special relativity - Does $p=mc$ hold for photons?

Known that $E=hf$, $p=hf/c=h/\lambda$, then if $p=mc$, where $m$ is the (relativistic) mass, then $E=mc^2$ follows directly as an algebraic fact. Is this the case?

Answer

As you may know, photons do not have mass.

Relating relativistic momentum and relativistic energy, we get:

$E^2 = p^2c^2+(mc^2)^2$.

where $E$ is energy, $p$ is momentum, $m$ is mass and $c$ is the speed of light.

As mass is zero, $E=pc$.

Now, we know that $E=hf$. Then we get the momentum for photon.

Note that there is a term called effective inertial mass. Photon does have it.