Einstein metric condition in f(R) gravity ($f(4\Lambda )=2\Lambda f^{\prime }(4\Lambda )$)

Every four-dimensional Einstein metric with $R_{ab}=\Lambda g_{ab}$ has constant Ricci scalar $R=4\Lambda$. The vacuum metric f(R) gravity tensor reduces to $(\Lambda f^{\prime }(4\Lambda )-f(4\Lambda )/2)g_{ab}$, giving the displayed necessary and sufficient condition for a fixed cosmological constant. Both function and derivative may vanish, so division by $f^{\prime }$ is unnecessary. Requiring this for all real $\Lambda$ and a smooth $f$ instead yields $f(R)=C{R}^2$.