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trying to figure if I understand the basics of $\Bbb Z_7$ field ${}=\{1,2,3,4,5,6\}$ on the following equation:

$6x^3$ = $5$

I tried to reduce the coefficients in the equation -

$6x^3$ = $5$ => $6*6x^3 = 6*5$ => $36x^3 = 30$ => mod7 => $x^3 = 2$

From this stage $x^3 = 2$. I don't know how to continue.

Thanks in advance.

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    $\begingroup$ For these kinds of questions, its a good idea to write down all the squares - or in this case, cubes - modulo the prime you are working with - in this case 7. $\endgroup$ – bounceback Oct 11 '18 at 17:17
  • $\begingroup$ Thanks for the good advice! I will definitely use it in the future. $\endgroup$ – OO1 Oct 11 '18 at 19:10
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Hint:

Clearly $0$ is not a solution. On the other hand, for any $x\ne0$, lil' Fermat says that $x^6=1$, so $(x^3)^2=1$. What can you deduce for $x^3$, knowing we're in a field?

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  • $\begingroup$ In that case, if I try to set any number between $0-6$ in $x$ , none of them will be equals to $2$. eventually, there is a conflict in that equation. $\endgroup$ – OO1 Oct 11 '18 at 19:13
  • $\begingroup$ Yes. You necessarily obtain $1$ or $-1$, since these are two obvious solutions of $y^2=1$, and in a field, a quadratic equation cannot have more than two solutions. $\endgroup$ – Bernard Oct 11 '18 at 19:24

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