codex 5.2 extra high attempt on Math-6106

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.