I am trying to solve a problem about bijections on finite fields, but I am stuck on this problem and don't see how can I solve it. So, here it is:
Given is a polynomial $g(y):=y^{3}+3y^{2}+3y+3\in \mathbb{Z}[y]$. Show that it defines a bijection from $\mathbb{F}_{11^{31415}}$ to itself.
To show the bijection $\mathbb{F}_{11^{31415}}\rightarrow \mathbb{F}_{11^{31415}}$, I've been thinking that I have to find an inverse map, but I don't know how to do it. I think this should be better than proving that the map is injective and surjective. I also noticed that this polynomial is irreducible over $\mathbb{Z}$ and that the number $31415=5\times 61\times 103$, but I don't know what that brings to the problem.
Observe that $g(y)=(y+1)^3+2$, so its inverse is something like $g^{-1}(x)=\sqrt[3]{x-2}\,-1$.
The only problematic part is the cubic root, so all you shoud show is that $x\mapsto x^3$ is a bijection in the given field. Since that is finite, its multiplicative group is cyclic, of order $11^{31415}-1$.
Since we have $11\equiv -1 \pmod3$, $\ 11^{31415}\equiv -1\pmod{3}$, so that the group order is coprime to $3$, and therefore $x\mapsto x^3$ is indeed a bijection:
By the Bezout identity, there are integers $a,b$ such that $a\cdot N+b\cdot 3=1$ where $N$ denotes the order of the group now. So that $x\mapsto x^b$ will be an inverse for $x\mapsto x^3$.