Given a prime number p > 3 and primitive root of p, r, I know (from some previous argument) that $$\frac{1-r^{p+1}}{1-r^2}$$
is a whole number. For context, I know this by the geometric series finite summation formula for the even powers of r up to $r^{p-1}$, including $r^0=1$).
I therefore want to reason that this term modulo p is equal to 1 because:
$$\frac{1-r^{p+1}}{1-r^2}\equiv\frac{1-r^{p-1}r^2}{1-r^2}\equiv\frac{1-r^2}{1-r^2}\equiv1\pmod p$$
where I argue I can perform this division modulo p since I know r is a primitive root and p>3 and therefore $1-r^2\neq0\pmod p$ and using the fact that $r^{p-1}\equiv1\pmod p$.
My question is whether performing such reductions modulo p is kosher even as part of a fraction such as this? in particular reducing the $r^{p-1}$ term to 1 while its part of such a fraction. Are there further assumptions I need to make (or prove) e.g. that the denominator $1-r^2$ is relatively prime to p or anything of that sort?
I suppose its automatically relatively prime given we know that it is not some multiple of p and p is prime. Is this indeed an assumption we need to use to perform this reduction or any other one?
