Adam, to understand why [math]c^2[/math] even appears in the equations, we will take a look at deriving time dilation. The following image illustrates a light clock at rest:
This clock measures time according to:
[math]\Delta t = \frac{2 L}{c}[/math]
where [math]L[/math] is the distance separating the reflective plates and [math]c[/math] is the speed of light. This gives us:
[math]\frac{\text{distance}}{\text{speed}}=\frac{\text{distance}}{\text{distance}/\text{time}} = \text{time}[/math]
We have twice the distance, [math]2L[/math], because the light must travel to the top plate and be reflected back to the bottom plate in order to measure time.
An observer in motion relative to the light clock will not see the path of light as a straight up and down path. Instead, they will see the following:
As we can see from the above image, an observer moving at a speed of [math]v[/math] relative to the light clock will see the path of the light elongated to a distance [math]R[/math], which is equal to the hypotenuse of the triangle:
Because the speed of light is the same regardless of one's frame of reference, an observer moving at a speed of [math]v[/math] relative to the light clock will see that the clock is measuring time according to:
[math]\Delta \tau = \frac{2 R}{c}[/math] where [math]R=\sqrt{\left(\frac{v \Delta \tau}{2}\right)^2+L^2}[/math] as determined by the distance formula (Pythagorean theorem).
Multiplying both sides of the equation by [math]c[/math] and squaring the result, we get:
[math]c^2 \Delta \tau^2 = 2^2 R^2 = 4 \left( \frac{v^2 \Delta \tau^2}{4} +L^2\right)=v^2 \Delta \tau^2+4L^2[/math]
Subtracting [math]v^2 \Delta \tau^2[/math] from both sides yield:
[math]c^2 \Delta \tau^2 - v^2 \Delta \tau^2 = 4L^2[/math]
Factoring [math]\Delta \tau^2[/math] on the left side yields:
[math]\Delta \tau^2 \left(c^2 - v^2 \right) = 4L^2[/math]
Now we can divide both sides by [math]\left(c^2 - v^2 \right)[/math]:
[math]\Delta \tau^2 = \frac{4L^2}{ \left(c^2 - v^2 \right)}[/math]
Factoring [math]c^2[/math] from the denominator on the right side and taking the square root of the result yields:
[math]\Delta \tau = \frac{2L}{ c \sqrt(1 - \frac{v^2}{c^2})}[/math]
Because [math]\Delta t = 2L/c[/math] we can substitute [math]\Delta t[/math] into the final result to yield the time dilation formula:
[math]\Delta \tau = \frac{\Delta t}{ \sqrt(1 - \frac{v^2}{c^2})}[/math]
As you can see from the above formula, [math]\Delta \tau = \gamma \Delta t[/math] where [math]\gamma = \frac{1}{ \sqrt(1 - \frac{v^2}{c^2})}[/math] is the Lorentz factor and it contains [math]v^2[/math] and [math]c^2[/math] (Gamma, [math]\gamma[/math], forms the basis for transformation in special relativity). Applying this to other equations, such as those for energy, yields: