$p_1^a - p_2^b g \equiv 0 \pmod{p_3}$ for some primitive root $g$, prime $p_3$, and fixed prime powers $p_1^a,p_2^b$

I was wondering whether one can prove for some fixed integer powers of fixed primes $p_1^a$ and $p_2^b$ that there exists some prime $p_3$ which has a primitive root g, such that:

$p_1^a - p_2^b g \equiv 0 \pmod{p_3}$

That is:

$\frac{p_1^a}{p_2^b} \equiv g \pmod{p_3}$

I was thinking that for $p_3 = 2p_4 + 1$ there should be a high likelihood of finding such a generator since $\varphi(p_3) = 2p_4$ in that case, thus limiting the possible sub-groups -- but I thought this seems like a problem for which there probably exists established theorems hence the question.

Thanks a lot!

• If $a$ and $b$ are both even, no such $p_3$ exists. – Will Fisher Jun 25 '17 at 1:48
• More generally, $p_3$ must satisfy $((a,b),p_3-1)=1$. – Will Fisher Jun 25 '17 at 1:58
• Thanks, I suppose this requires also that if $a = 1, b = 1$ and $p_1 \not\equiv 1 \pmod{p_3}$ that $p_2$ isn't the modular inverse of $g$. – Niklas Jun 25 '17 at 8:49