  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.
