Reused writeup on Math-6106 (gpt pro 5.2)

Erdős Problem 1052 --- collection record

PROVENANCE AND STATUS

Official problem: https://www.erdosproblems.com/1052

Saved collaborative-database status: open. Snapshot checked: 2026-09-23T17:22:24Z. Latest status and formalization links: https://teorth.github.io/erdosproblems/

This is an attributed collection of existing writeups. The model-folder names below identify their original attribution; placement in GPT_6_Astra_Ultra does not attribute their mathematical work to that model. All locally available source versions selected for this problem are reproduced below, without editing their text.

Proof claims, computations, literature statements, and completion estimates in the reused text have not been independently verified in this collection pass. A source claiming a solution does not change the saved upstream status.

Primary source (latest numbered version in the preferred model folder): attacks/erdos/gpt_pro_5.2/1052.tex.

REUSED SOURCE 1

Original attribution: codex_5.2_extra_high. Original file: attacks/erdos/codex_5.2_extra_high/1052.tex.

OUTPUT 1) FORMAL RESTATEMENT Let $n\ge 1$ be an integer. A divisor $d\mid n$ is unitary if $\gcd(d,n/d)=1$. Let \[ U(n) := \{d \mid n : \gcd(d,n/d)=1\} \] be the set of unitary divisors, and let \[ \sigma^*(n) := \sum_{d\in U(n)} d \] be the unitary divisor sum. Define $n$ to be unitary perfect if \[ \sum_{\substack{d\in U(n) \\ d<n}} d = n. \] Equivalently (see Lemma 1 below), $n$ is unitary perfect iff $\sigma^*(n)=2n$.

The open question is whether the set \[ \{n\in \mathbb{N} : n \text{ is unitary perfect}\} \] is finite.

2) QUICK LITERATURE/CONTEXT CHECK Browsing is not available. I only record what the problem statement itself claims: - No odd unitary perfect numbers exist. - Five examples are known: $6,60,90,87360,146361946186458562560000$. I do not verify those claims here.

3) ATTACK PLAN Proof track: 1. Characterize unitary divisors via prime factorization and derive the product formula for $\sigma^*(n)$. 2. Use the product formula to prove structural constraints (e.g., no odd unitary perfect numbers). 3. Classify small-parameter cases (e.g., exactly two distinct primes) by direct algebra.

Disproof track: 1. Attempt to construct an infinite family solving $\sigma^*(n)=2n$. 2. Seek parameterized solutions with fixed number of primes or fixed $2$-adic valuation.

Chosen path: provide fully proved structural lemmas and a complete classification when $n$ has exactly two prime factors; leave finiteness open.

4) WORK Lemma 1 (Proper vs. full unitary sum). For $n\ge 1$, let $U(n)$ be the set of unitary divisors and $\sigma^*(n)=\sum_{d\in U(n)} d$. Then \[ \sum_{\substack{d\in U(n) \\ d<n}} d = n \quad\Longleftrightarrow\quad \sigma^*(n)=2n. \]

Proof. Since $n\mid n$ and $\gcd(n,n/n)=\gcd(n,1)=1$, we have $n\in U(n)$. Hence \[ \sigma^*(n) = n + \sum_{\substack{d\in U(n) \\ d<n}} d. \] Therefore the proper-unitary sum equals $n$ iff $\sigma^*(n)=2n$. ∎

Lemma 2 (Unitary divisors of a prime power). If $n=p^a$ with $p$ prime and $a\ge 1$, then $U(n)=\{1,p^a\}$.

Proof. Every divisor of $p^a$ has the form $p^k$ with $0\le k\le a$. We have $\gcd(p^k,p^{a-k})=p^{\min(k,a-k)}$. This equals $1$ iff $\min(k,a-k)=0$, i.e., $k=0$ or $k=a$. Thus the only unitary divisors are $1$ and $p^a$. ∎

Corollary 3 (No prime power is unitary perfect). If $n=p^a$ with $a\ge 1$, then $n$ is not unitary perfect.

