Let's have the QED Lagrangian.
$$\mathcal{L} = -\frac{1}{4} F_{\mu\nu}F^{\mu\nu} + \bar{\Psi}(i\partial_\mu \gamma^\mu - m)\Psi + g\bar{\Psi}A_\mu \gamma^\mu \Psi.\tag{1}$$
The equations of motion are:
$$ \partial_\mu F^{\mu\nu} = g\bar{\Psi}\gamma^\nu \Psi \tag{2}$$
and
$$(i\partial_\mu \gamma^\mu - m)\Psi = -g A_\mu \gamma^\mu \Psi.\tag{3}$$
If we use the first EOM and replace the current (in the third term of the $\mathcal{L}$) by the derivative of $F_{\mu\nu}$ we find the following Lagrangian.
$$\mathcal{L} = \frac{1}{2} F_{\mu\nu}F^{\mu\nu} + \bar{\Psi}(i\partial_\mu\gamma^\mu - m)\Psi.\tag{4}$$
We can see that the interacting term between the fermion and the gauge boson disappeared. The main question is How is possible that two equivalent lagrangians (we can go from one to the other by EOM) have different Feynman diagrams?