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.

  • $\begingroup$ $8$ is a zero. (Indeed, $\pm 8$ are the only zeros.) You must have entered it into Wolfram Alpha incorrectly: Mod[8^86,29] yields 6. $\endgroup$ – Jim Ferry Apr 29 '18 at 4:34
  • $\begingroup$ Oh so what I did was I thought $(8^{86}-6)$/29 had to be an integer value, which wolfram alpha does not give. $\endgroup$ – kemb Apr 29 '18 at 4:36
  • $\begingroup$ I checked Wolfram and I got $0$ for $x^{86}-6\pmod{29}$ $\endgroup$ – marwalix Apr 29 '18 at 4:37
  • $\begingroup$ Thanks, I guess I was typing into wolfram alpha incorrectly. $\endgroup$ – kemb Apr 29 '18 at 4:39
  • $\begingroup$ I don't like denoting $Z_n$ for a cyclic group of order $n$, but for a field it looks far worse... and $F_{29}$ is the standard notation. $\endgroup$ – YCor May 1 '18 at 18:13

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.

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  • $\begingroup$ Hello, yes this is exactly what I got. However when I plug into wolfram alpha $(8^{86}- 6)$/ 29, to check my work, I don't get an integer value. $\endgroup$ – kemb Apr 29 '18 at 4:34
  • $\begingroup$ @kemb you should plug into wolfram « 8^86-6 mod 29 ». There must be rounding issues... $\endgroup$ – marwalix Apr 29 '18 at 4:41
  • $\begingroup$ I see thanks so much. $\endgroup$ – kemb Apr 29 '18 at 4:43

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