Proof. By Lemma 2, the proper unitary divisors sum to $1$, so Lemma 1 gives $\sigma^*(n)=1+p^a\ne 2p^a$ for $p^a>1$. Hence $n$ is not unitary perfect. ∎

Lemma 4 (Unitary divisors via prime factorization). Let $n=\prod_{i=1}^k p_i^{a_i}$ be the prime factorization with distinct primes $p_i$. A divisor $d\mid n$ is unitary iff for each $i$, the exponent of $p_i$ in $d$ is either $0$ or $a_i$. Consequently, \[ U(n)=\left\{\prod_{i\in S} p_i^{a_i} : S\subseteq \{1,\dots,k\}\right\}. \]

Proof. Write $d=\prod p_i^{b_i}$ with $0\le b_i\le a_i$. Then $\gcd(d,n/d)=\prod p_i^{\min(b_i,a_i-b_i)}$. This equals $1$ iff $\min(b_i,a_i-b_i)=0$ for all $i$, i.e., $b_i\in\{0,a_i\}$ for each $i$. This gives the stated characterization. ∎

Lemma 5 (Product formula for $\sigma^*$). With the notation of Lemma 4, \[ \sigma^*(n)=\prod_{i=1}^k (1+p_i^{a_i}). \]

Proof. By Lemma 4, unitary divisors correspond bijectively to subsets $S$ of $\{1,\dots,k\}$, with divisor $d_S=\prod_{i\in S} p_i^{a_i}$. Summing over all subsets gives \[ \sigma^*(n)=\sum_{S} \prod_{i\in S} p_i^{a_i} = \prod_{i=1}^k (1+p_i^{a_i}), \] by the distributive expansion of the product. ∎

Lemma 6 (No odd unitary perfect numbers). If $n$ is unitary perfect, then $n$ is even.

Proof. Suppose $n$ is odd and write $n=\prod_{i=1}^k p_i^{a_i}$ with odd primes $p_i$. Then each factor $1+p_i^{a_i}$ is even, so by Lemma 5, $\sigma^*(n)$ is divisible by $2^k$. If $k\ge 2$, then $\sigma^*(n)$ is divisible by $4$, but $\sigma^*(n)=2n$ has only a single factor of $2$. Hence $k=1$, so $n=p^a$ is an odd prime power, contradicting Corollary 3. ∎

Lemma 7 (Two-prime classification). If $n$ has exactly two distinct prime factors and is unitary perfect, then $n=6$.

Proof. Let $n=p^a q^b$ with distinct primes $p,q$ and $a,b\ge 1$. By Lemma 4, $U(n)=\{1,p^a,q^b,p^a q^b\}$, so the proper-unitary sum is $1+p^a+q^b$. Unitary perfectness gives \[ p^a q^b = 1+p^a+q^b. \] Rearrange: \[(p^a-1)(q^b-1)=2. \] Since $p^a-1$ and $q^b-1$ are positive integers, the only possibility is $(p^a-1,q^b-1)=(1,2)$ or $(2,1)$. Thus $(p^a,q^b)=(2,3)$ or $(3,2)$, so $n=6$. ∎

Lemma 8 (Multiplicativity of $\sigma^*$). If $\gcd(a,b)=1$, then $\sigma^*(ab)=\sigma^*(a)\sigma^*(b)$.

Proof. Write the prime factorization of $ab$ as the disjoint union of the prime factors of $a$ and $b$. By Lemma 5, $\sigma^*(ab)$ is the product over all prime powers dividing $ab$ of $(1+p^e)$. Since $\gcd(a,b)=1$, this product splits into the product for $a$ and the product for $b$, which equals $\sigma^*(a)\sigma^*(b)$. ∎

Lemma 9 (2-adic budget constraint). Let $n$ be unitary perfect and write $n=2^m \prod_{i=1}^r p_i^{a_i}$ with odd primes $p_i$. Then \[ \sum_{i=1}^r \nu_2(1+p_i^{a_i}) = m+1, \] and in particular $r \le m+1$.

Proof. By Lemma 5 and multiplicativity (Lemma 8), \[ \sigma^*(n)=(1+2^m)\prod_{i=1}^r (1+p_i^{a_i}). \] Since $m\ge 1$ (Lemma 6), the factor $1+2^m$ is odd, hence \[ \nu_2(\sigma^*(n)) = \sum_{i=1}^r \nu_2(1+p_i^{a_i}). \] Unitary perfectness gives $\sigma^*(n)=2n$, so $\nu_2(\sigma^*(n))=\nu_2(2n)=m+1$. Each odd $p_i$ satisfies $1+p_i^{a_i}$ even, so each summand is at least $1$, yielding $r\le m+1$. ∎

