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.

  • 1
    $\begingroup$ 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


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?

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.