I need help with the following question:
Let q be an odd prime power and function f is defined as $f(x)=x^{q+1}$.
Fix an element $a\in F^*_{q^n}$ i.e. $a\neq 0$. Let $D_a$ is defined as follows: $$D_a : F_{q^n} \rightarrow F_{q^n}$$ $$ x\rightarrow f(x+a)-f(x)$$
where $n\geq 2$
- Prove that if n is odd then $D_ a$ is 1-1 and onto?
- What would happen if n is even?
I begin with assuming that $D_a(x)=D_a(y)$ for arbitrary $x$ and $y$ in $F_{q^n}$ and try to see that $x=y$.
I could not see the result?
Because $(x+y)^q=x^q+y^q$ you have $(x+a)^{q+1}=(x+a)(x^q+a^q)$ and therefore the difference $$ L_a(x):=D_a(x)-a^{q+1}=ax^q+a^qx $$ is an $\Bbb{F}_q$-linear mapping. Therefore the question can be settled by studying the kernel of $L_a$. After all, from linear algebra we recall that a linear transformation is injective if and only if its kernel is trivial.
Let $b\neq0$ be an element of $\mathrm{ker}(L_a)$. Then $ab^q+a^qb=0$, or equivalently $(a/b)^{q-1}=-1.$ This implies that the order of $a/b$ is not a factor of $q-1$, but is a factor of $2(q-1)$. On the other hand, the order of $a/b$ is a factor of $q^n-1$ as that holds for all the elements of $\Bbb{F}_{q^n}$.
The parity of $n$ plays a role, because (recall that $q$ was assumed to be odd) $$ \frac{q^n-1}{q-1}=1+q+q^2+\cdots+q^{n-1}\equiv n\pmod 2. $$ If $n$ is odd, it follows that $\gcd(2(q-1),q^n-1)=q-1$, proving the non-existence of the above element $b\in\Bbb{F}_{q^n}$, and hence also injectivity of $L_a$ as well as $D_a$. As the domain and codomain have the same finite cardinality, injectivity is equivalent to bijectivity.
On the other hand, if $n$ is even, then $2(q-1)\mid q^n-1$. By cyclicity of the multiplicative group $\Bbb{F}_{q^n}^*$ there exists an element $\gamma\in\Bbb{F}_{q^n}$ of order $2(q-1)$. As $-1$ is the only element of order two, it follows that $\gamma^{q-1}=-1$. Therefore $$L_a(a\gamma)=a(a^q\gamma^q)+a^q(a\gamma)=a^{q+1}\gamma(\gamma^{q-1}+1)=0.$$ Hence $D_a(0)=D_a(a\gamma)$, and we have shown that $D_a$ is neither injective not surjective.