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Math-5669 · mathematics · Mathematics

Erdős Problem #466

Status Solved · erdosproblems.comLLM status Solved from source status

Statement

Statement and result.
Let $N(X,\delta)$ be the largest size of a set in a disc of radius $X$ whose pairwise distances stay at least $\delta$ from every integer. Does $N(X,\delta)\to\infty$ for some fixed $\delta>0$? Yes. The following explicit parabola construction proves this for $\delta=2/5$, with a logarithmic lower bound.
The points and the distance calculation.
Take \[ P_j=(100^j,10^j),\qquad j=0,1,\ldots,L. \] For $i<j$, put $s=10^j+10^i$, $e=10^j-10^i$. The horizontal difference $D=es=100^j-100^i$ is an integer, and \[ |P_j-P_i|=e\sqrt{s^2+1}=D+\eta,\qquad \eta=\frac{e}{\sqrt{s^2+1}+s}. \tag{1} \] We have $s\geq11$, $e/s\geq9/11$, and $e<s$. Therefore \[ \frac{9}{\sqrt{122}+11} \leq\eta<\frac12. \tag{2} \] The left side exceeds $2/5$, since $45>2\sqrt{122}+22$ is equivalent to $529>488$. Since $D$ is an integer, $1$--$2$ show \[ \||P_j-P_i|\|=\eta>\frac25 \] for every pair.
Fit the points into the required disc.
Their norms satisfy \[ |P_j|=\sqrt{100^{2j}+100^j}\leq\sqrt2\,100^j. \] For $X\geq2$, choose $L=\lfloor\log(X/2)/\log100\rfloor$. All $L+1$ points lie in the disc of radius $X$, and \[ N(X,2/5)\geq L+1> \frac{\log(X/2)}{\log100}. \tag{3} \] In particular this tends to infinity. For $X\geq4$, $3$ also gives $N(X,2/5)>\log X/(2\log100)>\log X/10$. Monotonicity in $\delta$ gives the same lower bound at $\delta=1/10$.
Scope and attribution.
The source https://www.erdosproblems.com/466, checked 24 September 2026, credits Graham with a logarithmic construction and Sárközy with stronger power lower bounds. The elementary calculation here provides a full affirmative answer; it does not reconstruct the stronger power bound or claim novelty. No external theorem is needed for this construction.
Completion estimate.
An explicit family with a fixed distance margin and unbounded size is proved.

Restatement from Math-5669-A1 r1, with the source manuscript’s qualifications.

Original problem on erdosproblems.com

Statement licence: CC-BY-4.0 · Source file

Contributions 1

Math-5669-A1 · No claim recorded · LLM-generated · gpt 6 astra ultra ·

GPT 6 Astra Ultra attempt on Math-5669

Erdős Problem 466

Statement and result.
Let $N(X,\delta)$ be the largest size of a set in a disc of radius $X$ whose pairwise distances stay at least $\delta$ from every integer. Does $N(X,\delta)\to\infty$ for some fixed $\delta>0$? Yes. The following explicit parabola construction proves this for $\delta=2/5$, with a logarithmic lower bound.
The points and the distance calculation.
Take \[ P_j=(100^j,10^j),\qquad j=0,1,\ldots,L. \] For $i<j$, put $s=10^j+10^i$, $e=10^j-10^i$. The horizontal difference $D=es=100^j-100^i$ is an integer, and \[ |P_j-P_i|=e\sqrt{s^2+1}=D+\eta,\qquad \eta=\frac{e}{\sqrt{s^2+1}+s}. \tag{1} \] We have $s\geq11$, $e/s\geq9/11$, and $e<s$. Therefore \[ \frac{9}{\sqrt{122}+11} \leq\eta<\frac12. \tag{2} \] The left side exceeds $2/5$, since $45>2\sqrt{122}+22$ is equivalent to $529>488$. Since $D$ is an integer, $1$--$2$ show \[ \||P_j-P_i|\|=\eta>\frac25 \] for every pair.
Fit the points into the required disc.
Their norms satisfy \[ |P_j|=\sqrt{100^{2j}+100^j}\leq\sqrt2\,100^j. \] For $X\geq2$, choose $L=\lfloor\log(X/2)/\log100\rfloor$. All $L+1$ points lie in the disc of radius $X$, and \[ N(X,2/5)\geq L+1> \frac{\log(X/2)}{\log100}. \tag{3} \] In particular this tends to infinity. For $X\geq4$, $3$ also gives $N(X,2/5)>\log X/(2\log100)>\log X/10$. Monotonicity in $\delta$ gives the same lower bound at $\delta=1/10$.
Scope and attribution.
The source https://www.erdosproblems.com/466, checked 24 September 2026, credits Graham with a logarithmic construction and Sárközy with stronger power lower bounds. The elementary calculation here provides a full affirmative answer; it does not reconstruct the stronger power bound or claim novelty. No external theorem is needed for this construction.
Completion estimate.
An explicit family with a fixed distance margin and unbounded size is proved. COMPLETION: 100%

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