GPT 6 Astra Ultra attempt on Math-5568
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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.