Let $(M,g)$ be a $d$-dimensional Riemannian manifold. Let $X \in \Gamma(TM)$ be a generator of conformal transformations. Blumenhagen shows in the book "Introduction to Conformal Field Theory" that such $X$ satisfies the equation
$$(d-1)\Box(\nabla \cdot X)=0.\tag{2.5}$$
where $\nabla\cdot X$ means $$\nabla \cdot X=\nabla^\mu X_\mu$$ with the summation convention understood and where $\Box$ is $$\Box = \nabla^\mu \nabla_\mu.$$
Then Blumenhagen says:
We note that Eq. (2.5) implies that $\nabla\cdot X$ is at most linear in $x^\mu$, i.e., $\nabla\cdot X = A + B_\mu x^\mu$ with $A$ and $B_\mu$ constant. Then it follows that $X_\mu$ is at most quadratic in $x^\nu$ and so we can make the ansatz: $$X_\mu = a_\mu + b_\mu x^\mu + c_{\mu\nu}x^{\mu} x^\nu.$$
In the above $x^\mu$ are the coordinate functions of a local chart $(x,U)$.
Now I find this argument quite weird. The equation $\Box f = 0$ is the Laplace equation, and we know of plenty solutions which are not just $A + B_\mu x^\mu$. In particular in $\mathbb{R}^d$ we have $f = 1/r$ such a solution on the domain $U = \mathbb{R}^d\setminus \{0\}$.
So what is behind Blumenhagen's argument? Why Eq. (2.5) implies $\nabla \cdot X$ linear in the coordinate functions?
I had the same question reading some pdfs from the internet. It seems the argument has become garbled from Di Francesco et al's CFT book, pp$96-7$. The equivalent of equation $(2.5)$ is $(4.8)$, which is fed back into a previous equation $(4.7)$ to give that the gradient of the gradient of div$X$ is zero - eq $(4.9)$. Then another equation $(4.5)$ is used to show the gradient of the gradient of vector X is a constant. This leads to the quadratic form for $X$. The main pdf I have been reading is https://arxiv.org/abs/1511.04074 In that document it makes a similar garbling. Your $2.5$ is essentially $2.9$ in that doc. To make the argument work you need to put $2.9$ into $2.8$ to eliminate the Laplace term. $(2.13)$ is the equivalent of $(4.5)$ in Di Fracesco et al.