 Solution

Substitute $T^a{}_{bc}={\ell }^a{g}_{bc}$ 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 obtainIf the two Ricci tensors agree, then $M_{bd}=S{g}_{bd}$. 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 $\boxed{{\tilde{R}}_{bd}=R_{bd}⟺{\ell }_{b;d}+{\ell }_b{\ell }_d=0}$.