Vtyjkyuk

Let $(M,g)$ be an $n$-dimensional complete Riemannian manifold. Consider the following equation
$$tag1 s=nc+Delta f,$$
where $s$ is the scalar curvature, $c$ is any real number and $fin C^2(M)$. It is given that by using contracted Binachi identity, the above equation implies that
$$tag2 |nabla f|^2+s+2cf=A_0,$$
for some constant $A_0$. I can not get the intermediate steps. Please help me to under stand this proof.

The above is found in p.376 of this paper.

This is in the context of Ricci soliton and (2) is a well-known formula proved by Hamilton which follows from the Ricci soliton equation instead of (1). The proof can be found in Lemma 1.1 in this survey. For completeness I'll record the proof here:

The Ricci soliton equation is
$$tag3 operatornameRc = c g + D^2 f,$$
or $R_jk = cg_jk + f_jk$. Taking covariant derivative to (3) and commuting the derivatives,
beginalignnabla _i R_jk &= nabla _i f_jk \
&= nabla_j f_ik + R_ijk^l nabla_l f \
&= nabla_j R_ik+R_ijk^l nabla_l f.
endalign
Taking the trace on $j$ and $k$, and using the contracted second Bianchi identity,
$$nabla_i s = nabla^j R_ij - R^l_i nabla_l f = frac 12 nabla_i s - R^l_i nabla_l f.$$
Thus $nabla _i s = -2 R_i^l nabla_l f$. So
beginalign
nabla_i(|nabla f|^2+s+2cf) &= 2f_ji nabla_j f + nabla_i s + 2cnabla_ i f \
&= 2(R_ij -c g_ij) nabla_j f - 2R_ij nabla _j f + 2c nabla_i f = 0.
endalign

Thus $|nabla f|^2+s+2cf$ is constant.