Asked by Barbeau [Ba76]. Can this be done if we drop the requirement that all $p\in P$ are prime and just ask for them to be relatively coprime, and similarly for $Q$?

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 $1\not\in P\cup Q$.

It is easy to see that, if $P$ and $Q$ are sets of primes, then $P$ and $Q$ are disjoint, and $\sum_{p\in P\cup Q}\frac{1}{p}\geq 2$, whence $\lvert P\cup Q\rvert \geq 60$.

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