<s>[INST] 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).
 [/INST] 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.
</s>