Difference Sets and Hyperovals Maschietti's theorem is as follows:
The $q+2$ set $D(x^k)$ is a hyperoval iff $D_k^*$ is a $(q-1,q/2-1,q/4-1)$ difference set in $GF(q)^*$. Where $q=2^d$, $2\leq q-2$ and 
$D_k=\{x+x^k\|x\in GF(q)\}$ regarded as image of the polynomial function $\tau:GF(q)\rightarrow GF(q)$, $\tau(x)=x+x^k$, and $D_k^* =$ Im $\tau$ \ $\{0\}$,
$D(x^k)=\{(1,x,x^k)\|x\in GF(q)\}$ $\cup$ $(0,1,0),(0,0,1)$.
To understand the theorem I made an example and got the idea. 
However,
in the paper "Gauss Sums, Jacobi Sums, and p-Ranks of Cyclic Difference Sets" "Theorem 5" the Maschietti's theorem proved briefly by using characters and Jacobi sums.
I need to understand this proof. Thanks for any help. 
Here is the proof:
Let $\chi$ be a multiplicative character of $GF(q)$. We know that since $D(x^k)$ is a hyperoval, $\tau: x \to x+x^{k}$ is a 2-to-1 map.  We have
$\chi(D_k^*)=\dfrac{1}{2} \sum_x \chi(x+x^k)=$ $\dfrac{1}{2} \sum_x \chi(x) \chi(1+x^{k-1}).$     (..............................Eq. 1)
We know that when $D(x^k)$ is a hyperoval then $k$ and $k-1$ are both relatively prime to $q-1$. Thus $(k-1,q-1)=1$. So there exist a multiplicative character $\phi$ of $F_q$ s.t. $\chi=\phi^{k-1}$. Hence, 
$\chi(D_k^*)=\dfrac{1}{2} \sum_x \phi(x^{k-1})\chi(1+x^{k-1})=\dfrac{1}{2} J(\phi,\chi).$ (...................................Eq. 2)
Noting that $\phi \chi =\phi ^k$ is nontrivial, we have $\chi(D_k^*)$ $\overline{\chi(D_k^*)}$ $=2^{d-2}$. Which comes from: if D is a $(v,k,\lambda)$ difference set, then $\chi(D)$ $\overline{\chi(D)}$ $=k-\lambda$. 
 A: We will assume that $D(x^{k}) = \{ (1,x,x^{k}) : x \in \mathbb{F}_{q}\} \cup \{(0,1,0), (0,0,1)\}$ is a hyperoval; by the earlier results in this paper, this implies that $x \mapsto x+x^{k}$ is a 2-to-1 map. We want to show that $D_{k}^{*} = \{ x+x^{k} : x \in \mathbb{F}_{q}^{*}\}$ is a $(q-1, \frac{q}{2}-1, \frac{q}{4}-1)$ difference set in $\mathbb{F}_{q}^{*}$. By Lemma 1.1, it is sufficient to show that $\chi(D_{k}^{*})\overline{\chi(D_{k}^{*})} = (\frac{q}{2}-1) - (\frac{q}{4}-1)$ for every nontrivial character $\chi$.
Let $\chi$ be any multiplicative character on $\mathbb{F}_{q}$. Since $x \mapsto x+x^{k}$ is 2-to-1, we have
$$\chi(D_{k}^{*}) = \sum_{a \in D_{k}^{*}} \chi(a) = \frac{1}{2} \sum_{x \in \mathbb{F}_{q}} \chi(x+x^{k}).$$
This is giving the definition of what we mean by a character evaluated on a set (the sum of character evaluations of the elements in that set), and we are able to rewrite it as a character sum over the elements of the field, which is super useful for when we want to apply techniques using Gauss of Jacobi sums.
Now we use the fact that $\chi$ is multiplicative to rewrite $\chi(x+x^{k})=\chi(x)\chi(1+x^{k})$. From here, we use the fact that the collection of multiplicative characters is isomorphic to $\mathbb{F}_{q}^{*}$, so in particular it is cyclic of order $q-1$.  So suppose that $\alpha$ is a generator for the cyclic group of multiplicative characters.  Since $(k-1,q-1) = 1$, $\alpha^{k-1}$ is also a generator, and so there is some $c$ such that $(\alpha^{k-1})^{c} = \chi$.  This means that $(\alpha^{c})^{k-1} = \chi$, showing that there exists $\phi (=\alpha^{c})$ with $\phi^{k-1} = \chi$.
Now the point of this is to be able to rewrite 
$$\chi(x)\chi(1+x^{k-1})=\phi^{k-1}(x)\chi(1+x^{k-1}) =\phi(x^{k-1})\chi(1+x^{k-1})$$
(again, we can do this since $\phi$ is a multiplicative character). This allows us to write $\chi(D_{k}^{*})$ as a Jacobi sum.
$$\chi(D_{k}^{*}) = \frac{1}{2}\sum_{x \in \mathbb{F}_{q}}\phi(x^{k-1})\chi(1+x^{k-1}) = \frac{1}{2}J(\phi,\chi)$$
Using this, we have that
$$\chi(D_{k}^{*}) \overline{\chi(D_{k}^{*})} = \frac{1}{4}J(\phi,\chi)\overline{J(\phi,\chi)},$$
and by (1.5) of the paper you reference, 
$$\frac{1}{4}J(\phi,\chi)\overline{J(\phi,\chi)} = \frac{1}{4}q = (\frac{q}{2} -1) - (\frac{q}{4}-1).$$
Using Lemma 1.1, this implies that $D_{k}^{*}$ is a $(q-1, \frac{q}{2}-1, \frac{q}{4}-1)$ difference set in $\mathbb{F}_{q}^{*}$.
