Showing that $f(x) = 4x^2 - 3x^7$ is a pemutation in $\mathbb{Z}_{11}$ and factoring it as a product of disjoint cycles I need to show that $f:\mathbb{Z}_{11}\to \mathbb{Z}_{11}$ given by $f(x) = 4x^2 - 3x^7$, where the operations are the usual in $\mathbb{Z}_{11}$, is a permutation of the elements of $\mathbb{Z}_{11}$. I also need to write $f$ as a product of disjoint cycles.
Given that I need to write it as product of disjoint cycles, I thought show the bijection by calculating $f$ in every point, but the direct calculations are long so I'm trying to use some modular arithmetic: the Fermat primality test gives that $x^{10}\equiv_{11} 1$ so that every $x^5$ is it's own inverse $\mod{11}$. So, once $11$ is prime, $x^5\equiv_{11} 1$ or $x^5\equiv_{11} -1$. So either $-3x^7\equiv_{11} -3x^2$ or $-3x^7\equiv_{11} 3x^2$, then we can simplify $f$ by:
$$
f(x) =
\begin{cases}
x^2,  & \text{if $x^5\equiv_{11} 1$} \\
7x^2, & \text{if $x^5\equiv_{11} -1$}
\end{cases}
$$
But still there's too much calculations to be done, there is another way or further simplifying on this one to have less calculations in the end? Thanks in advance.
 A: Let $R$ denote the set of nonzero quadratic residues modulo $11$, and $N$ the set of quadratic nonresidues.  Note that $2$ is a generator of the multiplicative group $(\mathbb{Z}/11\mathbb{Z})^\times$.  Since $f(x)=x^2$ if $x\in R$, we have $f|_R:R\to R$ and $f$ maps
$$2^2\mapsto 2^4\mapsto 2^8\mapsto 2^6\mapsto 2^2\;\;(4\mapsto 5\mapsto 3\mapsto 9\mapsto 4)$$
and
$$2^0\mapsto 2^0\;\;(1\mapsto 1)\,.$$
Since $$f(x)=7x^2=128x^2=2^7x^2=2^{-3}x^2$$ if $x\in N$, we conclude that $f|_N:N\to N$.  Note that $f$ maps
$$2^1\mapsto 2^9\mapsto 2^5\mapsto 2^7\mapsto 2^1\;\;(2\mapsto6\mapsto 10\mapsto 7\mapsto 2)\,,$$
and
$$2^3\mapsto 2^3 \;\;(8\mapsto 8)\,.$$
Along with $f(0)=0$, we conclude that $f$ is a permutation on $\mathbb{Z}/11\mathbb{Z}$ with the following cycle decomposition:
$$(0)\,(1)\,(8)\,(2\;\;6\;\;10\;\;7)\,(4\;\;5\;\;3\;\;9)\,.$$
I am quite curious how these polynomials are found (namely, polynomial functions that act like permutations on $\mathbb{Z}/m\mathbb{Z}$, where the positive integer $m$ tends to be prime).  Is there a construction, or are these polynomials results of brute-force searching?  
At least, if $p$ is a prime natural number such that $p\equiv 3\pmod{4}$, $r\neq 0$ is a quadratic residue modulo $p$, and $n$ is a quadratic nonresidule modulo $p$, then we can take 
$$f(x):=(r+n)\left(\frac{p+1}{2}\right)\,x^2+\Biggl(r\left(\frac{p+1}{2}\right)+n\left(\frac{p-1}{2}\right)\Biggr)\,x^{\frac{p+3}{2}}\,.$$
Then, $f$ is a permutation on $\mathbb{Z}/p\mathbb{Z}$.  The polynomial of this particular question comes from $p=11$, $r=1$, and $n=7$.
A: Let's proceed as in the comments of @hardmath and @AndreasBlass.
We'll just evaluate and see which (if any) permutation it is.
$f(0)=0, f(1)=1,f(2)=6,f(3)=9,f(4)=5,f(5)=3,f(-5)=10,f(-4)=2,f(-3)=8,f(-2)=4,f(-1)=7$.
Thus we have $(26\,10\,7)(4539)$.
