Prove that $π(x)=x^3$ is an even permutation Let $p$ be a prime number  $p\equiv  \text{2 (mod 3)}$ and let be $\pi (x) = x^3$ considered as a permutation of $\mathbb{Z}/p \mathbb{Z}$
Prove that $\pi$ is an even permutation iff $p\equiv \text{3 (mod 4)}$
I have no idea how to solve it. 
 A: $\mathbb{Z}/(p\mathbb{Z})^*=\langle g\rangle $ is a cyclic group with $p-1$ elements, $g^1,g^2,\ldots,g^{p-1}=1$. Since $p\equiv 2\pmod{3}$, the map $x\mapsto x^3$ is surjective over $\mathbb{Z}/(p\mathbb{Z})^*$ and the orbit of $g^a$ is given by $g^a,g^{3a},g^{9a},\ldots.$ $+1$ and $-1$ are mapped into themselves, and they are the only elements with such property, by factoring $x^3-x$ over $\mathbb{F}_p$. There are some transpositions if $\frac{x^9-x}{x^3-x}=(x^2+1)(x^4+1)$ has a root in $\mathbb{F}_p$, but $\Phi_4(x)$ has a root in $\mathbb{F}$ iff $p\equiv 1\pmod{4}$ and $\Phi_8(x)$ has a root in $\mathbb{F}_p$ iff $p\equiv 1\pmod{8}$. I leave to you to study when $4$-cycles, $6$-cycles and so on may occur, and how many of them we have. The parity of the wanted permutation is clearly related to the number of cycles with an even length.
There might be some slick way of exploiting Zolotarev's lemma by considering the permutation induced by the multiplication by $3$ in $\mathbb{Z}/((p-1)\mathbb{Z})^*$. It is well-known that for any prime $q\geq 5$, $-3$ is a quadratic residue $\!\!\pmod{q}$ iff $q\equiv 1\pmod{3}$ and $-1$ is a quadratic residue $\!\!\pmod{q}$ iff $q\equiv1\pmod{4}$.
A: This is not an answer. Just a collection of my observations. I couldn't settle this question with the structures inside extension fields $\Bbb{Z}/p\Bbb{Z}$, so I started out on a quest to settle the followig slightly more general question.

Let $m$ be a natural number not divisible by three. Let $\pi_m$ the permutation $x\mapsto 3x$ of $\Bbb{Z}/m\Bbb{Z}$. Determine the parity $\delta_m=\operatorname{sgn}(\pi_m)$ of $\pi_m$.

The fond hope is that this would succumb to a suitable Chinese Remainder Theorem style analysis. It is easy to see that knowing $\delta_{p-1}$ will settle the OP's question (see e.g. Jack's +1 answer for an explanation). So let's begin by looking at the cases when $m$ is an odd prime first. This is easy.
Observation 1. Assume that $q>3$ is a prime. Then $\delta_q=+1$ when 
$q\equiv\pm1\pmod{12}$, and $\delta_q=-1$ when $q\equiv\pm5\pmod{12}$. In other words, in these cases
$$
\delta_q=\left(\frac3q\right).
$$
Proof. Let $\ell$ be the multiplicative order of three modulo $q$. Thus $\ell\mid q-1$ and apart from the fixed point $0$ of $\pi_q$ the rest of $\Bbb{Z}/q\Bbb{Z}$ is partitioned into $\ell$-cycles. Obviously there are $s=(q-1)/\ell$ of those. 
If $q\equiv \pm5\pmod{12}$ then quadratic reciprocity tells us that $3$ is a quadratic non-residue modulo $q$. This means that $\ell$ is not a factor of $(q-1)/2$, implying that $\ell$ must be even and also that $s$ is odd. Therefore $\pi_q$ is a product of an odd number of cycles of an even length and therefore $\delta_q=-1$ in this case.
On the other hand, if $q\equiv\pm1\pmod{12}$, then $3$ is a quadratic residue. This means that $\ell$ is a factor of $(q-1)/2$. Consequently $s$ is even, and $\delta_q=+1$ because $\pi_q$ is the product of an even number of disjoint cycles of equal lengths. QED.
Observation 2. Whenever $m>2$ is a power of two we have $\delta_m=-1$.
Proof.  The claim $\delta_{2^n}=-1$ can be shown by induction on $n$. In the base case of $n=2$ we see that $\pi_4=(0)(13)(2)$ is an odd permutation.
The restriction of $\pi_{2^{k+1}}$ to the subgroup $2\Bbb{Z}_{2^{k+1}}$ is a "replica" of the permutation $\pi_{2^k}$. By induction hypothesis this restriction is an odd permutation. The unit group $U=\Bbb{Z}_{2^{k+1}}^*$ on the other hand is (as above) partitioned into disjoint $\ell$ cycles, where $\ell$ is the order of three in $U$. Because $|U|=2^k$ it follows that $\ell$ is a power of two. It is well known that $U$ is not cyclic when $2^{k+1}\ge8$. Therefore $\ell$ is a proper factor of $|U|$. Consequently there are an even number of disjoint $\ell$-cycles in the restricition of $\pi_{2^{k+1}}$ to $U$ (actually it is also well known that the order of $3$ is $|U|/2$ but we don't need that here). QED.
A wild guess. If $m,n$ are both coprime to $6$ then we have full multiplicativity
$$
\delta_{mn}=\delta_m\delta_n.
$$ Observe that $m$ and $n$ are allowed to have common factors.
I have tested this (aided by my new shiny upgrade of Mathematica) quite a bit, but not yet extensively. It seems to hold!
If this guess holds, then we can calculate $\delta_m$ whenever $m$ is odd.
We still need to get a way of combining this with Observation 2 to find a forula for $\delta_m$.
(to be continued, I hope. Possibly by somebody else.)
