Is $E=mc^2$ only partially true? I've obviously seen that $E=mc^2$. But I've also seen that the equation $E^2=(mc^2)^2+(pc)^2$ (where $p$ = momentum) is true.
If you square $E=mc^2$, and subtract the result from the other equation, you get $0=(pc)^2$. In other words, momentum doesn't exist!
I assume combining these equations is a mistake. Is $E=mc^2$ only something that holds in certain circumstances? Is $E^2=(mc^2)^2+(pc)^2$ the more complete, correct version of $E=mc^2$?
I know I'm wrong -- where, exactly, is my mistake?
 A: There are two ways to answer this, based on two different definitions of mass. In most modern courses, $m$ refers specifically to the rest energy of the object (meaning how much energy it has at rest). In that view, $E=mc^2$ only applies to objects at rest, while the full formula is $E^2=p^2c^2+m^2c^4$ (you can also say $E=\gamma mc^2$, where $\gamma$ is the Lorentz factor).
There is also another definition, used occasionally in older textbooks, in which $m$ refers to the relativistic mass of the object. In this view, the relativistic mass and the rest energy $m_0$ are related by $m=\gamma m_0$, where $\gamma$ is the Lorentz factor. So, under this set of definitions, $E=mc^2$ is the only valid equation. Let me stress that this view of mass is largely obsolete at this point (there are a few answers on this site that explain some reasons why), but it's useful to keep in mind that it was used in the past, especially since it was prominent when $E=mc^2$ first entered the public consciousness.
A: You have got it. $E=mc^2$ is only true for particles at rest - i.e. particles with no momentum $(p=0)$! The general equation is indeed $E^2=p^2c^2+m^2c^4$
A: The answer depends on what the symbol $m$ represents.
If $m$ is the (non-zero) invariant mass of a particle, then $E=mc^2$ holds in an inertial reference frame (IRF) in which the particle is at rest.  If the particle has speed $v$ in an IRF, then the expression for the energy is
$$E = \gamma mc^2 = \frac{mc^2}{\sqrt{1 - \frac{v^2}{c^2}}}$$
and this equation is consistent with the energy-momentum equation written as
$$E^2 = (pc)^2 + (mc^2)^2$$
However, and confusingly, it is sometimes the case that $m$ represents the  (more or less outdated) relativistic mass $\gamma m_0$ where $m_0$ is the rest mass thus 
$$E = mc^2 = \gamma m_0c^2 = \frac{m_0c^2}{\sqrt{1 - \frac{v^2}{c^2}}}$$
and then the energy-momentum equation is written as
$$E^2 = (pc)^2 + (m_0c^2)^2$$ 
