Special properties of binary Galois field Consider two finite field $A=GF(2^{2k})$ and $B=GF(2^{2n+1})$, where by $GF$ mean Galois field and $k$ and $n$ are two natural number. Assume the following equation
$$
x^2+x+1=0 \tag{1}
$$
First question: How to prove that the equation $(1)$ has exactly two solutions in the finite field $A$ and there is no solution for the equation $(1)$ in the Galois field $B$. 
Second question: Assume the following two condtions
$$
\begin{array}{ccc}
\sum_{i=1}^4\, \alpha_i=\alpha_1+\alpha_2+\alpha_3+\alpha_4=0 & &  \\
&&\tag{2} \\
\sum_{i=1}^4\, \alpha_i^5=\alpha_1^5+\alpha_2^5+\alpha_3^5+\alpha_4^5=0 & & 
\end{array}
$$
where $\alpha_i$, $1\leq i \leq 4$, are elements of a finite field. 
How to show that the conditions $(2)$ hold by the four elements of finite field  $A$ but there are no four elements in the finite field $B$ such that satisfy in the conditions $(2)$.
For example, consider the  Galois field $A=GF(2^4)$ that is constructed by the polynomial $\beta^4+\beta^3+1=0$. We can check that the following four elements of finite field $A=GF(2^4)$ hold in the conditions $(2)$:
$$
(\alpha_1,\alpha_2,\alpha_3,\alpha_4)=(1,\beta,\beta^2+1,\beta^2+\beta)\,.
$$
Thanks for any suggestion. 
Edition: 
One of the most important matrix in the cryptography is MDS matrix. An interesting method for construction of MDS matrix 
is by Vandermonde  matrix. A vandermonde matrix is defined as follows 
\begin{equation}
A=\left(
\begin{array}{ccccc}
 1  & a_1  & a_1^2 & \cdots  & a_1^{n-1}\\
 1  & a_2  & a_2^2 & \cdots  & a_2^{n-1}\\
\vdots    & \vdots    & \vdots  &  \vdots &\vdots    \\
\vdots    & \vdots    & \vdots  &  \vdots &\vdots    \\
 1  & a_{n-1}  & a_{n-1}^2 & \cdots  & a_{n-1}^{n-1}\\
 1  & a_n  & a_n^2 & \cdots  & a_n^{n-1}\\
\end{array}
\right)\, .
\end{equation}
where $a_i$, $1\leq i \leq n$, are elements of $GF(2^{q})$, that is denoted with 
$$
A=van(a_1,a_2,\cdots, a_n)\, .
$$
It is proved in this article that if
$$
 A=van(a_1,a_2,\cdots, a_n) \quad , \quad B=van(b_1,b_2,\cdots, b_n)\, .
$$
be two vandermonde matrix 
such that $a_i\neq b_j$ for $1\leq i,j \leq n$ then the matrices $A\,B^{-1}$ and $B\, A^{-1}$ are MDS matrices.
Now, consider the following vandermonde matrix of order $4$:
\begin{equation}
C=\left(
\begin{array}{cccc}
 1  & \alpha_1  & \alpha_1^2  & \alpha_1^3\\
 1  & \alpha_2  & \alpha_2^2  & \alpha_2^3\\
 1  & \alpha_3  & \alpha_3^2  & \alpha_3^3\\
 1  & \alpha_4  & \alpha_4^2  & \alpha_4^3\\
\end{array}
\right)\, .
\end{equation}
where $\alpha_i$, $1\leq i \leq n$, are elements of $GF(2^{2q})$ and satisfy in the condition $(2)$ and are distinct.
 It can be proved that the inverses of matrix $C$, denoted with $C^{-1}$, can be obtained in the following form 
\begin{equation}
C^{-1}=\left(
\begin{array}{cccc}
u\,\alpha_1^3 +u\,v  & u\,\alpha_2^3 + u\,v  & u\, \alpha_3^3 +u\, v   & u\, \alpha_4^3 + u\, v \\
u\, \alpha_1^2  & u\, \alpha_2^2  &u\,  \alpha_3^2   &u\,  \alpha_4^2\\
u\, \alpha_1  & u\, \alpha_2  & u\, \alpha_3 & u\, \alpha_4\\
 u  & u  & u   & u\\
\end{array}
\right)\, .
\end{equation}
where $u$ and $v$ are defined as follows 
$$
u=\sum_{i=1}^4\, \alpha_i^{-3}\quad , \quad v=\sum_{i=1}^4\, \alpha_i^{3}
$$
The last result about the form of $C^{-1}$ matrix,  is part of my research about MDS matrix. 
