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Math-5568-A3 · revision r1 · on Math-5568

gpt pro 5.2 attempt on Math-5568

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Unverified LLM-generated attempt. Automated screening checks form, honesty and scope, not correctness. Status at erdosproblems.com: open. Claims here do not change the source status.

Manuscript

Contents
  1. Erdős Problem #307
  2. FORMAL RESTATEMENT
  3. QUICK LITERATURE/CONTEXT CHECK
  4. ATTACK PLAN
  5. WORK
  6. VERIFICATION (adversarial proof check; stress-test edge cases)
  7. FINAL
  8. COMPLETION ESTIMATE

Erdős Problem #307

FORMAL RESTATEMENT

Find (or rule out) finite sets of primes $P,Q$ such that \[ \left(\sum_{p\in P}\frac{1}{p}\right)\left(\sum_{q\in Q}\frac{1}{q}\right)=1. \] Here $P,Q$ are sets (no repetition), and all primes are positive.

Stress points / ambiguity checks.

  • If a prime appears in both $P$ and $Q$, then the $p$-adic valuation of the product is $-2$, so equality to $1$ is impossible; hence any solution must have $P\cap Q=\varnothing$ (proved below).
  • The equation is verifiable: one explicit example would settle it affirmatively.
  • It is helpful to repackage the equation in an integral form using products of primes (also done below).

QUICK LITERATURE/CONTEXT CHECK

A quick web check (Erdős Problems site; MathOverflow) indicates that the existence of such prime sets is currently listed as open; weakened coprime versions allowing $1\in P\cup Q$ have explicit examples, but no example is known with all denominators prime. (See the sources cited in the accompanying chat message.)

ATTACK PLAN

  1. Derive structural necessities: disjointness; an exact integer system relating the products $\prod P$ and $\prod Q$; size lower bounds.
  2. Recast the problem as a search for a 2-cycle of a multiplicative arithmetic map $g$ on squarefree integers. This gives a crisp computational/structural target.
  3. (Unfinished) Attempt constructions/obstructions for 2-cycles of $g$.

WORK

Let \[ A:=\sum_{p\in P}\frac1p,\qquad B:=\sum_{q\in Q}\frac1q, \] so the desired identity is $AB=1$.

Lemma 1 (Disjointness). If $AB=1$ with $P,Q$ sets of primes, then $P\cap Q=\varnothing$.

Proof. Assume for contradiction that some prime $r\in P\cap Q$. Write \[ A=\frac{1}{r}+u,\qquad B=\frac{1}{r}+v \] where $u,v$ are sums of reciprocals of primes $\neq r$. Since $r\nmid$ denominators in $u$, we have $u\in\mathbb Z_r$ (the $r$-adic integers), and similarly $v\in\mathbb Z_r$. Then \[ A=\frac{1+ru}{r},\quad B=\frac{1+rv}{r}, \] and $1+ru\equiv 1\pmod r$, so $v_r(1+ru)=0$, hence $v_r(A)=-1$; similarly $v_r(B)=-1$. Therefore \[ v_r(AB)=v_r(A)+v_r(B)=-2\neq 0=v_r(1), \] contradiction. $\square$

Lemma 2 (Reduced-form denominators). For any finite set $P$ of primes, if we set $D_P:=\prod_{p\in P}p$ and \[ N_P:=\sum_{p\in P}\frac{D_P}{p}\in\mathbb Z, \] then \[ \sum_{p\in P}\frac1p=\frac{N_P}{D_P}\quad\text{and}\quad \gcd(N_P,D_P)=1. \]

Proof. The displayed fraction identity is just common-denominator addition. For $\gcd$, fix $r\in P$. Modulo $r$, every term $D_P/p$ with $p\neq r$ is divisible by $r$, while the term $D_P/r$ is not divisible by $r$. Hence $N_P\not\equiv 0\pmod r$. As this holds for every $r\in P$, no prime divisor of $D_P$ divides $N_P$, so $\gcd(N_P,D_P)=1$. $\square$

Corollary 3 (Integer "amicable" form). Assume $AB=1$ with $P\cap Q=\varnothing$. Let $D:=D_P=\prod_{p\in P}p$ and $E:=D_Q=\prod_{q\in Q}q$. Then necessarily \[ N_P=E\qquad\text{and}\qquad N_Q=D, \] i.e. \begin{equation}\label{wsref-0-65-71-3a-61-6d-69-63-61-62-6c-65} E=\sum_{p\in P}\frac{D}{p},\qquad D=\sum_{q\in Q}\frac{E}{q}. \end{equation}

