= Solution
The decay is energetically open, but two spin-zero <pions> have total <angular momentum> equal to their relative <orbital angular momentum> $L$. Conserving the parent <spin> requires $L=1$. Interchanging two identical <neutral pions> multiplies this orbital state by $(-1)^L=-1$, whereas their total state must be symmetric by Bose statistics. There is no different <spin> or species label to supply another minus sign. Thus \b[the two-neutral-pion final state is forbidden]. Its <parity> for $L=1$ would actually be $(-1)^2(-1)^1=-1$, matching the parent; the decisive obstruction is identical-boson <symmetry>, not the <energy> or the shared quark-wavefunction component.
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