How to solve $\ x^{12}\equiv 87\pmod{101}$ Can someone help me to solve the following congruent equation. I have tried to use Fermat's little theorem but failed to solve this:
$x^{12}=87(\mod 101)$

Numbers congruent to $87 \pmod {101}$ are $87,\; 87+101=188,\; 87+2\times 101=\color{red}{289}, ...$
and $87-101=-14,\;87-2\times 101=-115,\;87-3\times 101=\color{blue}{-216},...$
Note that therefore $17^2=289 \equiv87\pmod{101}$
and $(-6)^3=-216 \equiv 87 \pmod{101}.$
Furthermore $17\times6=102 \equiv 1 \pmod{101}.$
Therefore $17^5\equiv17^2/6^3\equiv -1 \pmod{101}.$
Thus $17^{12}=17^{10}\times17^2=(17^5)^2\times17^2\equiv87 \pmod {101},$
so $x=17$ is a solution to $x^{12}\equiv87\pmod{101}$.
