= Solution
Keeping the same two spin states but incorrectly imposing <boson> exchange symmetry would require a symmetric total <wavefunction>. The spin-antisymmetric <singlet state> would then accompany $\Phi_-$, while the spin-symmetric <triplet state> would accompany $\Phi_+$. Thus
$$
E_S^{\rm hypothetical}=E_{\rm one}+J-K,\qquad
E_T^{\rm hypothetical}=E_{\rm one}+J+K,
$$
and $\boxed{E_T^{\rm hypothetical}-E_S^{\rm hypothetical}=2K}$. The ordering reverses: the triplet would be destabilized by $2K$ relative to the singlet. A genuinely spinless <boson> would have no singlet/triplet spin distinction; this comparison changes exchange symmetry while retaining the question's spin degrees of freedom.
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