I'm trying to solve $$3^b \equiv 22 \mod 31$$ I know that this is of course equivalent to $$b \equiv L_3(2) + L_3(11) \mod 30$$ but I don't know how to solve those either. I can obviously just compute each exponent of $3$ by hand and hope to get lucky, but I'm looking for a shorter method.
2026-02-23 10:19:33.1771841973
I know that 3 is a primitive root of $31$. How can I solve $3^b \equiv 22$?
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Hint: $22\equiv -9\pmod{31}$, so it is enough to find $L_3(9)$ and $L_3(-1)$. For $L_3(-1)$, you can use the fact that $-1$ is the unique element of order $2$ in the multiplicative group.