Lemma 10 (Normalized product identity). With $n=2^m \prod_{i=1}^r p_i^{a_i}$ as above, unitary perfectness is equivalent to \[ \prod_{i=1}^r \left(1+\frac{1}{p_i^{a_i}}\right)=\frac{2^{m+1}}{1+2^m}. \]

Proof. Divide the identity $\sigma^*(n)=2n$ by $n$ and use Lemma 5 to get \[ \frac{\sigma^*(n)}{n} = \frac{1+2^m}{2^m}\prod_{i=1}^r \left(1+\frac{1}{p_i^{a_i}}\right)=2. \] Rearranging gives the stated formula. ∎

Lemma 11 (Odd part divisible by $1+2^m$). Let $n=2^m \prod_{i=1}^r p_i^{a_i}$ be unitary perfect. Then $1+2^m$ divides the odd part $\prod_{i=1}^r p_i^{a_i}$.

Proof. From Lemma 10, \[ (1+2^m)\prod_{i=1}^r (p_i^{a_i}+1)=2^{m+1}\prod_{i=1}^r p_i^{a_i}. \] Since $\gcd(1+2^m,2^{m+1})=1$, the odd factor $1+2^m$ must divide $\prod_{i=1}^r p_i^{a_i}$. ∎

Corollary 11a (If $m$ is odd, then $3\mid n$). If $n$ is unitary perfect and $m=\nu_2(n)$ is odd, then $3$ divides $n$.

Proof. If $m$ is odd, then $2^m\equiv 2 \pmod{3}$, so $1+2^m\equiv 0 \pmod{3}$. By Lemma 11, $1+2^m$ divides the odd part of $n$, hence $3\mid n$. ∎

Lemma 12 (One odd prime factor case). If $n=2^m p^a$ with $p$ odd prime is unitary perfect, then $n=6$.

Proof. Lemma 10 gives \[ 1+\frac{1}{p^a}=\frac{2^{m+1}}{1+2^m}, \] hence \[ p^a=\frac{1+2^m}{2^m-1}=1+\frac{2}{2^m-1}. \] Thus $2^m-1$ divides $2$, so $2^m-1\in\{1,2\}$ and $m=1$. Then $p^a=3$, giving $n=2\cdot 3=6$. ∎

Lemma 13 (Case $m=1$ with two odd primes). If $n=2\cdot 3^a q^b$ with $q$ an odd prime $q\ne 3$ is unitary perfect, then $n=90$.

Proof. Lemma 10 with $m=1$ gives \[ \left(1+\frac{1}{3^a}\right)\left(1+\frac{1}{q^b}\right)=\frac{4}{3}, \] so \[ (3^a+1)(q^b+1)=4\cdot 3^{a-1} q^b. \tag{*} \] Since $\gcd(3^a+1,3)=1$, the factor $3^a+1$ divides $4q^b$, hence $3^a+1=2^t q^b$ with $t\in\{1,2\}$ (because $3^a+1$ is even). Then $(*)$ becomes $2^t(q^b+1)=4\cdot 3^{a-1}$, so $q^b+1=2^{2-t}3^{a-1}$.

If $a$ is even, then $3^a+1\equiv 2\pmod{4}$, so $t=1$ and $(3^a+1)/2=q^b=2\cdot 3^{a-1}-1$. Equating gives $3^a=9$, hence $a=2$ and $q^b=5$.

If $a$ is odd, then $3^a+1\equiv 0\pmod{4}$, so $t=2$ and $(3^a+1)/4=q^b=3^{a-1}-1$, which forces $3^a=15$, impossible. Thus $a=2$, $q=5$, $b=1$, and $n=2\cdot 3^2\cdot 5=90$. ∎

Lemma 14 (Case $m=2$ with two odd primes). If $n=2^2\cdot 5^a q^b$ with $q$ an odd prime $q\ne 5$ is unitary perfect, then $n=60$.

Proof. By Lemma 11, $1+2^2=5$ divides the odd part, so one odd prime is $5$. Lemma 10 with $m=2$ gives \[ \left(1+\frac{1}{5^a}\right)\left(1+\frac{1}{q^b}\right)=\frac{8}{5}, \] hence \[ (5^a+1)(q^b+1)=8\cdot 5^{a-1} q^b. \tag{**} \] Since $\gcd(5^a+1,5)=1$, we have $5^a+1\mid 8q^b$, so $5^a+1=2^t q^b$ for some $t\le 3$. But $5^a\equiv 1\pmod{4}$, so $5^a+1\equiv 2\pmod{4}$ and $t=1$. Then $(**)$ gives $2(q^b+1)=8\cdot 5^{a-1}$, so $q^b+1=4\cdot 5^{a-1}$. Equating $q^b=(5^a+1)/2$ yields $5^a=5$, hence $a=1$ and $q^b=3$. Thus $n=2^2\cdot 5\cdot 3=60$. ∎

Lemma 15 (Case $m=3$ with two odd primes). If $n=2^3\cdot 3^a q^b$ with $q$ an odd prime $q\ne 3$ is unitary perfect, then no such $n$ exists.

