Units in time dilation calculation I am currently working through Brian Greene's "World Science U" course on special relativity, and I have a question regarding one of the calculations performed for an exercise on time dilation (MODULE 12: Time Dilation: Examples - Problem 1).  The question is as follows:

Imagine that today is your birthday and you decide that you want to celebrate your next birthday on planet Zaxtar, 100 million light-years away.  How fast would you need to travel if you leave immediately?

I have no difficulties with the setup of the equation:  If $t_E$ is the time it take the ship to travel to planet Zaxtar from the perspective of someone on Earth, $t_S$ is the time it takes the ship to travel to planet Zaxtar from the perspective of someone on the ship, and if the ship moves at a constant velocity $v$, then we should simply solve the equation:
$$
t_S =t_E /\gamma \implies 1 = \frac{10^8}{v}\sqrt{1 - \frac{v^2}{c^2}}.
$$
Here is my difficulty: If one solves this equation for $v$, then we see that:
$$\begin{align}
v^2 = 10^{16} (1 - v^2/c^2) &\implies v^2 = 10^{16} - 10^{16} v^2/c^2
\\
&\implies v^2 \left( 1 + \frac{10^{16}}{c^2}\right) = 10^{16}
\\
&\implies v^2 = \frac{c^2 \cdot 10^{16}}{c^2 + 10^{16}}
\\
&\implies v = c \frac{10^8}{\sqrt{10^{16} + c^2}}.
\end{align}$$
However, at the beginning of his solution (video available here) equation, Brian sets $c = 1$ by using appropriate units: $c = 1$ light-yr/yr.  He proceeds to solve for $v$, getting:
$$
v = \frac{10^8}{\sqrt{10^{16} + 1}},
$$
and then he argues using dimensional analysis, that we should have
$$
v  = c \frac{10^8}{\sqrt{10^{16} + 1}},
$$
which follows since the units for $10^8$ and $\sqrt{10^{16} + 1}$ are light-yr, so that they cancel.  
My question is this: what goes wrong with the first method used to solve the problem?  It is clear that if $c = 1$, then trivially $c^2 = 1$, but where is the mistake regarding the units?  It seems that the units are not quite right due to the fact that $10^{16}$ and $c^2$ have different units.
 A: They don't have different units. If you follow things carefully you see that the $10^{16}$ has units of $(lyr/yr)^2$, so provided you use the same units for $c^2$ then everything works. I would recommend explicitly sticking the units in at the beginning. Your first, equation, for example, should read
\begin{equation}
1 yr = (10^{8} lyr) \frac{1}{\gamma v}
\end{equation}
A: Good question.  You should certainly be able to solve this in SI units, where $c$ and $v$ have units and $c\ne 1$  The way to find the problem is to put units on things and make sure they are consistent.  The $10^8$ at the start is then in light-years, the 1 is in years, so the $v$ outside the square root should be in light-years/year, not in m/sec.  You can then convert the $10^8$ and $1$ to SI units, or measure that $v$ in the appropriate units.  You can solve that by making the equation $$1 = \frac{10^8c}{v}\sqrt{1 - \frac{v^2}{c^2}}.$$  Now all the numbers are dimensionless and if you follow it through it works out.  
Setting $c=1$ without units makes things easier, but it also removes one way to find errors in calculation.  You have one less unit to find a mismatch.