Thanks for all useful comments and answer Professor Jyrki Lahtonen. 
 A: This depends on Newton's identities relating certain symmetric polynomials to each other (alternatively you can just crank this out with pencil-and-paper work).
If we denote by $e_1,e_2,e_3,e_4$ the elementary symmetric functions, i.e. the coefficients of
$$
P(x)=(x-\alpha_1)(x-\alpha_2)(x-\alpha_3)(x-\alpha_4)=x^4+e_1x^3+e_2x^2+e_3x+e_4,
$$
and by $p_i, i\in\Bbb{N},$ the power sum
$$
p_i=\alpha_1^i+\alpha_2^i+\alpha_3^i+\alpha_4^i,
$$
then by the so called Freshman's dream (in characteristic two) we get
$$
\begin{aligned}
p_1&=e_1,\\
p_2&=\alpha_1^2+\alpha_2^2+\alpha_3^2+\alpha_4^2=(\alpha_1+\alpha_2+\alpha_3+\alpha_4)^2=e_1^2,\\
p_4&=e_1^4,
\end{aligned}
$$
and then a couple of applications of 
Newton's identities eventually give that
$$
p_5=e_1^5+e_2e_1^3+e_3e_1^2+e_2^2e_1+e_2e_3+e_1e_4.\qquad(*)
$$
Your system $(2)$ states that $p_1=e_1=0$ and that $p_5=0$. Plugging in 
$e_1=0$ into $(*)$ then gives the simple consequence
$$
0=p_5=e_2e_3.
$$
So we can conclude that if $(\alpha_1,\alpha_2,\alpha_3,\alpha_4)$ is a solution of $(2)$ then either $e_3=0$ or $e_2=0$.
Let us first look at the case $e_3=e_1=0$. Then we have
$$
\begin{aligned}
P(x)&=x^4+e_1x^3+e_2x^2+e_3x+e_4\\
&=x^4+e_2x^2+e_4\\
&=(x^2+\sqrt{e_2}x+\sqrt{e_4})^2,
\end{aligned}
$$
by the Freshman's dream and the fact that in any finite field of characteristic two every element has a (unique) square root. This shows that all the roots of $P(x)$ are double roots contradicting the assumption that the $\alpha_i$s were to be distinct. Observe that this argument works equally well for the field $A$ as well as the field $B$.
The other case $e_2=e_1=0$ is different. This time
$$
P(x)=x^4+e_3x+e_4.
$$
Let us fix the field $K=GF(2^m)$.
The following trick from the theory of linearized polynomials allows us to make progress. Let $L(x)=x^4+e_3x$. Again by Freshman's dream we have
$$
L(a+b)=L(a)+L(b)\qquad(**)
$$
for all $a,b\in K$. If $\alpha_i,i=1,2,3,4,$ are four distinct zeros of $P(x)$, then $L(\alpha_i)=P(\alpha_i)+e_4=e_4$. By $(**)$ this implies that
for all $i=2,3,4$ we have
$$
L(\alpha_i-\alpha_1)=L(\alpha_i)+L(\alpha_1)=e_4+e_4=0.
$$
As $L(0)=0$ the (linearized) polynomial $L(x)$ also has four zeros.
But
$$
L(x)=x(x^3+e_3),
$$
so the non-zero roots of $L(x)$ are exactly the cubic roots of $e_3$. This explains why we get different behavior according to parity of $m=[K:GF(2)]$.
Namely:


*

*If $m$ is odd, then $3\nmid 2^m-1$. As the group $K^*$ is cyclic of order $2^m-1$ this implies that every element of $K$ has a unique cube root in $K$. Applying this to $e_3\in K$ implies that $L(x)$ has exactly two roots in $K$, and therefore $P(x)$ cannot have four roots in $K$ either.

*On the other hand id $m$ is even, then $3\mid 2^m-1$, and the cubes of $K^*$ form a subgroup of index three - each with three cube roots. In this case the polynomial $L(x)$ has four distinct roots in $K$ whenever $e_3$ is a non-zero cube in $K$. Therefore for some choice of $(e_3,e_4)$ the polynomial $P(x)$ has the required four distinct solutions. Consequently system $(2)$ also has solutions of the required type. An easy example is when $\alpha_i,i=1,2,3,4,$ range over the elements of the subfield $GF(4)$. In that case $\alpha_i^5=\alpha_i^2$ and it is easy to check that you get a solution.

