Erdős Problem #307

Erdős Problem #307 (Round 2)

1) ROUND-2 OBJECTIVE

Path C (structural obstruction / correction of the search space).

Round 1 reduced the problem to finding a nontrivial squarefree $2$-cycle of \[ g(n):=\sum_{p\mid n}\frac{n}{p}. \] That reformulation is correct but still very broad. The most promising Round-2 direction is therefore to prove local congruence constraints that any solution must satisfy, thereby ruling out whole classes of constructions and making any future search much more rigid.

2) Round-1 FOUNDATION USED

I use the following Round-1 results without reproving them.

• If \[ 1=\left(\sum_{p\in P}\frac1p\right)\left(\sum_{q\in Q}\frac1q\right), \] then $P\cap Q=\emptyset$.
• Writing \[ D:=\prod_{p\in P}p,\qquad E:=\prod_{q\in Q}q, \] one has the exact identities \[ E=\sum_{p\in P}\frac{D}{p}=g(D),\qquad D=\sum_{q\in Q}\frac{E}{q}=g(E). \] In particular $D,E$ are squarefree and coprime.
• The simple reciprocal-sum argument gives the necessary bound \[ |P\cup Q|\ge 59. \]

3) NEW INSIGHT / TOOL (ROUND-2)

The genuinely new ingredient is a cross-congruence lemma: for every $q\in Q$, \[ \sum_{p\in P} p^{-1}\equiv 0 \pmod q, \] and symmetrically for every $p\in P$, \[ \sum_{q\in Q} q^{-1}\equiv 0 \pmod p, \] where inverses are taken in the relevant finite field.

This turns the problem into a family of simultaneous local zero-sum conditions. It immediately yields parity and residue-class balance laws, and rules out any construction in which one side lies entirely in one nonzero residue class modulo a prime from the other side unless the cardinality is a multiple of that prime.

4) ATTACK PLAN (ROUND-2)

The main Round-1 gap was that the $2$-cycle reformulation did not explain why such cycles should be rare. The plan is:

1. derive exact local congruence conditions from $E=g(D)$ and $D=g(E)$;
2. extract concrete consequences for the primes $2$ and $3$;
3. rule out an entire class of "residue-homogeneous" search strategies.

This does not solve the problem, but it cuts the candidate space much more sharply than the Round-1 product reformulation alone.

5) WORK (ROUND-2)

Let \[ A:=\sum_{p\in P}\frac1p=\frac{E}{D},\qquad B:=\sum_{q\in Q}\frac1q=\frac{D}{E}. \]

Lemma 5.1. No finite nonempty set of distinct primes has reciprocal sum equal to an integer.

Proof. For a nonempty prime set $R$, let \[ M_R:=\prod_{r\in R} r, \qquad N_R:=\sum_{r\in R}\frac{M_R}{r}. \] By the Round-1 denominator lemma, $\gcd(N_R,M_R)=1$, so the fraction \[ \sum_{r\in R}\frac1r=\frac{N_R}{M_R} \] is already in lowest terms with denominator $M_R>1$. Hence it cannot be an integer. $\square$

Corollary 5.2. In any solution of Problem #307 one has $D\neq E$. Equivalently, neither of the two factors can equal $1$.

Proof. If $D=E$, then $A=E/D=1$ and $B=D/E=1$, contradicting Lemma 5.1 on either side. $\square$

Thus one factor is strictly larger than $1$ and the other strictly smaller than $1$.

Lemma 5.3 (cross-congruence). For every $q\in Q$, \[ \sum_{p\in P} p^{-1}\equiv 0 \pmod q, \] and for every $p\in P$, \[ \sum_{q\in Q} q^{-1}\equiv 0 \pmod p. \]

Proof. Fix $q\in Q$. Since $q\mid E=g(D)$ and $q\notin P$, we have $q\nmid D$, so $D$ is invertible modulo $q$. Reducing \[ E=\sum_{p\in P}\frac{D}{p} \] modulo $q$ and multiplying by $D^{-1}$ gives \[ 0\equiv \sum_{p\in P} p^{-1}\pmod q. \] The second statement is the same argument with the roles of $P$ and $Q$ swapped. $\square$

Corollary 5.4 (parity law).

• If $2\in Q$, then $|P|$ is even.
• If $2\in P$, then $|Q|$ is even.
• If $2\notin P\cup Q$, then both $|P|$ and $|Q|$ are odd.

Proof. Modulo $2$, every nonzero residue is $1$, so Lemma 5.3 gives \[ \sum_{p\in P} p^{-1}\equiv |P|\equiv 0\pmod 2 \] whenever $2\in Q$. This proves the first claim; the second is symmetric. If $2\notin P\cup Q$, then $D$ and $E$ are odd, so from the Round-1 identities \[ E=\sum_{p\in P}\frac{D}{p},\qquad D=\sum_{q\in Q}\frac{E}{q} \] each side is a sum of odd integers; hence $E\equiv |P|\pmod 2$ and $D\equiv |Q|\pmod 2$. Since $D,E$ are odd, both $|P|$ and $|Q|$ are odd. $\square$

Corollary 5.5 (the $3$-balance law). If $3\in Q$, then \[ \#\{p\in P: p\equiv 1\pmod 3\} \equiv \#\{p\in P: p\equiv 2\pmod 3\} \pmod 3. \] The analogous statement with $P,Q$ swapped also holds.

