Erdős Problem 307 #
Reference: erdosproblems.com/307
Are there two finite set of primes $P$ and $Q$ such that
$$ 1 = \left( \sum_{p \in P} \frac{1}{p} \right) \left( \sum_{q \in Q} \frac{1}{q} \right) $$ ?
Asked by Barbeau [Ba76].
[Ba76] Barbeau, E. J., Computer challenge corner: Problem 477: A brute force program.
Instead of asking for sets of primes, ask only that all elements in the sets be relatively coprime.
Cambie has found several examples when this weakened version is true. For example, $$ 1=\left(1+\frac{1}{5}\right)\left(\frac{1}{2}+\frac{1}{3}\right) $$ and $$ 1=\left(1+\frac{1}{41}\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{7}\right). $$
There are no examples known of the weakened coprime version if we insist that $1\not\in P\cup Q$.
A machine-checked barrier for Erdős 307 (Bonfioli, 2026): any solution with Q nonempty uses at
least 59 primes in total, and (∏_{p ∈ P} p)² ≥ 4·10¹¹² — i.e. ∏_{p ∈ P} p ≥ 2·10⁵⁶ (and, by
symmetry, the same for ∏ Q); so no solution lies below a prime-product of 2.09·10⁵⁶. The full
sorry-free proof is in the linked repository (Closed.lean at tag v1.0.0): the left conjunct is
card_ge_59, the right is erdos307_barrier_closed. The only non-logical input is a native_decide
evaluation of the first 59 primes; the axioms are propext, Classical.choice, Quot.sound together
with that native_decide.