Getting from $E^2 - p^2c^2 = m^2c^4$ to $E = \gamma mc^2$ What is each mathematical step (in detail) that one would take to get from:
$E^2 - p^2c^2 = m^2c^4$ 
to
$E = \gamma mc^2$,
where $\gamma$ is the relativistic dilation factor.
This is for an object in motion.
NOTE: in the answer, I would like full explanation. E.g. when explaining how to derive $x$ from $\frac{x+2}{2}=4$, rather than giving an answer of "$\frac{x+2}{2}=4$, $x+2 = 8$, $x = 6$" give one where you describe each step, like "times 2 both sides, -2 both sides" but of course still with the numbers on display. (You'd be surprised at how people would assume not to describe in this detail).
 A: First you set $c=1$.
$$ E^2 - p^2 = m^2$$
Then you think about it, it is saying that the relativistic length of the energy momentum vector is "m". The ratio of p to E is the velocity, since that's what happens to a four-vector under a boost, it gets space-components and time component whose ratio is the velocity. From $|p|=v|E|$, you substitute,
$$ E^2(1-v^2) = m^2 $$
And 
$$ E= { m\over\sqrt{1-v^2}}$$
Done.
It is generally a sign of total incompetence to not set c to 1, it just makes ridiculously trivial geometrical formulas, which, as you can see above, are absolutely transparent, look like they are sophisticated or complicated.
A: Starting with relativistic momentum
$$p^2 = \left( \gamma m v \right)^2 = \frac{m^2 v^2}{1 - \frac{v^2}{c^2}}$$
one than gets
$$E =  \pm \sqrt{ m^2 c^4 + p^2 c^2 } = \pm \sqrt{ m^2 c^4 + \frac{m^2 v^2 c^2}{1 - \frac{v^2}{c^2}} } = \pm mc^2 \sqrt{\frac{1- \frac{v^2}{c^2}}{1- \frac{v^2}{c^2}} + \frac{\frac{v^2}{c^2}}{1- \frac{v^2}{c^2}}}  = \pm \gamma mc^2$$
A: Starting with your given equation, we add $p^2 c^2$ to both sides to get
$$ E^2=m^2 c^4 + p^2 c^2$$
now using the definition of relativistic momentum $p=\gamma m v$ we substitute that in above to get
$$E^2 = m^2 c^4 +(\gamma m v)^2 c^2=m^2 c^4 +\gamma^2 m^2 v^2 c^2$$
Now, factoring out a common $m^2 c^4$ from both terms on the RHS in anticipation of the answer we get
$$E^2=m^2 c^4 (1+\frac{v^2}{c^2}\gamma^2)$$
Now using the definition of $\gamma$ as 
$$\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$
and substituting this in for $\gamma$ we get
$$E^2=m^2 c^4 \left(1+\frac{\frac{v^2}{c^2}}{1-\frac{v^2}{c^2}}\right)$$
and making a common denominator for the item in parenthesis we get
$$E^2=m^2 c^4 \left( \frac{1}{1-\frac{v^2}{c^2}} \right)=m^2 c^4 \gamma^2$$
Taking the square root of both sides gives
$$E=\pm \gamma mc^2$$
Hope this helps.
