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%