= Solution
An <S wave (quantum scattering)> has $L=0$ and positive <parity>. The <deuteron>'s $J=1$ therefore requires a <spin-one-half triplet state>, $S=1$. The two-nucleon <isospin> description makes this the antisymmetric $T=0$ channel. Its existence proves sufficient attraction to bind in the $\,{}^3S_1$-dominated channel.
A <spin one-half> pair could also form $S=0$. The absence of a bound $\,{}^1S_0$ <deuteron> indicates that the singlet interaction does not bind, not that it is everywhere repulsive. Together with triplet binding, this shows that a spin-independent central attraction is insufficient. No radial excited bound level also constrains the strength and range: the attraction supports only a shallow low-energy <bound state>.
For identical <neutrons> or <protons>, two-particle <isospin> has $T=1$ and is symmetric. At even $L$, spatial exchange is symmetric, so the <Pauli exclusion principle> requires an antisymmetric <spin> state, $S=0$. A low-energy $nn$ or $pp$ pair cannot occupy the <deuteron>'s $S=1,T=0$ channel. With approximate <charge independence of nuclear forces>, their singlet <strong interaction> corresponds to the unbound neutron-proton singlet; proton-proton electrostatic repulsion further disfavors binding. \b[There is no bound <dineutron> or <diproton> in these physical low-energy channels.] The inference uses charge independence and singlet-channel behavior; the <deuteron>'s quantum numbers alone do not constrain every imaginable potential.
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