= Solution
Choose the <Riemann curvature tensor> convention
$$
[\nabla_\mu,\nabla_\nu]\psi=\frac14R_{\mu\nu ab}\gamma^{ab}\psi,\qquad
\{\gamma^\mu,\gamma^\nu\}=2g^{\mu\nu},
$$
with positive <scalar curvature> for a round sphere. The connection preserves the gamma matrices. If $D=\gamma^\mu\nabla_\mu$, its square is
$$
D^2\psi=\nabla^2\psi+\frac12\gamma^{\mu\nu}[\nabla_\mu,\nabla_\nu]\psi
=\nabla^2\psi+\frac18\gamma^{\mu\nu}R_{\mu\nu ab}\gamma^{ab}\psi.
$$
Here $\nabla^2$ is the <rough Laplacian>, including the connection on the derivative index. In the product of two antisymmetric gamma pairs, the fully antisymmetric four-gamma term vanishes by the <first Bianchi identity>. The two-gamma contractions vanish because the <Ricci tensor> is symmetric, while the scalar contractions give $-2R$. Thus the <Lichnerowicz spinor-square formula> is
$$
\boxed{D^2=\nabla^2-\frac R4.}
$$
Multiplying $(D+m)\psi=0$ by $-D+m$, with constant $m$, proves
$$
\boxed{-\nabla^2\psi+\frac R4\psi+m^2\psi=0.}
$$
In Riemannian signature $D$ in this positive-Clifford convention is formally <anti-self-adjoint>; the more usual <self-adjoint> Dirac operator $iD$ has square $-\nabla^2+R/4$. Stating this convention avoids an apparent sign conflict between the two forms of the formula.
Now let $D\psi=0$ on the compact Riemannian <spin manifold>. Use the positive-definite spinor <inner product> and integrate the squared equation. <Integration by parts> has no boundary contribution, giving
$$
0=\int_M\left(\langle\psi,-\nabla^2\psi\rangle+\frac R4|\psi|^2\right)dV
=\int_M\left(|\nabla\psi|^2+\frac R4|\psi|^2\right)dV.
$$
Both terms are nonnegative. Smoothness then forces $\nabla\psi=0$ pointwise. Consequently \b[every harmonic spinor is parallel], and $R|\psi|^2=0$. A nonzero parallel spinor has constant positive norm on its connected component, so its <scalar curvature> vanishes there. This is <compact harmonic-spinor rigidity under nonnegative scalar curvature>.
The full Ricci conclusion follows from integrability, not merely from vanishing <scalar curvature>. Since a parallel spinor has zero curvature for its <spin connection>, contract its <Ricci identity> with $\gamma^\nu$:
$$
0=\gamma^\nu[\nabla_\mu,\nabla_\nu]\psi
=\frac14R_{\mu\nu ab}\gamma^\nu\gamma^{ab}\psi
=-\frac12R_{\mu\nu}\gamma^\nu\psi.
$$
The three-gamma term vanishes by the <Bianchi identity>; the remaining contractions give the displayed Ricci term. For each fixed $\mu$, put $v_\nu=R_{\mu\nu}$. <Clifford multiplication> twice gives
$$
0=(v_\nu\gamma^\nu)^2\psi=|v|^2\psi.
$$
Positive definiteness and $\psi\ne0$ imply $v=0$. Thus the <Riemannian parallel-spinor Ricci-flatness> result is
$$
\boxed{R_{\mu\nu}=0.}
$$
On a connected manifold a nontrivial parallel solution is nowhere zero, so this holds everywhere. On a disconnected manifold it holds on every component supporting a nonzero spinor; the stated nowhere-vanishing hypothesis ensures all components are covered. The positive-definite <metric tensor> is essential to the last inference.
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