gpt pro 5.4 attempt on Math-5568

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.