Why the Lagrangian of a free particle cannot depend on the position or time, explicitly? On p. 5 in $\S$3 pf the book of Mechanics by Landau & Lifshitz, it is claimed that

[...] for a free particle, the homogeneity of space and time implies
  that Lagrangian cannot depend on position or time, explicitly.[...]

However, in my understanding, Lagrangian of a system is the function that determines the equation of motion, i.e given a the initial conditions and the Lagrangian of the system, we can determine the future configuration of the system, as in the case Newton's second law.
However, we also do know that adding a constant to our Lagrangian, or a time derivative of a function of position and time, the equation of motion does not change, hence, we get an "equivalent" Lagrangian in the sense that both functions lead us to the same conclusion about the dynamics of the system at hand.
Given this, I cannot understand why the Lagrangian of a free particle cannot depend on the position or time, explicitly. 
I mean it is clear that if that is the case, we have a simple Lagrangian that satisfy all the properties that you would expect it to have; however, why this is the only case that a free particle can have as a Lagrangian. 
Note: I have read this question, but I still cannot understand why does the origin would have a privileged status in that case.
 A: A free particle has no external forces acting on it. Therefore, momentum and energy are conserved. By Noether's Theorem this means the system has spatial and temporal translation symmetry. If the Lagrangian has explicit position or time dependence, then this cannot be the case.
Also note that adding a constant to the Lagrangian is not the same thing as adding in explicit position or time dependence.
Ultimately, it seems like the book is making a physical argument rather than a mathematical one. There do exist transformations that end up making the same equations of motion, but if you want to interpret the Lagrangian as the difference between the kinetic and potential energy, then you don't want explicit position and time dependence. If there was this dependence, then this would mean the particle is no longer free.
A: I will try to avoid jargon: if the Lagrangian is explicitly independent on a coordinate, like position or time, then the corresponding canonical momenta of that coordinate is conserved. So, for a free particle there are no external forces acting on it, so the Lagrangian only has a kinetic energy term which depends only on the velocity of the particle. Thus, energy is conserved because time does not appear in the Lagrangian, and linear momentum is conserved because position does not appear in the Lagrangian. 
As you said, you may add a time derivative of a function that satisfies the Euler-Lagrange equations, but this does not grant you the privilege of imposing explicit time or position dependence.

however, why this is the only case that a free particle can have as a Lagrangian. 

I think this is a far deeper question than you might realize: as it turns out, Lagrangians are not mathematically unique generally, but in terms of producing the correct equations of motion for some system only certain Lagrangians work. It's an open problem of theoretical physics to answer "why these Langrangians for these systems?" and there's no obvious answer to it currently other than that it works in producing the correct equations of motion (verified experimentally).
