An equation for momentum, $$p = \gamma mv, \tag{1}$$ then for photon we get some indeterminate momentum which is 0/0.
But the formula $E=(m^2c^4+p^2c^2)^{1/2}$ gives for photon $$p=E/c .\tag{2}$$ And from equation $E=hc/\lambda$, we get an energy which is finite and then using (2) we get the momentum.
If equation (1) gives photon's momentum in an indeterminate form which is not true, then should (1) be termed valid for all particles?
I am not an expert of this subject but my understanding is that the equations like
$p=\gamma m_0v$
or
$E=\gamma m_0c^2$
are not considered to give the correct idea but they give quick estimations for the massive particles.
I think that equation 2 is the correct form and equation 1 is the compact form that is derived from equation 2 for massive particles.
Or is it for photon (1) doesn't work?
The four-momentum for a massive particle is
$$p^\mu = mU^\mu$$
where
$$U^\mu=\frac{dx^\mu}{d\tau}$$
is the (time-like) four-velocity of the particle (the unit tangent vector of the particle's time-like world line). The components of the four-momentum are
$$m(\gamma c, \gamma \vec v) = (E/c, \gamma m \vec v)$$
Note carefully that the 'length' of the (time-like) four-momentum is (proportional to) the rest energy - the time component of the four-momentum in a coordinate system in which the particle is at rest
$$p_\mu p^\mu = m^2\left((\gamma c)^2 - (\gamma v)^2 \right) = (mc)^2 = (E_0/c)^2$$
from which we get the relativistic energy-momentum relationship
$$(E/c)^2 - \gamma^2 m^2 v^2 = (E_0/c)^2 \rightarrow E^2 = (pc)^2 + (E_0)^2 = (pc)^2 + (mc^2)^2 $$
However, the world line of a photon is light-like (null) and thus there is no four-velocity for a photon; $U^\mu$ does not exist for a null world-line.
Nonetheless, the (light-like) four-momentum for a photon exists and is given by
$$p^\mu = (E/c,\vec p)$$
and since it is light-like, its 'length' is zero (null)
$$p_\mu p^\mu = (E/c)^2 - p^2 = 0 \rightarrow E^2 = (pc)^2$$
