special relativity - Trivial Solution for Energy Momentum Equation

In special relativity we define momentum as

$$\frac{m v } {\sqrt{1 - v^2 / c^2}} \tag{1} \label{1}$$ and energy as $$E = \frac{m c^2 } {\sqrt {1 - v^2 / c^2}} \tag{2} \label{2}$$ So with these, we can derive relation for momentum and energy: $$\begin{align} E^2 - p^2 c^2 =& \frac{ m^2 c^4 - m^2 v^2 c^2} {1 - v^2 / c^2} \\ = & \frac{ m^2 c^4 ( 1 - v^2 / c^2)} { 1 - v^2 c^2} \\ = & (m c^2)^2 \tag{3}\label{3} \end{align}$$ Physicist say for massless particle (photon) $E$ is indeterminate $0/0$ and the same also for momentum (you can have zero mass, provided you have same velocity with $c$ ). So equation $\ref{3}$ become
$$E = p c \tag{4} \label{4}$$ But how we can say equation $\ref{4}$ is true whereas it is derived from (depend on) result in equation $\ref{3}$. If $m = 0$ then $E^2 = 0$ and $p^2 c^2 = 0$ too and it is become trivial identity $0 = 0$.

And if we still forcing to use equation $\ref{4}$ we must remember that for photon, energy and momentum are indeterminate and can take any values, $E = 0/0$ and $p = 0/0$ so equation $\ref{4}$ become a strange form

$$0/0 = 0/0$$ LHS can be filled with any values, and also RHS, and it is inconsistencies in mathematical formulation. I can say that $5= 3$, $100 = 200$, $45 = 23$, etc.