Proof. By Lemma 2 we may write \[ A=\frac{N_P}{D},\qquad B=\frac{N_Q}{E} \] with $\gcd(N_P,D)=\gcd(N_Q,E)=1$. The equation $AB=1$ becomes \[ N_PN_Q=DE. \] Since $P$ and $Q$ are disjoint, $\gcd(D,E)=1$; also $\gcd(N_P,D)=1$ implies every prime dividing $D$ must divide $N_Q$, hence $D\mid N_Q$. Similarly $E\mid N_P$. Write $N_Q=D\cdot t$ and $N_P=E\cdot s$ with $s,t\in\mathbb Z_{>0}$. Then $N_PN_Q=DE\cdot st=DE$ forces $st=1$, hence $s=t=1$, i.e. $N_P=E$ and $N_Q=D$. Substituting back yields (eq:amicable). $\square$

Reformulation via an arithmetic map. For squarefree $n$ define \[ g(n):=\sum_{p\mid n}\frac{n}{p}. \] If $n=\prod_{i=1}^k p_i$ then $g(n)=\sum_{i=1}^k \prod_{j\neq i} p_j$. By Lemma 2, $\gcd(n,g(n))=1$ for squarefree $n$.

If (eq:amicable) holds, then with $D=\prod_{p\in P}p$ and $E=\prod_{q\in Q}q$ we have $E=g(D)$ and $D=g(E)$. Thus Problem #307 is equivalent to asking whether $g$ has a 2-cycle on squarefree integers: does there exist squarefree $D\neq 1$ such that $E:=g(D)$ is squarefree and $g(E)=D$?

A simple size obstruction. From $AB=1$ we have $A+B\ge 2$ (AM--GM), hence \[ \sum_{r\in P\cup Q}\frac1r = A+B \ge 2. \] Since the reciprocal sum is maximized by taking the smallest primes, one obtains a lower bound on $|P\cup Q|$. A direct computation gives \[ \sum_{i=1}^{58}\frac{1}{p_i}\approx 1.9987400431<2 \quad\text{and}\quad \sum_{i=1}^{59}\frac{1}{p_i}\approx 2.0023501514>2, \] where $p_i$ denotes the $i$th prime. Therefore any solution must have \[ |P\cup Q|\ge 59. \]

VERIFICATION (adversarial proof check; stress-test edge cases)

  • Lemma 1: The key check is $u,v\in\mathbb Z_r$; this holds because each term $1/p$ with $p\neq r$ is an $r$-adic integer. Also $1+ru\equiv 1\pmod r$ ensures $v_r(A)=-1$ with no cancellation.
  • Lemma 2: The congruence argument mod $r$ is robust because exactly one term survives mod $r$.
  • Corollary 3: The divisibility deductions use coprimality $\gcd(N_P,D)=1$ and $\gcd(D,E)=1$; both are proved/obvious from Lemma 2 and disjointness.
  • The "$|P\cup Q|\ge 59$" bound uses numerical computation of prime reciprocal sums; it is not logically necessary for Corollary 3, but it is a correct necessary condition given the displayed computed inequalities.

FINAL

UNRESOLVED.
  • What is proved here. Any solution must have $P\cap Q=\varnothing$, and with $D=\prod_{p\in P}p$, $E=\prod_{q\in Q}q$ it must satisfy the exact integer system \[ E=\sum_{p\in P}\frac{D}{p},\qquad D=\sum_{q\in Q}\frac{E}{q}, \] equivalently $E=g(D)$ and $D=g(E)$ for $g(n)=\sum_{p\mid n}n/p$ on squarefree $n$. Also necessarily $|P\cup Q|\ge 59$.
  • Key obstacle. No method here constructs a 2-cycle of $g$ (and none is ruled out).
  • Most promising next step. Use the 2-cycle formulation to guide a targeted search: seek squarefree $D$ with many prime factors such that $g(D)$ is squarefree and $g(g(D))=D$; try to incorporate congruence constraints and size heuristics to prune.
  • Deliverables if continuing. Either an explicit $(P,Q)$ (settling the problem affirmatively), or a structural obstruction proving $g$ has no 2-cycles on squarefree integers.

COMPLETION ESTIMATE

COMPLETION: 35%