Proof. If $3\in Q$, then $3\notin P$ because $P\cap Q=\emptyset$. Hence every $p\in P$ is congruent to $1$ or $2$ mod $3$, and in $\mathbf F_3$ one has $1^{-1}=1$ and $2^{-1}=2\equiv -1$. Lemma 5.3 therefore gives \[ 0\equiv \sum_{p\in P} p^{-1} \equiv \#\{p\equiv 1\pmod 3\}-\#\{p\equiv 2\pmod 3\} \pmod 3, \] which is exactly the claim. $\square$

Corollary 5.6 (residue-homogeneous constructions are heavily restricted). Suppose $q\in Q$, and all primes in $P$ lie in one fixed nonzero residue class $a\pmod q$. Then \[ q\mid |P|. \] Symmetrically, if $p\in P$ and all primes in $Q$ lie in one fixed nonzero residue class $b\pmod p$, then \[ p\mid |Q|. \] In particular, if $q\in Q$ and $q>|P|$, then $P$ cannot be contained in a single nonzero residue class modulo $q$.

Proof. If every $p\in P$ satisfies $p\equiv a\pmod q$, then every inverse satisfies $p^{-1}\equiv a^{-1}\pmod q$. Lemma 5.3 gives \[ 0\equiv \sum_{p\in P} p^{-1}\equiv |P|\,a^{-1}\pmod q. \] Since $a\not\equiv 0\pmod q$, also $a^{-1}\not\equiv 0\pmod q$, and therefore $q\mid |P|$. The second claim is identical with the roles reversed. $\square$

6) ADVERSARIAL VERIFICATION

• The cross-congruence lemma uses the invertibility of $D$ modulo $q$ (or $E$ modulo $p$). This is valid because Round 1 already proved $P\cap Q=\emptyset$.
• Lemma 5.1 is only a corollary of the Round-1 reduced-denominator fact; no hidden cancellation is possible.
• Corollary 5.4 splits into two cases. The first two bullets come from Lemma 5.3 modulo $2$; the third bullet uses the explicit oddness of the summands when neither side contains $2$.
• These new local conditions do not yet improve the Round-1 lower bound $|P\cup Q|\ge 59$, and I am not claiming a proof of the stronger bound "$\ge 60$" appearing on the current webpage discussion.

7) FINAL

UNRESOLVED (BUT STRICTLY ADVANCED).

The strict advance over Round 1 is the local zero-sum structure of Lemma 5.3 and its consequences:

• neither factor can equal $1$;
• every prime on one side imposes a modular vanishing condition on the other side;
• parity and mod-$3$ balance laws follow immediately;
• residue-homogeneous search patterns are ruled out unless the opposite-side cardinality is divisible by the controlling prime.

This does not settle Problem #307, but it sharply narrows the allowed architecture of any future solution.

8) COMPLETION ESTIMATE

COMPLETION: 47%

9) REFERENCES

1. P. Erdős and R. Graham, Old and New Problems and Results in Combinatorial Number Theory, problem source for #307.
2. T. F. Bloom, Erdős Problem #307, Erdős Problems website, accessed 2026-03-07.
3. T. F. Bloom, Erdős Problem #307 -- Discussion thread, Erdős Problems website, accessed 2026-03-07.

10) OUTPUT FORMAT

This file is already in drop-in LaTeX format with no preamble.

Erdős Problem #307: independent continuation

Date: 23 September 2026. Source versions read in full: 307.tex in gpt_pro_5.4.

1) OBJECTIVE AND SOURCE AUDIT

We seek prime sets whose reciprocal sums multiply to one. The source's squarefree two-cycle reformulation and cross-congruences are valid. I turn them into an exact one-sided search specification and check the small-cardinality reciprocal bound with rational arithmetic.

2) WORK: ONE SIDE DETERMINES THE OTHER

For squarefree $D>1$ define $g(D)=\sum_{p\mid D}D/p$. For every $p\mid D$, $g(D)\equiv D/p\not\equiv0\pmod p$, so $(g(D),D)=1$. Consequently a fixed finite nonempty prime set $P$, with product $D$, can belong to a solution if and only if $E=g(D)>1$ is squarefree and $g(E)=D$. The other side is then uniquely the prime factors of $E$. Since $(D,E)=1$, the two prime sets are disjoint. This follows because reduced fractions for the two reciprocal sums must be mutual reciprocals. It is also sufficient by direct multiplication.

For a bounded search over $P$, an exact rejection sequence is therefore: compute $D,E$; reject $E=1$; factor $E$; reject a repeated prime factor; finally test the integer identity $g(E)=D$. Merely passing the source's local congruences is not a substitute for the last equality.

The two reciprocal sums have product one, so their sum is at least two by the arithmetic--geometric mean inequality. Disjointness makes this sum the reciprocal mass of their union.

Here is a useful bound on one finite boundary case. Let $s_j$ be the sum of reciprocals of the first $j$ primes. Exact fraction arithmetic gives $s_{58}<2<s_{59}$. If a solution had exactly 59 distinct primes in its union, and largest prime $q$, then \[ 2\le\sum_{p\in P\cup Q}1/p\le s_{58}+1/q, \] so \[ q\le\left\lfloor\frac1{2-s_{58}}\right\rfloor=793. \] Thus that boundary case is a genuinely finite search among primes at most 793; it is not made infinite by an unbounded last prime. I have computed this bound but have not enumerated all such partitions.

3) VERIFICATION AND FINAL

The official page records the stronger lower bound 60 on the union size; I do not claim to improve it, nor to reprove it merely from $s_{58}<2$. The finite-bound calculation and exact two-cycle test are independent checks that avoid the erroneous inference that cross-congruences alone force a solution. UNRESOLVED. No nontrivial two-cycle was found or excluded in general. The target and literature comparison were checked at https://www.erdosproblems.com/307.

8) COMPLETION ESTIMATE

COMPLETION: 5% This is a subjective estimate of progress toward the entire stated problem, not a probability that the conjecture or an unverified proof is correct.

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%