Remeslennikov’s conjecture on profinite rigidity of free groups

Prove that if a residually finite group $G$ has the same finite quotients (equivalently, the same profinite completion) as a free group of rank at least 2, then $G$ is free.

Background

Profinite rigidity questions aim to recover group properties from profinite completions. For free groups this is a central open conjecture with implications for parafree groups and profinite classification within the one-relator domain.