= Solution
Put $\kappa=8\pi G/c^4$ and $T=T_\nu{}^\nu$. The <contracted Bianchi identity> and <metric compatibility> give
$$
\left(\frac12-\alpha\right)\partial_\mu R
=-\kappa\nabla_\nu T_\mu{}^\nu.
$$
Taking the <trace> of the field equation in four dimensions gives $(1-4\alpha)R=-\kappa T$. For $\alpha\ne1/4$, eliminate $R$:
$$
\boxed{\nabla_\nu T_\mu{}^\nu
=\frac{\tfrac12-\alpha}{1-4\alpha}\,\partial_\mu T.}
$$
Ordinary covariantly coupled matter obeys <stress-energy conservation>, $\nabla_\nu T_\mu{}^\nu=0$. Its <trace> need not be constant, so the <trace coefficient fixed by stress-energy conservation> is
$$
\boxed{\alpha=\frac12.}
$$
This restores the <Einstein tensor> and is the only coefficient compatible with arbitrary conventional matter without adding a new constraint or changing the matter conservation law. Other coefficients can work for specially restricted sources with constant <trace>, but do not give the same universal matter coupling.
The excluded case $\alpha=1/4$ must be treated before dividing: its <trace> equation requires $T=0$. If stress-energy is also conserved, its divergence equation requires $\partial_\mu R=0$. It can describe a restricted traceless-source sector with constant scalar curvature, not general matter such as ordinary dust. In particular it is not the general <trace>-free Einstein equation with source $T_{\mu\nu}-g_{\mu\nu}T/4$, which is a different equation.
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