Erdős Problem 823: ratios within a divisor-sum fiber
1. Statement and declared input
Given $\alpha\ge1$, find positive integers $n_j,m_j$ with $\sigma(n_j)=\sigma(m_j)$ and $n_j/m_j\to\alpha$. We reconstruct Pollack's affirmative proof. Its explicitly external deep input is the bounded-prime-clusters theorem for affine forms: there is an absolute integer $K\ge2$ such that every admissible collection of $K$ distinct, pairwise nonproportional integral affine forms with positive leading coefficients has two forms simultaneously prime for infinitely many positive integer arguments. This is the Maynard form of the Zhang--Maynard theorem, stated as Proposition 1 in Pollack's paper. Admissible means that no prime divides the product of the forms for every integer argument. The divergence of the sum of reciprocals of the primes is the only other standard input. All fiber constructions and the passage to arbitrary ratios are proved below.
2. Approximating abundancy logarithms
Fix a finite forbidden prime set $\mathcal P$. The positive numbers $\log(1+1/p)$ for primes $p\notin\mathcal P$ tend to zero and have divergent sum. Given $t>0$ and $\eta>0$, discard enough primes that every remaining summand is less than $\eta$, then add successive summands until their sum first reaches $t$. The resulting squarefree product $A$, avoiding $\mathcal P$, satisfies \[
t\le\log\frac{\sigma(A)}A<t+\eta. \tag{1}
\] We can also require all primes used to exceed any prescribed bound. In particular $A$, and hence $\sigma(A)$, can be arbitrarily large. This lets us choose a finite list of such $A$'s with distinct divisor sums, which will ensure that the affine forms below really are distinct. For finite $\mathcal P$ let $L_{\mathcal P}$ be the closure in $\mathbb R$ of \[
\left\{\log\frac mn:\ \sigma(m)=\sigma(n),\quad
\gcd\left(mn,\prod_{p\in\mathcal P}p\right)=1\right\}.
\]
3. A difference in every $K$-point set
We claim that for any real numbers $t_1<\cdots<t_K$, some positive difference $t_b-t_a$ belongs to $L_{\mathcal P}$. Translate all $t_i$ by the same constant so that they are positive. For each integer $j$, use $1$ to choose squarefree $A_{i,j}$ avoiding $\mathcal P$ with \[
\log(\sigma(A_{i,j})/A_{i,j})\longrightarrow t_i
\] and with $\sigma(A_{1,j}),\ldots,\sigma(A_{K,j})$ distinct. The forms \[
\sigma(A_{i,j})X-1\qquad(1\le i\le K)
\] are admissible: at $X=0$ their product is $(-1)^K$. Their determinants are nonzero because their leading coefficients are distinct. The declared prime-clusters theorem therefore supplies a pair $a_j<b_j$ and a positive $u_j$ such that \[
p_j=\sigma(A_{a_j,j})u_j-1,\qquad
q_j=\sigma(A_{b_j,j})u_j-1
\] are primes larger than $j$, every $A_{i,j}$, and every forbidden prime. We may impose these lower bounds because the theorem gives infinitely many arguments. Multiplicativity now gives the exact identity \[
\sigma(p_jA_{b_j,j})=(p_j+1)\sigma(A_{b_j,j})
=(q_j+1)\sigma(A_{a_j,j})=\sigma(q_jA_{a_j,j}).
\] Both arguments avoid $\mathcal P$. A fixed pair $(a,b)$ occurs infinitely often. Along that subsequence, \[
\log\frac{q_jA_{a,j}}{p_jA_{b,j}}
=\log\frac{\sigma(A_{b,j})/A_{b,j}}{\sigma(A_{a,j})/A_{a,j}}+o(1)
\longrightarrow t_b-t_a.
\] This proves the claim. The $o(1)$ is justified by $p_j,q_j\to\infty$, through replacing each prime by that prime plus one.
4. Building an arbitrary target by coprime small steps
Let $T=\log\alpha$. For any $\varepsilon>0$, apply the claim to the $K$ points $0,\varepsilon/K,\ldots,(K-1)\varepsilon/K$. For every finite forbidden prime set there is an actual fiber ratio with logarithm in \[
\left(\frac{\varepsilon}{2K},\varepsilon\right). \tag{2}
\] Indeed one of the indicated differences belongs to the closure, and a sufficiently close actual ratio lies in this open interval. Choose pairs $(a_i,b_i)$ with equal divisor sums and logarithms in $2$, successively forbidding every prime dividing any earlier $a_jb_j$. Thus different pairs have pairwise coprime supports. Their logarithms are positive and bounded below, so their partial sums eventually reach $T$. At the first such index $J$, \[
T\le\sum_{i=1}^J\log(a_i/b_i)<T+\varepsilon.
\] Set $n=\prod_{i=1}^Ja_i$, $m=\prod_{i=1}^Jb_i$. Coprimality across pairs, rather than any unwarranted general multiplicativity, gives \[
\sigma(n)=\prod_i\sigma(a_i)=\prod_i\sigma(b_i)=\sigma(m).
\] The ratio lies in $[\alpha,\alpha e^\varepsilon)$. Taking $\varepsilon=1/j$ gives the requested sequence. For $\alpha=1$ one may simply take $n_j=m_j=1$. Inverting ratios also covers any positive target below one.
5. Sources and assessment
Sources:
https://www.erdosproblems.com/823; P. Pollack,
Remarks on fibers of the sum-of-divisors function, Section 2, author manuscript
https://www.pollack-math.net/preimages-maier3.pdf, published DOI
https://doi.org/10.1007/978-3-319-22240-0_18. The affine-form prime theorem is imported as explicitly stated above; its analytic proof is not reproduced.
SOLVED relative to the declared prime-clusters input. The density lemma, exact equal-divisor-sum construction, admissibility and distinctness checks, and coprime accumulation argument are fully supplied. This reconstructs Pollack's known solution and claims no new result or local formal verification.
Completion rate estimate: 100%.