 3. Search for better constructions than grids: the grid lower bound is only $n^{c/\log\log n}$. Try finite-field or algebraic constructions (e.g., projections of higher-dimensional lattice shells, or structured sets with many equal norms) that keep every point “rich”.

(iv) What a minimal counterexample would likely look like (if $f(n)\le n^{o(1)}$ is false)

To refute $f(n)\le n^{o(1)}$, one would need a family $A_n$ where every point is the center of a circle containing $n^\varepsilon$ other points, with $\varepsilon>0$ fixed. That would force:

* a very dense incidence pattern between points and circles, * many points lying on many circles with controlled intersections, * likely a highly algebraic configuration (lattice- or finite-geometry-like), because random sets won’t produce $n^\varepsilon$-rich circles for all centers.

That’s the current boundary between what is proved (incidence bound $n^{4/11}$) and what’s conjectured (subpolynomial).

[1]: https://www.erdosproblems.com/92 " Erdős Problem #92 " [2]: https://mathworld.wolfram.com/SumofSquaresFunction.html "Sum of Squares Function -- from Wolfram MathWorld" [3]: https://en.wikipedia.org/wiki/Prime_number_theorem "Prime number theorem - Wikipedia"

OUTPUT 1) FORMAL RESTATEMENT Definitions/Conventions. For a finite set $A\subset\mathbb R^2$ and a point $x\in A$, define \[g_A(x):=\max_{r>0}\big|\{y\in A\setminus\{x\}: |x-y|=r\}\big|,\] the maximum number of other points at a common distance from $x$. Define \[f(n):=\max\left\{t:\ \exists A\subset\mathbb R^2,\ |A|=n,\ \text{and } g_A(x)\ge t \text{ for all }x\in A\right\}.\]

Question. Is it true that $f(n)\le n^{o(1)}$ (or even $f(n)<n^{O(1/\log\log n)}$)?

2) QUICK LITERATURE/CONTEXT CHECK Web browsing is not available in this session. I only use the "trivial" $f(n)\ll n^{1/2}$ claim from the statement and re-prove it below.

3) ATTACK PLAN Proof track ideas: 1. For upper bounds, translate the condition "each $x$ has many equidistant neighbors" into a point-circle incidence problem and apply extremal graph bounds. 2. For lower bounds, build structured configurations (lattices, circular constructions) where each point has many equidistant neighbors.

Disproof track ideas: 1. If one could construct $A$ with $g_A(x)\ge n^\delta$ for some fixed $\delta>0$ for all $x$, that would refute $n^{o(1)}$.

Chosen path in this attempt: prove the clean $O(\sqrt n)$ upper bound via a Kővári--Sós--Turán argument, and provide a simple universal lower bound.

4) WORK Lemma 1 (Trivial upper bound $f(n)=O(\sqrt n)$). For all $n\ge 1$, \[f(n)\le 2\sqrt n+2.\]

Proof. Let $A$ be a set of $n$ points witnessing $f(n)$, so that $g_A(x)\ge f(n)$ for all $x\in A$. For each $x\in A$, choose a radius $r_x>0$ such that \[\big|\{y\in A\setminus\{x\}: |x-y|=r_x\}\big| \ge f(n).\] Consider a bipartite graph $B$ with left part $L=A$ (centers) and right part $R=A$ (points), and put an edge from $x\in L$ to $y\in R$ iff $|x-y|=r_x$. Then every left vertex has degree at least $f(n)$, so \[e(B)\ge n f(n).\tag{1}\]

Geometric input: two circles intersect in at most two points. For distinct centers $x\ne x'$, the circles with radii $r_x$ and $r_{x'}$ (centered at $x$ and $x'$) intersect in at most two points. Thus $x$ and $x'$ have at most two common neighbors in $R$. Equivalently, $B$ is $K_{2,3}$-free with the 2-vertex side in $L$.

We now bound $e(B)$ directly. Let $d(r)$ be the degree of a right vertex $r\in R$. Count length-2 paths $\ell-r-\ell'$ with $\ell\ne \ell'$ and $r\in R$: \[P:=\sum_{r\in R}\binom{d(r)}{2}.\] For a fixed unordered pair $\{\ell,\ell'\}\subseteq L$, the two circles (centered at $\ell$ and $\ell'$ with radii $r_\ell$ and $r_{\ell'}$) intersect in at most two points, so $\ell,\ell'$ have at most $2$ common neighbors in $R$. Thus $P\le 2\binom{n}{2}<n^2$.

Also, \[P=\frac12\sum_r d(r)^2-\frac12\sum_r d(r)\ge \frac12\cdot\frac{e(B)^2}{n}-\frac12 e(B)\] by Cauchy--Schwarz and $\sum_r d(r)=e(B)$. Therefore $\frac{e(B)^2}{2n}-\frac{e(B)}{2}\le n^2$, which implies $e(B)\le 2n^{3/2}+2n$. Combine with $1$ to obtain \[n f(n)\le 2n^{3/2}+2n,\] so $f(n)\le 2\sqrt n+2$. $\square$

Lemma 2 (A simple lower bound $f(n)\ge 2$ for $n\ge 3$). For every $n\ge 3$, one has $f(n)\ge 2$.

Proof. Let $A$ be the vertex set of a regular $n$-gon. For each vertex $x\in A$, its two adjacent vertices are at the common distance equal to the side length. Thus $g_A(x)\ge 2$ for all $x$, so $f(n)\ge 2$ by definition. $\square$

FAST REALITY CHECK Lemma 1 matches the "trivial" bound $f(n)\ll n^{1/2}$ mentioned in the problem text, with an explicit constant. The problem asks whether one can improve this all the way to $n^{o(1)}$, which is far beyond this elementary argument.

5) VERIFICATION - Lemma 1 uses only the circle intersection fact and the KST extremal bound for $K_{2,3}$-free bipartite graphs. - Lemma 2 uses a concrete configuration (regular polygon) and checks $g_A(x)\ge 2$.

6) FINAL UNRESOLVED

(i) Strongest fully proved partial result: I proved the explicit upper bound $f(n)\le 2\sqrt n+2$ (Lemma 1) and a universal lower bound $f(n)\ge 2$ for $n\ge 3$ (Lemma 2).

(ii) First gap: Improve the upper bound from $\Theta(\sqrt n)$ to $n^{o(1)}$ (or find a construction with $f(n)\ge n^\delta$ for some fixed $\delta>0$).

(iii) Top 3 next moves (concrete): 1. Replace the crude $K_{2,3}$-free extremal bound with sharper point-circle incidence theorems to push the exponent below $1/2$. 2. Explore whether additional structure forced by the "for every $x$" condition can yield stronger global constraints than pairwise circle intersections. 3. Try to explicitly construct sets where each point is a center of a circle containing many other points (e.g. lattice-based constructions) and quantify $g_A(x)$.

(iv) What a minimal counterexample would likely look like (structure/parameters): A family of point sets $A_n$ with $|A_n|=n$ and $g_{A_n}(x)\ge n^\delta$ for all $x\in A_n$ for some fixed $\delta>0$, showing that $f(n)$ is polynomially large and disproving $f(n)\le n^{o(1)}$.