Erdős Problem 1001
Statement and scope.
Let $A>0,c>1$, and let $S(N,A,c)$ be the measure of the $\alpha\in(0,1)$ for which \[
\left|\alpha-\frac pq\right|<\frac A{q^2}
\quad\text{for some }N\leq q\leq cN,\quad (p,q)=1.
\] The problem asks for the existence and explicit value of its limit for arbitrary fixed $A,c$. This write-up proves the exact formula \[
S(N,A,c)=\frac{12A\log c}{\pi^2}
+O_{A,c}\!\left(\frac{\log N}{N}\right)
\quad\text{if }0<A\leq\frac{c}{1+c^2}.
\tag{1}
\] It also gives rigorous bounds for all $A$. The overlapping-interval regime, and therefore the general problem, remains unresolved in this write-up.
When the approximation intervals are disjoint.
For two different reduced fractions $p/q,r/s$, \[
\left|\frac pq-\frac rs\right|
=\frac{|ps-rq|}{qs}\geq\frac1{qs}.
\] If $N\leq q,s\leq cN$, their ratio lies between $1/c$ and $c$, so \[
A\left(\frac1{q^2}+\frac1{s^2}\right)
\leq\frac1{qs}
\quad\text{whenever}\quad A\leq\frac1{c+1/c}.
\] The open approximation intervals are consequently disjoint; equality can at most make their endpoints touch.
For $N>\max(A,1)$, every relevant centre in $(0,1)$ is $p/q$ with $1\leq p<q$, and its radius is smaller than its distance from both endpoints. Fractions outside $(0,1)$ cannot contribute for such $N$: $0,1$ are not reduced fractions with denominator greater than one, and other fractions are at distance at least $1/q>A/q^2$ from that interval. Therefore disjointness gives the exact identity \[
S(N,A,c)=2A\sum_{N\leq q\leq cN}\frac{\varphi(q)}{q^2}.
\tag{2}
\]
Evaluate the weighted totient sum.
The elementary divisor identity $\varphi(q)/q=\sum_{d\mid q}\mu(d)/d$ yields \[
\sum_{N\leq q\leq cN}\frac{\varphi(q)}{q^2}
=\sum_{d\leq cN}\frac{\mu(d)}{d^2}
\sum_{N/d\leq r\leq cN/d}\frac1r.
\] For $d\leq N$, comparison of a harmonic sum with its integral gives the inner sum as $\log c+O_c(d/N)$, uniformly in $d$. The total error is \[
O_c\!\left(\frac1N\sum_{d\leq N}\frac1d\right)
=O_c(\log N/N).
\] For $N<d\leq cN$, the inner sum is bounded by a constant depending only on $c$, and the total contribution is $O_c(1/N)$. Finally \[
\sum_{d\leq N}\frac{\mu(d)}{d^2}
=\frac1{\zeta(2)}+O(1/N)=\frac6{\pi^2}+O(1/N).
\] Absolute convergence justifies the Euler-product identity here. Inserting these estimates in $2$ proves $1$, including the boundary value $A=c/(1+c^2)$.
Bounds without disjointness.
The union bound and the same totient estimate give, for arbitrary fixed $A>0$, \[
\limsup_{N\to\infty}S(N,A,c)
\leq\min\left(1,\frac{12A\log c}{\pi^2}\right).
\] Monotonicity in $A$, followed by $1$ with $A_0=\min(A,c/(1+c^2))$, gives \[
\liminf_{N\to\infty}S(N,A,c)
\geq\frac{12\log c}{\pi^2}
\min\left(A,\frac{c}{1+c^2}\right)>0.
\] These bounds do not prove that the upper and lower limits coincide for larger $A$. If $c=1$, at most one denominator occurs and the total interval length is $O_A(1/N)$; if $0<c<1$, no denominator occurs. The limit is zero in both cases.
Remaining gap and attribution.
The small-$A$ formula is the classical result of Erdős, Szüsz, and Turán. The current page
https://www.erdosproblems.com/1001, checked 24 September 2026, records the general existence theorem of Kesten--Sós and explicit treatments by Boca and by Xiong--Zaharescu. Those general theorems are not being substituted for a proof here. The missing step is a uniform treatment of overlapping intervals sufficient to evaluate the union measure and prove its convergence for every $A,c$. The proof above establishes the stated regime and bounds only; no novelty or local formal verification is claimed.
Completion estimate.
COMPLETION: 45%