Parity of discrete logarithms is independent of base By "parity" I mean the residue with respect to modulus 2, that is simply even or odd.
The problem is stated as follows:
"Assume p is an odd prime, suppose $r_1$ and $r_2$ are primitive roots mod p and $a\in \{x|gcd(x,p)=1 \land 0\lt x \lt p\}$. Show that $log_{r_1}a\equiv log_{r_2}a \pmod{2}$"
The author doesn't say so but I assume the logs are with respect to modulus p, not modulus 2.
Because $r_1$ is a primitive root, we have $log_{r_1}r_2=q$ for $q\gt 1$ and $log_{r_1}a \equiv q\cdot log_{r_2}a\pmod{p}$ because $a\equiv r_2^{log_{r_2}a}=(r_1^{log_{r_1}r_2})^{log_{r_2}a}$
$\equiv r_1^{log_{r_1}r_2\cdot log_{r_2}a}$
If q is odd, we are done since multiplication by an odd number does not change parity. I do not know how to demonstrate that q is odd. I think it has something to do with the fact that the modulus p is prime, which I have not used.
 A: I'm quite certain you're correct the discrete logs are wrt modulus $p$ rather than $2$. Also, you have the right idea for how to prove the result. To finish the proof, for simpler algebra, let
$$b = \log_{r_1}a, \; \; c = \log_{r_2}a \tag{1}\label{eq1A}$$
This gives
$$a \equiv r_1^b \equiv r_2^c \pmod{p} \tag{2}\label{eq2A}$$
First, assume $b$ is odd. Since $r_1$ is a primitive root, there exists an integer $d$ where $r_2 \equiv r_1^d \pmod{p}$. Thus, \eqref{eq2A} becomes
$$r_1^b \equiv (r_1^d)^c \pmod{p} \implies r_1^{dc - b} \equiv 1 \pmod{p}  \tag{3}\label{eq3A}$$
Being a primitive root, $r_1$'s multiplicative order is $\phi(p) = p - 1$, and this multiplicative order must divide $dc - b$, i.e.,
$$dc - b \equiv 0 \pmod{p - 1} \implies dc - b = k_1(p - 1), \; k_1 \in \mathbb{N} \tag{4}\label{eq4A}$$
It's stated $p$ is an odd prime, so $p - 1$ is even. Thus, with $b$ being odd, then $dc$ is odd so $c$ must also be odd. Thus, you have $b \equiv c \equiv 1 \pmod{2}$ in this case.
Next, consider $b$ to be even. Since $r_2$ is a primitive root, there exists an integer $e$ where $r_1 \equiv r_2^e \pmod{p}$. Thus, \eqref{eq2A} becomes
$$(r_2^e)^b \equiv r_2^c \pmod{p} \implies r_2^{eb - c} \equiv 1 \pmod{p} \tag{5}\label{eq5A}$$
Similar to before, we get
$$eb - c \equiv 0 \pmod{p} \implies eb - c = k_2(p - 1) \; k_2 \in \mathbb{N} \tag{6}\label{eq6A}$$
Since $p - 1$ and $b$ are even, then $eb$ is even, so $c$ must also be even. This gives $b \equiv c \equiv 0 \pmod{2}$ in this case.
This shows that whether $b$ is odd or even, you have what you're asked to prove, i.e.,
$$\log_{r_1}a \equiv \log_{r_2}a \pmod{2} \tag{7}\label{eq7A}$$
