Converse of the Lagrangian form-invariance The form-invariance of the Lagrange equations implies the existence of a function $\ A( q_k, t)$ so that
$\ \begin{equation} L' (q_k,  v_k, t) -L(q_k,  v_k, t) = \frac d {dt} A( q_k, t) \end{equation}$
Is there any way of proving the converse? That is, asuming the existence of the funcion $A$ prove the form-invariance of the Lagrangre equations. 
 A: An elegant proof of the converse is obtained by using the action formulation.  Recall that for a given Lagrangian $L$, the action is a functional of paths defined as follows:
\begin{align}
  S_L[q] = \int_{t_a}^{t_b} dt\, L_q(t)
\end{align}
Now suppose that two Lagrangians differ by a total time derivative;
\begin{align}
  L'_q(t)-L_q(t) = \frac{dA_q}{dt}(t),
\end{align}
then we immediately find that
\begin{align}
  S_{L'}[q] - S_{L}[q] = A_q(t_b) - A_q(t_a)
\end{align}
If, in addition, $A$ is a local function (see Definition of Local Function) of paths $q$, then this implies that
\begin{align}
  \delta S_{L'}[q] - \delta S_L[q] = 0
\end{align}
for all variations keeping the endpoints of paths fixed.  Therefore, for all such variations, $q$ is a stationary point of $S_{L'}$ if and only if $q$ is a stationary point of $S_L$.  Now, if $L$ and $L'$ are local functions that depend only on $q$, it's first derivative, and time, namely if
\begin{align}
  L_q(t) &= L(q(t), \dot q(t), t) \\
  L'_q(t) &= L'(q(t), \dot q(t), t),
\end{align}
then a path $q$ is a stationary point of the action for all such variations if and only if it satisfies the corresponding Euler-Lagrange equations.  Combining these facts, we see that if $q$ satisfies the EL equations for $L$, then it is a stationary point of $S_L$, so it is also a stationary point of $S_{L'}$ and therefore satisfies the EL equations for $L'$, as desired.
Addendum.
We show that the variation of a function $A$ that is local in paths $q:[t_a,t_b]\to \mathbb R$ and their first derivatives vanishes at the endpoints $t_a$ and $t_b$ for all variations that keep path endpoints fixed.
Let a function $A$ that is local in paths $q$ and their time derivatives be given.  Namely, $A$ is such that there exists a function $\alpha$ for which
\begin{align}
  A_q(t) = \alpha(q(t), \dot q(t), t)
\end{align}
for all paths $q$.  Now let a path $q$ defined on $[t_a, t_b]$ be given, and let $\delta q$ be a variation of $q$, then if $\delta A_q$ denotes the variation of $A_q$ induced by $\delta q$, then we have
\begin{align}
  \delta A_q(t) = \frac{\partial \alpha}{\partial q}(q(t), \dot q(t), t)\delta q(t) + \frac{\partial\alpha}{\partial\dot q}(q(t), \dot q(t),t) \delta\dot q(t)
\end{align}
Now, for a sufficiently smooth variation that keeps the endpoints $q(t_a), q(t_b)$ of the path fixed, one has
\begin{align}
  \delta q(t_a) &= \delta q(t_b)=0\\ \delta \dot q(t_a) &= \delta \dot q(t_b) = 0,
\end{align}
which you should be able to convince yourself of (see https://physics.stackexchange.com/a/93290/19976 for mathematical details of variations that will help) from which it follows that
\begin{align}
  \delta A_q(t_a)= \delta A_q(t_b) = 0.
\end{align}
as desired.
