How does one interpret $m=p/c$? In A.P. French's Special Relativity, page $20$, the author wrote,

For photons we have $$\tag{1}E=pc$$
  and $$\tag{2}m=\frac{E}{c^2}$$
  (the first experimental, the second based on Einstein's box). Combining these we have $$\tag{3}m=\frac{p}{c}$$

He didn't give any justification for combining equation $1$ (valid only for photons) and equation $2$ (which can't be applied to photons since they are massless) into equation $\textbf{3}$.
To which particles does equation $\textbf{3}$ apply to: massless or non-massless?
I couldn't find a satisfactory answer in this post.
 A: Unfortunately, you’re trying to learn from a 50-year-old book that presents the concepts in what we now consider to be a very confusing way. 
Now, we (generally) use “mass” m to denote the rest mass of a particle: the mass that’s characteristic of any electron, for example. Then in the modern convention:
$$ (mc^2)^2+ (cp)^2 = E^2$$
French uses “mass” to denote a form of total relativistic energy. (I'm describing the following, not defending it; this is not a good idea) He's using a concept that we now call "relativistic mass", a "mass" that increases with the gamma factor: a moving body "has more mass".  Mathematically, this is like saying that $mc^2 = E$ always, so as the body gets more energy $E$ with the addition of momentum $p$, the mass goes up.  (I agree this is confusing; that's my point) All of this goes completely sideways with photons, where the rest mass is zero, the gamma factor is infinite, and their product has to miraculously work.
Learning from this book is going to be difficult because you’ll be learning a terminology and form of calculation that will cause confusion when you ask questions, and probably isn’t consistent with what you already know. 
