We derive the exact $T=0$ seniority-zero eigenstates of the isovector pairing Hamiltonian for an even number of protons and neutrons. Nucleons are supposed to be distributed over a set of non-degenerate levels and to interact through a pairing force with constant strength. We show that these eigenstates (and among them, in particular, the ground state) are linear superpositions of products of $T=1$ collective pairs arranged into $T=0$ quartets. This grouping of protons and neutrons first into $T=1$ collective pairs and then into $T=0$ quartets represents the distinctive feature of these eigenstates. This work highlights, for the first time on the grounds of the analytic expression of its eigenstates, the key role played by the isovector pairing force in the phenomenon of nuclear quarteting.
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