Why can't we use the law of cosines to prove Fermat's Last Theorem? In investigating approaches to Fermat's Last Theorem I came across the following and I can't figure out where I am going wrong.  Any input would be greatly appreciated.
We want to show that $a^n + b^n = c^n$ cannot hold for odd $n>1$ and pairwise relatively prime $a$, $b$, and $c$.  Assuming by way of contradiction that we have $a^n + b^n = c^n$ we must have $a$, $b$, and $c$ forming the sides of a triangle since $(a+b)^n > c^n$ so $a+b>c$.  Therefore the law of cosines can apply and we can write:
$$c^2 = a^2+b^2 - 2ab{\cos{C}}$$
where $C$ is the angle opposite to side $c$.  If we add and subtract $2ab$ on the right-hand side we get
$$c^2 = {(a+b)}^2 -2ab(\cos{C}+1)$$
Now, $a+b$ and $c$ share a common factor since $(a+b) | (a^n+b^n)$ for odd $n$ and $c^n = a^n+b^n$.  (Here $x | y$ means as usual, "$x$ divides $y$").Therefore, they share the same factor with $2ab(\cos{C}+1)$.
  Now, $\cos{C} + 1$ must be a rational number since $a$, $b$, and $c$ are all integers.  So let $\cos{C} +1 = \frac{r}{s}$ where $r$ and $s$ are integers and $(r,s)=1$. (i.e. $\frac{r}{s}$ is a reduced fraction).  (Here, $(r,s)$ means as usual the greatest common divisor of $r$ and $s$.)
Now assuming $a$, $b$, and $c$ are relatively prime we must have $(ab) |s$ for otherwise $c$ and $2ab$ would share a common factor.  Even moreso we must have $ab=s$ since otherwise $\frac{2abr}{s}$ would not be an integer.  (Since $c - a - b$ is even, we don't need $2 | s$).  So we can write:
$$\cos{C}+1 = \frac{r}{ab}$$ or equivalently $$\cos{C} = \frac{r - ab}{ab}$$
Now we had from the law of cosines:
$$c^2 = a^2+b^2 - 2ab{\cos{C}}$$
so making the substitution $\cos{C} = \frac{r - ab}{ab}$ we get
$$c^2 = a^2 + b^2 - 2r + 2ab$$
If we subtract $a^2$ to both sides and factor out the $b$ on the right-hand side, we get:
$$c^2 - a^2 = b(b + 2a) - 2r$$
Now, $(c - a) | (c^2 - a^2)$ and also $(c-a) | (c^n - a^n)$. Then we must have $((c-a),b) >1$ since $b^n = c^n - a^n$.  From the equation above, we must therefore also have $(b,2r) > 1$.  Similarly we can show that we must have $(a,2r) > 1$.
However, both of these conclusions are problematic since $r$ was initially assumed to be relatively prime to $s = ab$.  The only other option is that $a$ and $b$ are both even, but this is also problematic since $a$ and $b$ are assumed to be relatively prime.
Thus we cannot have $a^n + b^n = c^n$  for odd $n>1$ and pairwise relatively prime $a$, $b$, and $c$.
I'm sure someone has thought of this approach before so where am I going wrong?

As Maik points out in the accepted answer we don't necessary need $ab=s$ in order to ensure that $a$, $b$, and $c$ are pairwise relatively prime.  We do however require that $(c,\dfrac{2abr}{s})>1$ since $(c, (a+b) )>1$.  
Now we need $s|ab$ because otherwise we would not get an integer value for $\dfrac{2abr}{s}$. Also we cannot have $(c,2ab)>1$ because this would imply that $a$, $b$, and $c$ share a common factor.  (As noted in some of the comments to my original question we could have $c$ being even if $a$ and $b$ are both odd, but then $4|c$, $4|{(a+b)}^2$ so we would need $4|2ab$ implying that either $a$ or $b$ is even, a contradiction.)
What I forgot was that we can still have $(c,r)>1$ and thus avoid any contradictions with $a$, $b$ and $c$ being pairwise relatively prime.
