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Source. Stijn Cambie, Resolution of Erdős' problems about unimodularity, arXiv:2501.10333v1 (17 January 2025), Theorem 1 and proof, PDF pp. 2--3 (the conclusion is on p. 3).

Bears on. #692.

Statement

For every integer $m\geq3$, the density $\delta_1(1,m)$ of integers having exactly one divisor in $\{2,\ldots,m-1\}$ is non-increasing as $m$ grows. Consequently it is unimodal.

Rewritten proof

For fixed $m$, let $p$ range over the primes at most $\sqrt{m-1}$ and let $q$ range over the primes in $[\sqrt m,m-1]$. Put

$$L_m=\prod_{p\leq\sqrt{m-1}}p^2 \prod_{\sqrt m\leq q\leq m-1}q.$$

The modulus has enough prime powers to determine whether an integer has zero or exactly one divisor in $\{2,\ldots,m-1\}$. A multiple of $p^2$ already has the two divisors $p$ and $p^2$, while every larger prime can occur only to the first power in this interval. Let $A_m$ count the residue classes modulo $L_m$ having exactly one such divisor, and let $\Phi_m$ count those having none. Then

$$\delta_1(1,m)=\frac{A_m}{L_m}, \qquad \delta_0(1,m)=\frac{\Phi_m}{L_m}=\frac{\varphi(L_m)}{L_m}.$$

The second equality holds because a residue has no divisor in the set if and only if it is coprime to $L_m$.

We prove by induction that $A_m/L_m$ is non-increasing and $A_m\geq\Phi_m$. For $m=3$, $L_3=2$ and $A_3=\Phi_3=1$.

If $m-1$ is neither a prime nor a prime square, adjoining it changes nothing: a multiple of a number with at least two distinct prime factors already has two old divisors, and a multiple of $p^e$ with $e\geq3$ already has both $p$ and $p^2$. It remains to consider the two possible changes.

A new prime

Let $m-1=p$. The prime $p$ was absent from $L_{m-1}$, so

$$L_m=pL_{m-1},\qquad \Phi_m=(p-1)\Phi_{m-1},\qquad A_m=(p-1)A_{m-1}+\Phi_{m-1}.$$

For each old residue there are $p-1$ lifts not divisible by $p$ and one lift divisible by $p$. Therefore

$$\frac{A_m}{\Phi_m} =\frac{A_{m-1}}{\Phi_{m-1}}+\frac1{p-1},$$

and, using $\Phi_{m-1}\leq A_{m-1}$,

$$\frac{A_m}{L_m} =\frac{(p-1)A_{m-1}+\Phi_{m-1}}{pL_{m-1}} \leq\frac{A_{m-1}}{L_{m-1}}.$$

A new prime square

Let $m-1=p^2$. The old modulus contains $p$ exactly once, hence

$$L_m=pL_{m-1},\qquad \Phi_m=p\Phi_{m-1}.$$

Every old exactly-one residue has $p$ lifts. The only lifts that cease to be exactly-one are those coming from an old residue whose unique divisor was $p$; there are

$$\varphi(L_{m-1}/p)=\frac{\varphi(L_{m-1})}{p-1}$$

such residues. Consequently

$$A_m=pA_{m-1}-\frac{\Phi_{m-1}}{p-1},$$

so $A_m/L_m<A_{m-1}/L_{m-1}$ and

$$\frac{A_m}{\Phi_m} =\frac{A_{m-1}}{\Phi_{m-1}}-\frac1{p(p-1)}.$$

It remains to check that the ratio never falls below $1$. Write

$$u_p=\frac1{p-1},\qquad v_p=\frac1{p(p-1)}.$$

Starting at $m=3$, each prime event $p\geq3$ adds $u_p$ to $A/\Phi$, and each square event $p^2$ subtracts $v_p$. The first three possible square losses, at $4$, $9$ and $25$, are covered by the source's two elementary comparisons

$$u_3=v_2, \qquad u_5>v_3+v_5 \quad\left(\frac14>\frac16+\frac1{20}\right).$$

For every prime $p\geq7$, the prime event $u_p$ occurs before the square event $v_p$ and satisfies $u_p>v_p$. At any finite stage, omit unneeded positive prime events and pair each square loss with the corresponding available positive term in these groups. The total change from the initial ratio $A_3/\Phi_3=1$ is therefore non-negative. Thus

$$\frac{A_m}{\Phi_m}\geq1,$$

which is the required $A_m\geq\Phi_m$. The inequality is strict once $m\geq6$. Both induction assertions now hold, proving that $\delta_1(1,m)=A_m/L_m$ is non-increasing.

Source correction. The source's last line on p. 2 prints $A/\varphi(L)\geq0$; the induction hypothesis and the stated goal require $A/\varphi(L)\geq1$. The event bookkeeping above supplies the intended stronger inequality.