how to prove $X+X^5+X^7$ in $F_{2^m}$ where $m \neq 0 \space \pmod 3$ is a permutation The Dickson polynomial of degree $7$:  $X+X^5+X^7$ in $F_{2^m}$ where $m \neq 0 \space \pmod 3$ , I checked the equation through implementation in $F_{2^8}$ and I found it is a permutation with differential uniformity of 6.
I understand the in monomial polynomial $x^d$ is a permutation in $F_{2^m}$ with a condition $\gcd(d,2^m-1)$, but I do not understand how to prove in binomial and trinomial polynomial.
How to prove this trinomial polynomial ($X+X^5+X^7$) is permutation ?
 A: Let's denote $D(x)=x^7+x^5+x$. The key property of this Dickson polynomial is the following:
$$
D(x+\frac1x)=x^7+x^{-7}.
$$
Verifying this is a straightforward computation using the binomial formula. You do need the bits: $\binom 7k$ is odd for all $k, 0\le k\le 7$, but $\binom 5k$ is odd only when $k\in\{0,1,4,5\}$.
Let $F=\Bbb{F}_{2^m}$ and $K=\Bbb{F}_{2^{2m}}$ the quadratic extension. I record the following fact:
Lemma. Every element $a\in F$ can be written in the form
$$a=\alpha+\alpha^{-1}\qquad(*)$$ for some $\alpha\in K^*$. There are two choices for the element $\alpha$. If $\alpha$ is one then $\alpha^{-1}$ is the other (the two choices are equal iff $\alpha=1$ iff $a=0$).
Proof. The equation $(*)$ is equivalent to $\alpha^2+2a\alpha+1=0$. This is a quadratic, so it has a solution in either $F$ or $K$. It is a quadratic, so there are then exactly two solutions, and the reciprocal obviously also works.
The main claim follows from this rather quickly. Assume that $D(a)=D(b)$ for some elements $a,b\in F$. According to the Lemma we can write $a=\alpha+\alpha^{-1}$
and $b=\beta+\beta^{-1}$ for some $\alpha,\beta\in K$. By the basic property of Dickson polynomials we thus have
$$
\alpha^7+\alpha^{-7}=D(a)=D(b)=\beta^7+\beta^{-7}.
$$
By applying the Lemma (to the element $D(a)=D(b)\in F$) we have either $\alpha^7=\beta^7$ or $\alpha^7=\beta^{-7}$.
Then comes the final ingredient (mentioned in the question). If $\gcd(7,2^{2m-1})=1$ then raising to the seventh power is a bijection on $K$. This is because $K^*$ is cyclic of order $2^{2m}-1$. The gcd constraint holds unless $3\mid m$, so we can conclude that
either $\alpha=\beta$ or $\alpha=\beta^{-1}$. Both of these cases imply $a=b$, proving injectivity of $D: F\to F$. Because $F$ is finite, an injective function $F\to F$ is automatically a permutation.
