Solving $x^d \equiv a \pmod{29}$ using primitive roots Can anyone please detail the general approach to questions of the form $x^d \equiv a \pmod{29}$? For example, Wolfram Alpha states that $x^5 \equiv 8 \pmod{29}$ has one solution, $x^4 \equiv 4 \pmod{29}$ has no solution, and $x^7 \equiv 12 \pmod{29}$ has many solutions, but I have no idea how I'd go about proving that no solutions exist, or that any solutions I find are comprehensive.
I know that $2$ is a primitive root $\bmod 29$, so $2^{28} \equiv 1 \pmod{29}$. For the first one I can manipulate to get $(2^{23})^5 \equiv 8 \pmod{29}$ to give $x \equiv 2^{23} \equiv 10 \pmod{29}$ as a solution, but this doesn't prove that no other solutions exist, does it? Also, I don't know how this approach would work with the third problem because $12$ cannot be written $2^n$.
Furthermore, I'm unsure of the general technique to show that the second is not soluble?

$(x^5)^{17}\equiv x^{85}\equiv (x^{28})^3x\equiv x\bmod 29$,
so if $x^5\equiv8\bmod29$, then $x\equiv8^{17}=2^{51}=2^{28}2^{23}\equiv2^{23}\equiv10\bmod29$.
If $x^4\equiv4=2^2\bmod29$ had solutions, then $1\equiv x^{28}\equiv (x^4)^7\equiv(2^2)^7=2^{14}\bmod29$, so $2$ would not be a primitive root.
$2^7\equiv12\bmod29$, so if $x^7\equiv12\bmod29$, then $x^7\equiv2^7\bmod29$,
so $x\equiv2a\bmod29$, where $a^7\equiv1\bmod29$.
$a^7\equiv1\bmod29$ when $a\equiv2^0, 2^4, 2^8, 2^{12}, 2^{16}, 2^{20}, $ or $2^{24}$; i.e., $a\equiv1, 16, 24, 7, 25, 23, $ or $20$;
so $x\equiv2, 3, 19, 14, 21, 17, $ or $11\bmod29$.
