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$x$ is a cubic residue mod p if it is of the form $a^3\pmod{p}$ for some residue $a$. Show if $p\equiv 1\pmod{3}$, then $x\pmod{p}$ is a cubic residue iff $x^{(p-1)/3} \equiv 1\pmod{p}$. Also, show if $p\equiv 2\pmod{3}$, then all $x\pmod{p}$ are cubic residues.

For the forward implication, I can do this easily by assuming $x\equiv a^3\pmod{p}$ for some $1\leq a\leq p-1$. Then $x^{(p-1)/3}\equiv (a^3)^{(p-1)/3} \equiv a^{p-1} \equiv 1\pmod{p}$ by Fermat's Theorem since p is prime so $\textrm{gcd}\,(a,p)=1$.

For the reverse implication, this is what I have so far. If $p\equiv 1\pmod{3}$, then $p=3k+1$ for some integer $k$, so if $x^{(p-1)/3} \equiv1\pmod{p}$, then $x^{(3k+1-1)/3}\equiv1\pmod{p}$, so $x^k \equiv1\pmod{p}$. But this is where I'm stuck. I don't know if what I've done is useful or not.

Finally, I am stuck on the last part as well where $p\equiv2\pmod{3}$. I think I may want to use the fact that for some r, I can say that the set $\{r,r^2,r^3,...,r^{p-1}\equiv1\}$ is the same set $\pmod{p}$ as $\{1,2,3,...,p-1\}$, because then this set represents all possible values of $x$. But I'm not sure how to use this.

Ice Tea
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mmm
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  • Do you know the fact that multiplicative group of finite field is cyclic? – userabc Apr 08 '17 at 22:04
  • I'm trying to do this more from a number theoretic approach rather than using ideas from group theory if possible. – mmm Apr 08 '17 at 22:12

2 Answers2

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For $p \equiv 1 ($mod $3)$. Use the fact that $-3$ is a quadratic residue for $p$ to show that there are exactly three solutions for $x^3-1 \equiv 0 ($mod $p)$. The key step is to make use of the identity $4(x^2+x+1)=(2x+1)^2+3$. Then we can conclude there are exactly $(p-1)/3$ non-zero cubic residues. All of them have to satisfy the equation $x^{(p-1)/3}-1 \equiv 0 ($mod $p)$. Since there are at most $(p-1)/3$ solutions for $x^{(p-1)/3}-1 \equiv 0 ($mod $p)$, the set of non-zero cubic residues coincides with the set of the solutions.

For $p \equiv 2 ($mod $3)$, Bezout's lemma gives us $a,b \in \mathbb{Z}$ such that $1=3a+(p-1)b$. For any $r$, $r \equiv r^{(3a+(p-1)b)} \equiv r^{3a} \equiv (r^a)^3 $mod $p$.

userabc
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    I am confused about the connection between the number of nonzero cubic residues and the equation $x^3 - 1 \equiv 0$ mod p. Why is the number of solutions relevant? – mmm Apr 09 '17 at 00:33
  • Suppose $\omega \neq 1$ is a cube root of unity mod $p$, then $a^3 \equiv b^3$ mod $p$ if and only if $a=b$ or $a=\omega b$ or $a= \omega^2 b$. So the set of cubic residue is ${1^3, 2^3, ..., (p-1)^3}$, which has $(p-1)/3$ elements. – userabc Apr 09 '17 at 00:51
  • Okay that makes sense, but I'm having trouble getting to the conclusion that it has (p - 1)/3 elements. I know that $x^3 - 1 \equiv (x - 1)(x^2 + x + 1)$. Which I'm guessing is where I need to use that identity, but how do I get from (x - 1) to 4 being multiplied out front? – mmm Apr 09 '17 at 00:59
  • The identity shows the existence of the cube root of unity. Since $y^2 \equiv -3 ($mod $ p)$ has solutions, then so does $(2x+1)^2 \equiv -3 ($mod $ p)$ and $x^2+x+1 \equiv -3 ($mod $ p)$. – userabc Apr 09 '17 at 01:04
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Hint: On a finite abelian group of order $n$, the map $x \mapsto x^3$ is a bijection iff $\gcd(3,n)=1$.

This is proved easily by considering the kernel.

Since $3$ is prime, $\gcd(3,n)=1$ iff $n \not \equiv 0 \bmod 3$. In our case, $n=p-1$.

lhf
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