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Claim. The answer to Problem 97 is no: there is a convex polygon each of whose vertices has at least four other vertices at distance exactly $1$ from it, so every vertex has four other vertices equidistant from it and no vertex is of the kind the question asks for. More generally, for every integer $k\ge2$ there is a convex polygon each of whose vertices has at least $k$ other vertices at distance $1$, which refutes the general form of the question, in which some fixed $k$ replaces four. The claim is a deduction made in this corpus from the accepted disproof of Problem 96 by Liam Kruer and Jensen Kohlmeyer, whose explanation states no consequence for this problem. The polygon's vertices are a subset of their set with $d=13$, so by their Theorem 1.1 it has at most $N_{13}=2\cdot13^{13}\,2^{13^{14}}$ vertices; the proof gives no coordinates.

Scope. Full: the polygon has no vertex of the kind the question asks for.

Depends on. The accepted claim on Problem 96, where Theorem 1.1 of Kruer and Kohlmeyer supplies, with $d=13$, a finite set in strictly convex position with more than three unit pairs per point; the negation of the bound $O(n)$ accepted there, applied with the constant $3$, supplies such a set as well. The deletion step and the passage to a polygon are proved below and use nothing else.

Proof. For a finite planar set $S$, write $u(S)$ for the number of unordered pairs of points of $S$ at distance exactly $1$, and $\deg_S(p)$ for the number of points of $S$ at distance $1$ from a point $p\in S$. Call $S$ in strictly convex position when no point of $S$ lies in the convex hull of the other points of $S$.

1. The input. Theorem 1.1 of Kruer and Kohlmeyer, as the claim page on Problem 96 records it, gives for every integer $d\ge2$ a set $V_d$ in strictly convex position with $|V_d|=N_d=2d^d\,2^{d^{d+1}}$ and $u(V_d)\ge\frac d4N_d$. With $d=13$ the set $V=V_{13}$ satisfies $u(V)\ge\frac{13}4|V|>3|V|$.

2. Deletion. Set $V_0=V$. While the current set $V_i$ has a point $p$ with $\deg_{V_i}(p)\le3$, choose one such point and set $V_{i+1}=V_i\setminus\{p\}$. Removing $p$ removes exactly the $\deg_{V_i}(p)$ unit pairs that contain $p$ and leaves every other pair, so $u(V_{i+1})=u(V_i)-\deg_{V_i}(p)\ge u(V_i)-3$, while $|V_{i+1}|=|V_i|-1$. Hence $\varphi(S)=u(S)-3|S|$ satisfies $\varphi(V_{i+1})\ge\varphi(V_i)$, and since $\varphi(V_0)>0$, every $V_i$ has $\varphi(V_i)>0$. Each step removes a point, so the process stops after at most $|V|$ steps, at a set $W$. Then $u(W)>3|W|\ge0$, so $W$ is nonempty, and by the stopping rule every point of $W$ has at least four other points of $W$ at distance $1$; in particular $|W|\ge5$.

3. Convex position. $W$ is a subset of $V$. If a point $p\in W$ lay in the convex hull of $W\setminus\{p\}$, it would lie in the convex hull of the larger set $V\setminus\{p\}$, against the strictly convex position of $V$. So $W$ is in strictly convex position. In particular no three points of $W$ are collinear, since the middle one would lie in the segment joining the other two, so the convex hull $P$ of $W$ is a convex polygon with nonempty interior. Every vertex of $P$ is a point of $W$, and every point of $W$ is a vertex of $P$: if a point of $W$ were not a vertex, it would be a convex combination of the vertices, which are other points of $W$. So the vertices of $P$ are exactly the points of $W$.

4. Conclusion. Every vertex of $P$ has at least four other vertices at the common distance $1$, so no vertex of $P$ has no four other vertices equidistant from it, and $P$ refutes the question. The common distance is $1$ at every vertex, which the question does not require.

5. The general form. For an integer $k\ge2$, take $d=4k-3\ge5$ in Theorem 1.1, so that $u(V_d)\ge\frac{4k-3}4|V_d|>(k-1)|V_d|$, and delete points with at most $k-1$ others at distance $1$, with $\varphi(S)=u(S)-(k-1)|S|$. The same argument leaves a set of at least $k+1\ge3$ points in strictly convex position, the vertex set of a convex polygon each of whose vertices has at least $k$ other vertices at distance $1$. The case $k=4$ is the polygon above.

The negation of the bound $O(n)$ that the accepted Lean proof establishes gives the input of step 1 directly: applied with the constant $3$, it gives some $n$ for which the most unit pairs among $n$ points in strictly convex position exceeds $3n$, and since that maximum is taken over counts at most $\binom n2$, some $n$-point set in strictly convex position has more than $3n$ unit pairs; steps 2 to 4 then apply unchanged.

Relation to the sources. The erdosproblems.com remarks on Problems 96 and 97 note that a positive answer to this problem, for four equidistant vertices, would give by induction at most $3n$ unit pairs among $n$ points in convex position. The deduction here is the contrapositive of that remark, applied to the accepted theorem of Problem 96, with the induction written out as the deletion step above. The converse fails: a counterexample to this problem need not have many unit pairs, since this problem lets the common distance vary with the vertex. Kruer and Kohlmeyer's explanation of their disproof of Problem 96 does not apply the remark to their theorem, so the claim is the project's: that application, written out in full, with the polygon a subset of their set with $d=13$ and the common distance $1$ at every vertex. The same deduction is recorded in brief on the claim page on Problem 96. The deletion step is also stated in general form in the manuscript of Kruer, Kohlmeyer and Price of 13 September 2026, in its Lemma 4.1 and Remark 4.5, as its claim page records. That manuscript is unrefereed and a pending claim here, and it derives its own disproof of this problem from its own construction; this page relies neither on its lemma nor on its construction.

Standing. The deduction is the project's own and awaits the project's review, so the claim is claimed and lists no evidence, and the problem's standing is claimed through it and through the manuscript of Kruer, Kohlmeyer and Price. No Lean proof checks it: the accepted Lean proof of Problem 96, linked above as the input's formalization, proves only the negation of the bound $O(n)$, and the deletion step and the passage to a polygon are not formalized. The deduction was not submitted to erdosproblems.com, which labeled the problem falsifiable.