Proof. Lemma 10 with $m=3$ gives \[ \left(1+\frac{1}{3^a}\right)\left(1+\frac{1}{q^b}\right)=\frac{16}{9}, \] so \[ (3^a+1)(q^b+1)=16\cdot 3^{a-2} q^b. \tag{***} \] If $a=1$, then the right-hand side equals $16q^b/3$, which is not an integer unless $q=3$, contrary to $q\ne 3$. Hence $a\ge 2$.

Since $\gcd(3^a+1,3)=1$, the factor $3^a+1$ divides $16q^b$, so $3^a+1=2^t q^b$ for some $t\le 4$. Cancelling $q^b$ in $(***)$ yields $2^t(q^b+1)=16\cdot 3^{a-2}$, so $q^b+1=2^{4-t}3^{a-2}$. Substituting back gives \[ 3^a+1=2^t(2^{4-t}3^{a-2}-1)=16\cdot 3^{a-2}-2^t, \] so \[ 2^t+1=7\cdot 3^{a-2}. \tag{****} \] If $a$ is even, then $3^{a-2}\equiv 1\pmod{8}$ and the right-hand side is $7\pmod{8}$, but $2^t+1\equiv 1,3,5\pmod{8}$, impossible. If $a$ is odd, then $3^{a-2}\equiv 3\pmod{8}$ so the right-hand side is $5\pmod{8}$, forcing $2^t+1\equiv 5\pmod{8}$ and thus $t=2$. Then $(****)$ gives $7\cdot 3^{a-2}=5$, impossible. Therefore no such $n$ exists. ∎

Lemma 16 (Case $m=4$ with two odd primes). If $n=2^4\cdot 17^a q^b$ with $q$ an odd prime $q\ne 17$ is unitary perfect, then no such $n$ exists.

Proof. Lemma 10 with $m=4$ gives \[ \left(1+\frac{1}{17^a}\right)\left(1+\frac{1}{q^b}\right)=\frac{32}{17}, \] so \[ (17^a+1)(q^b+1)=32\cdot 17^{a-1} q^b. \tag{*****} \] Since $\gcd(17^a+1,17)=1$, we have $17^a+1\mid 32 q^b$, hence $17^a+1=2^t q^b$ for some $t\le 5$. Cancelling $q^b$ in $(*****)$ yields $2^t(q^b+1)=32\cdot 17^{a-1}$, so $q^b+1=2^{5-t}17^{a-1}$. Substituting back gives \[ 17^a+1=2^t(2^{5-t}17^{a-1}-1)=32\cdot 17^{a-1}-2^t, \] so \[ 2^t+1=15\cdot 17^{a-1}. \tag{******} \] Since $17^{a-1}\equiv 1\pmod{8}$, the right-hand side is $7\pmod{8}$, while $2^t+1\equiv 1,3,5\pmod{8}$ for $t\ge 1$. This is impossible, so no such $n$ exists. ∎

Example (checked by hand): For $n=6$, the unitary divisors are $1,2,3,6$; the proper sum is $1+2+3=6$, hence $6$ is unitary perfect. This uses no computation beyond listing divisors.

5) VERIFICATION Edge cases: - $n=1$: $U(1)=\{1\}$, proper sum $=0\ne 1$, so $1$ is not unitary perfect. - $n=p$ prime: Corollary 3 applies (proper sum $=1$). - $n=6$: verified above as unitary perfect.

Quantifiers: All lemmas apply to $n\ge 1$ with explicit prime factorizations where used. Lemma 6 relies only on Lemma 5 and parity, Lemma 7 uses only algebra on two primes, and Lemmas 8--16 follow from Lemma 5 plus basic properties of $\nu_2$, parity, and coprime factorizations.

No outside theorems beyond unique factorization and basic divisor facts are used.

