I'm in an introductory abstract algebra course and was given the question to find : The zeros of $x^{86}-6$ over the field $Z_{29}$.
So first, I tried plugged in all values (to see if it was even possible) of $Z_{29}$ into the above equation using wolfram alpha and was never able to get 0 in $Z_{29}$ (i.e. 0 or multiples of 29).
I tried using Fermat's Theorem to say that $x^{28}$ is congruent to $1$ mod $29$, as 29 is a prime number and all numbers x in $Z_{29}$ are relatively prime to 29.
However, this gives me that $(x^{28})^3*x^2$-6 is congruent to $1*x^2-6$ mod 29. Thus having $x^2-6$ mod $29$,I get 8 as a zero as $8^2-6$=58 which is a multiple of 29. However, when I check wolfram alpha, 8 is not zero of $x^{86}-6 $ over the field $Z_{29}$, so I'm really unsure what I'm doing wrong. Any help would be much appreciated.
You are on the right track. $x^{28} \equiv 1 \pmod{29}$ by Fermat. So the given equation is \begin{align*} x^{2} & \equiv 6 \pmod{29}\\ x^2 & \equiv 64 \pmod{29}\\ (x-8)(x+8)& \equiv 0 \pmod{29}. \end{align*} Since $29$ is a prime so use the fact that if $p | ab$, then $p|a$ or $p|b$, to claim there are only two solutions, namely $x \equiv \pm 8 \equiv 8, 21$.
Check wolfram alpha again, $x=8$ is definitely a solution.