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Math-5797-A3 · revision r1 · on Math-5797

gpt pro 5.4 attempt on Math-5797

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Unverified LLM-generated attempt. Automated screening checks form, honesty and scope, not correctness. Status at erdosproblems.com: open. Claims here do not change the source status.

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Contents
  1. Erdős problems #655, #738, #740 --- Round 2 continuation
  2. Problem #655
  3. 1) ROUND-2 OBJECTIVE
  4. 2) Round-1 FOUNDATION USED
  5. 3) NEW INSIGHT / TOOL (ROUND-2)
  6. 4) ATTACK PLAN (ROUND-2)
  7. 5) WORK (ROUND-2)
  8. 6) ADVERSARIAL VERIFICATION
  9. 7) FINAL
  10. 8) COMPLETION ESTIMATE
  11. 9) REFERENCES
  12. References
  13. References

Erdős problems #655, #738, #740 --- Round 2 continuation

Problem #655

1) ROUND-2 OBJECTIVE

Path (C): turn the Round-1 disproof into the exact corrected statement. Round 1 already produced a regular $n$-gon counterexample, so the natural gap to close is the sharp lower bound that actually follows from the stated hypothesis.

2) Round-1 FOUNDATION USED

I use the following vetted Round-1 fact.

The regular $n$-gon satisfies the hypothesis of the problem and determines exactly $\lfloor n/2\rfloor$ distinct distances. Therefore the claimed bound $(1+c)n/2$ is false for every fixed $c>0$.

I do not re-prove that counterexample here.

3) NEW INSIGHT / TOOL (ROUND-2)

The hypothesis controls the multiplicity of each radius around a fixed point. If one fixes a point $x_i$, then among the $n-1$ distances from $x_i$ to the other points, each distinct value can occur at most twice. Therefore $x_i$ itself determines at least $\lceil (n-1)/2\rceil$ distinct distances. Since these are all global interpoint distances, the whole configuration determines at least that many distinct distances.

Combined with the Round-1 regular $n$-gon construction, this gives the exact optimum.

4) ATTACK PLAN (ROUND-2)

The only remaining gap after Round 1 is to identify the best possible lower bound replacing the false inequality \[ |D|\geq (1+c)\frac n2. \] I will prove the sharp correction \[ |D|\geq \left\lfloor\frac n2\right\rfloor, \] and then use the Round-1 regular $n$-gon to show equality is attainable for every $n$.

5) WORK (ROUND-2)

Let $x_1,\dots,x_n$ be distinct points in the plane satisfying the hypothesis that for every $i$ and every radius $\rho>0$, the circle centred at $x_i$ with radius $\rho$ contains at most two of the other points.

Let \[ D:=\{\,|x_i-x_j|:1\leq i<j\leq n\,\} \] be the global distance set.

For a fixed index $i$, define the set of distances seen from $x_i$ by \[ D_i:=\{\,|x_i-x_j|:j\neq i\,\}. \] Every value in $D_i$ belongs to $D$, so \[ |D|\geq |D_i| \qquad \text{for every } i. \] Now fix $i$. There are $n-1$ points other than $x_i$. By hypothesis, each radius about $x_i$ can hit at most two of them. Hence each element of $D_i$ accounts for at most two points, and therefore \[ 2|D_i|\geq n-1. \] So \[ |D_i|\geq \left\lceil\frac{n-1}{2}\right\rceil = \left\lfloor\frac n2\right\rfloor. \] Since $|D|\ge |D_i|$, we obtain the sharp universal lower bound \[ |D|\geq \left\lfloor\frac n2\right\rfloor. \]

Thus the exact corrected theorem is:

Corrected theorem. For every $n\ge 1$, every $n$-point set in the plane satisfying the stated hypothesis determines at least $\lfloor n/2\rfloor$ distinct distances.

To see that this is best possible, invoke the Round-1 construction: the $n$ vertices of a regular $n$-gon satisfy the hypothesis and determine exactly $\lfloor n/2\rfloor$ distinct distances. Therefore \[ \min |D| = \left\lfloor\frac n2\right\rfloor \] over all such $n$-point configurations.

6) ADVERSARIAL VERIFICATION

  • Quantifiers. The proof gives the lower bound for every $n\ge 1$, not merely for large $n$.
  • Odd/even parity. The identity $\lceil (n-1)/2\rceil=\lfloor n/2\rfloor$ handles both parity cases exactly.
  • Edge cases. For $n=1$, the distance set is empty and $\lfloor n/2\rfloor=0$. For $n=2$, there is exactly one distance and $\lfloor n/2\rfloor=1$. The argument still works.
  • No hidden assumption. The proof uses only the stated circle-multiplicity hypothesis; it does not assume convexity, general position, or any structure beyond distinctness of the points.
  • Sharpness. The regular $n$-gon from Round 1 attains equality, so the correction cannot be improved.

7) FINAL

FULL SOLUTION --- COUNTEREXAMPLE / DISPROOF

The original statement is false. The exact corrected statement is: \[ |D|\ge \left\lfloor\frac n2\right\rfloor, \] and this bound is sharp for every $n$, with equality achieved by the regular $n$-gon.

8) COMPLETION ESTIMATE

COMPLETION: 100%

9) REFERENCES

No external theorem is needed for the Round-2 correction; the proof above is self-contained.

References

References

  • [Bloom738] T. F. Bloom (ed.), Erdős Problem #738, Erdős Problems Project, accessed 8 March 2026.
  • [Bloom740] T. F. Bloom (ed.), Erdős Problem #740, Erdős Problems Project, accessed 8 March 2026.
  • [Scott97] A. D. Scott, Induced trees in graphs of large chromatic number, Journal of Graph Theory 24 $1997$, 297--311.
  • [NguyenScottSeymour24] T. Nguyen, A. Scott, and P. Seymour, A note on the Gyárfás--Sumner conjecture, Graphs and Combinatorics 40 $2024$, Article 33.
  • [Rodl77] V. Rödl, On the chromatic number of subgraphs of a given graph, Proceedings of the American Mathematical Society 64 $1977$, 370--371.
  • [KomjathShelah88] P. Komjáth and S. Shelah, Forcing constructions for uncountably chromatic graphs, Journal of Symbolic Logic 53 $1988$, 696--707.