6) FINAL UNRESOLVED

(i) Strongest fully proved partial result: No odd unitary perfect numbers exist (Lemma 6), no prime power is unitary perfect (Corollary 3), and if a unitary perfect number has exactly two distinct primes then it is $6$ (Lemma 7). We also have structural constraints from Lemmas 9--11 (2-adic budget, normalized product identity, and $1+2^m$ dividing the odd part), and explicit classifications in small cases: if $n=2^m p^a$ then $n=6$ (Lemma 12), if $m=1$ and there are exactly two odd primes then $n=90$ (Lemma 13), if $m=2$ and there are exactly two odd primes then $n=60$ (Lemma 14), and there are no solutions with exactly two odd primes when $m=3$ or $m=4$ (Lemmas 15--16).

(ii) First gap: A global bound on either the number of distinct odd prime factors of a unitary perfect number or its $2$-adic valuation; such a bound would imply finiteness via known strategies and is not achieved here.

(iii) Top 3 next moves: 1. Prove a uniform bound on the number of distinct odd prime factors of a unitary perfect number. 2. Prove a uniform bound on $\nu_2(n)$ for unitary perfect $n$. 3. Extend the fixed-$m$ two-odd-prime analysis beyond $m=4$ using Lemma 11 and modular constraints.

(iv) Minimal counterexample structure (if finiteness is false): An infinite sequence of even unitary perfect numbers with growing numbers of distinct odd primes and rapidly increasing $2$-adic valuation, consistent with the product formula constraints.

REUSED SOURCE 2

Original attribution: gpt_pro_5.2. Original file: attacks/erdos/gpt_pro_5.2/1052.tex.

1) FORMAL RESTATEMENT

For an integer $n\ge 1$, a divisor $d\mid n$ is called a unitary divisor (notation $d\parallel n$) if $\gcd(d,n/d)=1$. Define the sum of unitary divisors \[ \sigma^*(n):=\sum_{d\parallel n} d. \] A number $n\ge 1$ is called unitary perfect if \[ \sigma^*(n)=2n. \] Question: Are there only finitely many unitary perfect numbers?

2) QUICK LITERATURE/CONTEXT CHECK

The provided text states: - There are no odd unitary perfect numbers. - Five unitary perfect numbers are known: $6,60,90,87360,146361946186458562560000$. I will not assume any deeper classification, but I will verify computationally the small ones and prove some standard structural lemmas.

3) ATTACK PLAN

Proof track: - Derive a multiplicative formula for $\sigma^*(n)$ (Lemma 4.1). - Translate the unitary perfect condition into a product equation $\prod (1+1/p^k)=2$. - Prove simple structural consequences (evenness, constraints on number of prime factors).

Construction/disproof track: - Perform a finite computational search up to a moderately large bound to confirm known solutions and look for new ones.

4) WORK

FAST REALITY CHECK (computation)

I searched all $2\le n\le 10^7$ by factoring each $n$ via a smallest-prime-factor sieve and evaluating $\sigma^*(n)$ using Lemma 4.1. Result: the only unitary perfect numbers in this range are \[ 6,\ 60,\ 90,\ 87360. \] (No new examples up to $10^7$.)

Lemma 4.1 (Formula for $\sigma^*(n)$)

Let $n=\prod_{i=1}^r p_i^{\alpha_i}$ be the prime factorization of $n$. Then the unitary divisors of $n$ are exactly the numbers \[ d=\prod_{i=1}^r p_i^{\varepsilon_i\alpha_i}\quad\text{with }\varepsilon_i\in\{0,1\}, \] and therefore \[ \sigma^*(n)=\prod_{i=1}^r (1+p_i^{\alpha_i}). \] In particular, $\sigma^*$ is multiplicative.

Proof. A divisor $d\mid n$ can be written uniquely as $d=\prod p_i^{\beta_i}$ with $0\le \beta_i\le \alpha_i$. Then $n/d=\prod p_i^{\alpha_i-\beta_i}$. The condition $\gcd(d,n/d)=1$ holds iff for every $i$ we do not have $p_i$ dividing both $d$ and $n/d$, which is equivalent to: for each $i$, either $\beta_i=0$ or $\beta_i=\alpha_i$. Thus unitary divisors correspond to choosing, independently for each prime, whether to include the full prime power $p_i^{\alpha_i}$ or not. Summing over all $2^r$ choices factors: \[ \sigma^*(n)=\prod_{i=1}^r (1+p_i^{\alpha_i}). \] Multiplicativity follows from the product form. $\square$

Corollary 4.2 (Unitary perfect condition as a product equation)

If $n=\prod p_i^{\alpha_i}$ is unitary perfect, then \[ \prod_{i=1}^r \left(1+\frac{1}{p_i^{\alpha_i}}\right)=2. \]

Proof. Divide the identity $\sigma^*(n)=2n$ by $n=\prod p_i^{\alpha_i}$ and use Lemma 4.1. $\square$

Lemma 4.3 (No odd unitary perfect numbers)

There is no odd integer $n$ with $\sigma^*(n)=2n$.

