the energy-momentum relation
How do a particle's energy, momentum, and mass fit together when it moves near light speed? The old E = p^2/2m is only a low-speed approximation. The energy-momentum relation is the exact, relativistic bookkeeping — a single equation linking energy, momentum, and rest mass that holds from rest to ultra-relativistic speeds, and even for massless light.
The relation is E^2 = (pc)^2 + (m c^2)^2, equivalently E^2 - (pc)^2 = (m c^2)^2. It is nothing but the invariant magnitude of the four-momentum: p·p = (E/c)^2 - |p|^2 = m^2 c^2, so it is Lorentz invariant — the same in every frame. Two limits check it: at rest p = 0 gives E = m c^2 (rest energy); expanding for small p gives E ~ m c^2 + p^2/2m, recovering the Newtonian kinetic energy plus the rest-energy offset. For a massless particle m = 0 it reduces to E = pc, the photon's relation.
This is the 'mass-shell' (or 'on-shell') condition that every real, freely propagating particle must satisfy; in quantum field theory the internal (virtual) lines of a Feynman diagram are explicitly OFF-shell, i.e. they violate E^2 = (pc)^2 + (mc^2)^2, which is allowed only because they are not observable propagating particles. Caveat: the m in this relation is the invariant rest mass, not a velocity-dependent 'relativistic mass'; and the relation is exact only for a free particle — in a potential you add the interaction energy separately.
A proton (m c^2 = 938 MeV) with momentum pc = 938 MeV has E = sqrt(938^2 + 938^2) = 1327 MeV, so its kinetic energy is 389 MeV — nowhere near the naive p^2/2m, because at this momentum it is genuinely relativistic.
Energy and momentum combine in quadrature with the rest energy, not linearly.
E = m c^2 alone is only the rest-frame special case (p = 0). The full, frame-independent statement is E^2 = (pc)^2 + (mc^2)^2; quoting E = mc^2 for a moving particle is a common but real error.