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.
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: