E^2 = (mc^2)^2 + (pc)^2
Four dimensional space-time has a metric analogous to but not quite the same as euclidean space: the time part has the opposite sign from the space parts.
ds^2 = (c dt)^2 - (dx^2 + dy^2 + dz^2)
Just as you can start with distance and then build up to momentum and energy in classical physics, in relativistic mechanics you can start with this metric and build up vectors in 4-space for velocity and energy/momentum. (Turns out that the time part is an energy while the space parts are momentum.) The upshot is that
(m_0 c^2)^2 = E^2 - (p c)^2
What's peculiar is that the temporal entry of the four vector gets the "opposite sign" for the Pythagorean theorem. That is, if you have a vector (t, x, y, z), then the "hypotenuse" (called an invariant) is t^2 - x^2 - y^2 - z^2 (up to a conventional choice of overall sign). What's surprising is that the hypotenuse is the thing that's the same for different observers. So, if you do a Lorentz transformation [3], t and the spatial entries will change, but the invariant combination won't.
Everybody knows E = mc^2 is the relationship between a particle's mass and its energy. But what is its energy if it's moving? Certainly the particle gains energy the faster it moves, right? Yes. E = mc^2 is the 0-momentum version of a relativistic expression. Since E is temporal and momentum p is spatial, the four vector can be written (E, pc), which has an invariant E^2 - (pc)^2. We call this invariant the rest mass[4], up to some factors of c. Algebraically moving things around gives us the "Pythagorean" form.
[0] https://en.wikipedia.org/wiki/Energy%E2%80%93momentum_relati...
[1] https://en.wikipedia.org/wiki/Four-vector
[2] https://en.wikipedia.org/wiki/Noether%27s_theorem
(mc^2)^2 = E^2 - (pc^2)^2
I know of no one doing modern research in relativity using this obsolete concept.
Now consider the momentum, p, of a photon. We have:
this gives us p=0*infinity, which is indeterminate, so we cannot use this equation to determine the momentum of a photon.Instead, we can use the engery-momentum relationship, which states:
(This is a generalization of the famous E=mc^2 equation to also consider the momentum). In this equation m refers to the rest mass, not inertial mass. Because we are dealing with a photon, we have m=0, which gives us: Indicating that the momentum of a massless object is proportional to its energy.