How to Solve $x^a=b$ in $\mathbb{Z}/n$
I'm preparing for an exam by looking over old exams and have to exam
questions of the same type but solved slightly differently.
a) $x^{13} = 3 \bmod{47}$, given that $3^{37}=14$. (Answer: $x=32$)
b) $x^{29} = 3 \bmod{184}$. (Answer: $x=75$)
The solutions are:
a) $\varphi(47)=46$. $1=2 \cdot 46-7 \cdot 13$, whence $37^{-1}=46-7=39$.
We then have $x^{13} = 3 \Rightarrow x^{13 \cdot 39}=3^{39}
\Leftrightarrow x=3^{39}$, etc.
b) $\varphi(184)=88$. $1=88-3 \cdot 29$. We then have $x^{29}=3
\Rightarrow x^{3 \cdot 29}=3^3 \Leftrightarrow x^{88-1} = 27
\Leftrightarrow x^{-1}=27 \Rightarrow x=27^{-1}$, etc.
Now, solving (b) by following the solution to (a) works fine (if you're
comfortable with calculating $3^{85}$), but when I try solving (a) using
(b) I run into trouble:
Again, $1=2 \cdot 46 - 7 \cdot 13$. Following (b), we should raise LHS and
RHS by $7$, which gives: $x^{7 \cdot 13}=3^7 \Leftrightarrow
x^{91}=3^7=25$.
Is it possible to proceed from here? If so, how? I thought maybe I could
start over and solve $x^{91}=25$, but I'm not sure how to do that; perhaps
it is correct that $x^{91}=x^{44}$, but then $\varphi(47)$ and $44$ are
not coprime whence we cannot write $1=44m+46n$, or we calculate $91^{-1}$
in $\mathbb{Z}/47$ and get $x=17$, which is false.
My real question here is, is there a general method to solve $x^a=b$ in
$\mathbb{Z/n}$ which would tackle both the above questions? (And also if
there is an efficient way to calculate $a^b$ modulo $n$, $b<n$...)