Degrees of freedom of a constrained vector I have to handle with this lagrangian of a real vector $\chi^\mu$
$$
\mathcal{L} = -\frac{1}{4}F_{\mu\nu}^2 + B^\mu \square \chi_\mu + C\, \partial_\mu \chi^\mu + \mathcal{L}_{int}
$$
where $B^\mu$ and $C$ are Lagrange multipliers and $F_{\mu\nu}$ is the field strength of $\chi^\mu$.
Here $\mathcal{L}_{int}$ includes all the interactions that are constructed according to the symmetry 
$$
\chi^\mu \rightarrow\chi^\mu +\xi^\mu
$$
where $\partial_\mu \xi^\mu=\square\xi^\mu=0$.
Let's turn off all the couplings controlling the interactions and just focus on the quadratic part of the action.
Question: I have the feeling this theory is sick or ill-defined. How many degrees of freedom are actually propagating? How does the answer change if we turn on the interactions?
---EDIT after answer----
My confusion might be clearer by discussing the problem in the path-integral formalism. The partition function is
$$\begin{align}
Z[0] &= \int \mathcal{D\,}B^\mu \mathcal{D}\,C\, \mathcal{D}\,\chi^\mu e^{i\int d^d x  -\frac{1}{4}F_{\mu\nu}^2 + B^\mu \square \chi_\mu + C\, \partial_\mu \chi^\mu + \mathcal{L}_{int}} \cr &\propto \int \mathcal{D}\,\chi^\mu e^{i\int d^d x  -\frac{1}{4}F_{\mu\nu}^2 + \mathcal{L}_{int}}\delta\left(\square\chi_\mu\right)\delta\left(\partial^\mu\chi_\mu\right) \cr &= \int \mathcal{D}\,\chi^\mu e^{i\int d^d x \mathcal{L}_{int}}
\end{align}$$
where I performed the integration over the Lagrange multipliers. It looks the kinetic term is not there. 
 A: *

*For starters, OP's system is over-constrained: 4 gauge fields $\chi_{\mu}(x)$ is constrained by 4+1=5 Lagrange multiplier fields $B^{\mu}(x)$ and $C(x)$. This leads to an ill-defined product $$\prod_x\delta^4(\Box\chi_{\mu}(x))~\delta(\partial^{\nu}\chi_{\nu}(x))$$ of delta distributions in the path integral. 

*Put in a different way: There is another gauge symmetry
$$ \delta \chi_{\mu}=0, \qquad \delta B^{\mu}~=~\partial^{\mu}\varepsilon,\qquad \delta C~=~\Box\varepsilon, $$
which needs to be gauge-fixed. After that, there will be no propagating DOF.
A: Just considering the free lagrangian.
If you compute the equations of motion (by using Euler-Lagrange formalism for higher-derivative lagrangians) you obtain the Maxwell's theory equation
$$ \partial_\mu F^{\mu\nu}=0.$$
Of course, the terms included by Lagrange multipliers are just constraints on $\chi_\mu$, so apart from the eoms, we have that
$$ \partial^\mu\chi_\mu=0.$$
This condition is fixing (partially) the gauge on $\chi_\mu$ (the so-called Lorentz condition).
The other constraint $\partial^2\chi_\mu$ is automatically satisfied from the previous ones:
$$\partial^\mu F_{\mu\nu} = \partial^2\chi_\nu- \partial_\nu (\partial^\mu\chi_\mu)=\partial^2\chi_\nu=0,$$
so I would say that the (free) lagrangian is just the Maxwell lagrangian written in a particular gauge. The answer to the question would be: there are 2 physical degrees of freedom propagating (in 4d) as in the Maxwell theory.
With respect to the interacting lagrangian, I do not have an answer.
