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This is from Paolo Aluffi's book "Algebra: Chapter 0". First, find the order of $[9]_{31}$ in the group $( \mathbb{Z}/31\mathbb{Z})^*$. Then, does the equation $x^3 - 9 = 0$ have any solutions in $\mathbb{Z}/31\mathbb{Z}$?

The order of $9$ is $15$: Repeated squaring shows $9^{16} = 9$. The order has to divide the order of the group, which is $30$, so that looks fine.

But I get stuck trying to prove $x^3=9$ does not have a solution. I tried to say something like:

"$9^{16} = 9$ but $3 \nmid 16$ so there is no solution"

... but that does not seem correct. Can I get a hint (not solution) on how to solve this? I am learning Group Theory, and have not yet gotten to Rings, Fields, or Modules.

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    $\begingroup$ What's $9^{10}$? $\endgroup$ Jun 1 '18 at 17:17
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Knowing that the (multiplicative) order of $9$ modulo $31$ is $15$, suppose there is an element $x$ such that $x^3 = 9$. What will its order be? You should find a contradiction.

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  • $\begingroup$ This and Lord Shark The Unknown helped out a lot. Since $x^{30} = 9^{10} \neq 1$, but $a^{30} = 1 $ for all $a$ in the group, there cannot be a solution for $x$ in this group. $\endgroup$
    – Larry B.
    Jun 1 '18 at 17:38

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