Erdős Problem 697: a threshold for congruent divisors
Statement and status.
Let $\delta(m,\alpha)$ be the density of integers divisible by some $d\equiv1\pmod m$ with $1<d<\exp(m^\alpha)$. The question asks for a constant $\beta>1$ separating limits zero and one as $m\longrightarrow\infty$. Hall's published resolution gives $\beta=1/\log2$, as recorded at
https://www.erdosproblems.com/697; the primary reference is R. R. Hall,
On some conjectures of Erdős in Astérisque, I,
https://doi.org/10.1016/0022-314X(92)90096-8. This attempt proves the elementary subcritical range, an exact finite formula, and a fixed-modulus limiting result. The sharp threshold argument remains unresolved in this reconstruction; the known value is attributed rather than presented as a new proof.
Existence and exact computation of the density.
For fixed $m,\alpha$, the admissible divisor set \[
D=\{1+jm:j\geq1,\ 1+jm<\exp(m^\alpha)\}
\] is finite. If it is empty the density is zero. Otherwise divisibility by a member of $D$ is periodic modulo $L=\operatorname{lcm}(D)$, and \[
\delta(m,\alpha)=
\sum_{\varnothing\ne S\subseteq D}
\frac{(-1)^{|S|+1}}{\operatorname{lcm}(S)}. \tag{1}
\] This follows from inclusion--exclusion and the density $1/\operatorname{lcm}(S)$ of common multiples. In particular there is no question about the existence of the density before taking the outer limit in $m$.
A rigorous subcritical bound.
Let $J=|D|$. If $J\geq1$, the union bound gives \[
\delta(m,\alpha)\leq\sum_{j=1}^J\frac1{1+jm}
<\frac1m\sum_{j=1}^J\frac1j
\leq\frac{1+\log J}{m}
<\frac{1+m^\alpha}{m}. \tag{2}
\] The same final bound holds when $J=0$. For every fixed $\alpha<1$, $2$ tends to zero. This proves a genuine part of the requested phase transition, but leaves the interval $1\leq\alpha<1/\log2$ untreated.
What can be proved when the modulus is fixed.
For each fixed $m\geq2$, the sum of reciprocals of primes $q\equiv1\pmod m$ diverges. This is a classical consequence of the prime number theorem in the fixed progression: partial summation gives $\sum_{q\leq z,\ q\equiv1\ (m)}1/q
=(\log\log z)/\varphi(m)+O_m(1)$. Using only such primes as admissible divisors and applying the Chinese remainder theorem yields \[
\delta(m,\alpha)\geq
1-\prod_{\substack{q<\exp(m^\alpha)\\q\ {\rm prime}\\q\equiv1\ (m)}}
\left(1-\frac1q\right).
\] Therefore, for each fixed $m\geq2$, \[
\lim_{\alpha\to\infty}\delta(m,\alpha)=1. \tag{3}
\] The proof uses the fixed-progression prime number theorem as an explicit external input. It does not give a bound uniform in $m$, and so does not justify exchanging the two limits in $3$.
Finite upper and lower certificates.
For completeness, let $\mu=\sum_{d\in D}1/d$ and \[
V=\mu+2\sum_{\substack{d,e\in D\\d<e}}\frac1{\operatorname{lcm}(d,e)}.
\] On residues modulo $L$, put $X=\sum_{d\in D}{\bf1}_{d\mid n}$. Then $\mathbb EX=\mu$, $\mathbb EX^2=V$, and Cauchy--Schwarz applied to $X{\bf1}_{X>0}$ gives \[
\frac{\mu^2}{V}\leq\delta(m,\alpha)\leq\min(1,\mu)
\quad(D\ne\varnothing). \tag{4}
\] Both $1$ and $4$ are exact finite certificates that do not assume independence of composite-divisor events.
Remaining gap.
The uniform distribution of these highly dependent composite divisors is the missing step. Neither $2$, $3$, nor the second-moment bound $4$ proves Hall's threshold or the limit one for any fixed $\alpha>1/\log2$ as $m$ grows. This attempt remains unresolved beyond the partial results stated.
Completion estimate.
COMPLETION: 40%