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Let $p$ be a prime number, and let $\mathbb{F}_p$ be the field with $p$ elements. How many elements of $\mathbb{F}_p$ have cube roots in $\mathbb{F}_p$?

I had this question on an exam and after reviewing I am still not sure. Any help would be appreciated.

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Hint: $x \mapsto x^3$ is a homomorphism from the multiplicative group of $\mathbb{F}_p$ to itself. –  Chris Eagle Jul 21 '12 at 23:53
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1 Answer

up vote 9 down vote accepted

1) If $\,p=3\,$ then $\,a^3=a\,\,\,,\,\forall a\in\Bbb F_p\,$, by Fermat's Little Theorem

2) If $\,3\nmid (p-1)\,$ then $\,f:\Bbb F_p^*\to \Bbb F_p^*\,\,\,,\,f(x):=x^3\,$ is an automorphism (can you see why? Check $\,\ker f\,$...)

3) Finally, if $\,3\mid (p-1)\,$ then the map $\,f\,$ above cannot be an automorphism, and since $\,\Bbb F_p^*\,$ is a cyclic group it then has one single subgroup of any order divinding $\,p-1\,$, so... (optional: add $\,0\,$)

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