= Solution
Substitute $T^a{}_{bc}=\ell^ag_{bc}$ into the preceding identity, using metric compatibility. Set $M_{bd}=\ell_{b;d}+\ell_b\ell_d$ and $S=g^{bd}M_{bd}=\nabla_a\ell^a+\ell_a\ell^a$. Contract the first and third curvature indices, in the stated curvature convention, to obtain
$$
\widetilde R_{bd}-R_{bd}=Sg_{bd}-M_{bd}.
$$
If the two <Ricci tensors> agree, then $M_{bd}=Sg_{bd}$. Taking the trace in dimension $n>1$ gives $S=nS$, so $S=0$ and $M=0$. Conversely $M=0$ makes the Ricci difference vanish. Thus \b[$\boxed{\widetilde R_{bd}=R_{bd}\iff\ell_{b;d}+\ell_b\ell_d=0}$].
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