Proof. Assume $n$ is odd and write $n=\prod_{i=1}^r p_i^{\alpha_i}$ with all $p_i$ odd. Then each factor $1+p_i^{\alpha_i}$ is even. Hence $\sigma^*(n)=\prod (1+p_i^{\alpha_i})$ is divisible by $2^r$. But $2n$ has exact $2$-adic valuation $v_2(2n)=1$. Therefore we must have $r=1$. So $n=p^\alpha$ for some odd prime $p$. Then $\sigma^*(n)=1+p^\alpha$, and $\sigma^*(n)=2n$ becomes $1+p^\alpha = 2p^\alpha$, i.e. $p^\alpha=1$, impossible. Thus no odd unitary perfect number exists. $\square$

Lemma 4.4 (Constraint relating $v_2(n)$ to the number of odd prime factors)

Let $n=2^a m$ with $m$ odd, and write $m=\prod_{i=1}^r p_i^{\alpha_i}$ with distinct odd primes $p_i$. If $n$ is unitary perfect then necessarily \[ r\le a+1. \]

Proof. Using Lemma 4.1, \[ \sigma^*(n)=(1+2^a)\prod_{i=1}^r (1+p_i^{\alpha_i}). \] The factor $(1+2^a)$ is odd. Each $(1+p_i^{\alpha_i})$ is even, hence contributes at least one factor of $2$. Therefore \[ v_2(\sigma^*(n))\ge r. \] But $\sigma^*(n)=2n$ implies $v_2(\sigma^*(n))=v_2(2n)=a+1$. Hence $r\le a+1$. $\square$

5) VERIFICATION

- Lemma 4.1: checked by direct prime-exponent analysis; no missing cases. - Lemma 4.3: the key step is comparing 2-adic valuations, which is exact and elementary. - Computation: sieve-based factorization up to $10^7$ found exactly the four known small examples. (This is a sanity check, not a proof of finiteness.)

6) FINAL

UNRESOLVED

(i) Strongest proved partial result. - Structural formula $\sigma^*(n)=\prod (1+p^\alpha)$ and the equivalent product equation $\prod (1+1/p^\alpha)=2$. - No odd unitary perfect numbers (Lemma 4.3). - Exhaustive search up to $10^7$ finds only $6,60,90,87360$.

(ii) First gap (crisp statement). Prove that the Diophantine equation \[ \prod_{i=1}^r \left(1+\frac{1}{p_i^{\alpha_i}}\right)=2 \] has only finitely many solutions in distinct primes $p_i$ and exponents $\alpha_i\ge 1$.

(iii) Top 3 next moves. 1. Use the inequality $\log(1+x)\le x$ to show $\sum 1/p_i^{\alpha_i} \ge \log 2$, then combine with lower bounds on $p_i^{\alpha_i}$ to constrain the set of possible prime powers. 2. Improve Lemma 4.4 by controlling higher 2-adic valuations of $1+p^\alpha$ for odd primes, potentially forcing stronger restrictions on the exponents and the number of odd primes. 3. Implement a branch-and-bound search using the product equation (Corollary 4.2) to enumerate all possible prime-power combinations below a given size bound, potentially extending verified search far beyond $10^7$.

(iv) Minimal counterexample structure. If infinitely many unitary perfect numbers exist, there must be an infinite family with growing sets of prime powers $p_i^{\alpha_i}$ such that the product $\prod(1+1/p_i^{\alpha_i})$ stays exactly $2$. This would require a delicate balance between introducing new small prime-power reciprocals and adjusting exponents so as not to overshoot; one would expect many large prime powers with very small reciprocals, and tight 2-adic constraints from Lemma 4.4.

COMPLETION ESTIMATE

The primary source supplies no extractable numerical completion estimate. Its mathematical progress is not numerically assessed here.

Newly verified progress in this collection pass: $0\%$. This measures new verification work, not the amount of mathematics achieved in the reused text.