Let $\mathbb{F}_{q}$ be the finite field with $q$ elements, where $q$ is a prime power.
Let $n \geq 1$ be an integer and consider $\mathbb{F}_{q^n}|\mathbb{F}_{q}$.
There is a theorem that says the following:
Theorem: There is always an element $\alpha \in \mathbb{F}_{q^n}$ that is primitive and normal over $\mathbb{F}_{q}$.
We say that one can prescribe the norm and the trace of a primitive and normal (over $\mathbb{F}_{q}$) element $\alpha \in \mathbb{F}_{q^n}$ if, for every $a,b \in \mathbb{F}_{q}^\ast$, with $b$ primitive, there is a primitive and normal element $\alpha \in \mathbb{F}_{q^n}$ such that $Tr_{\mathbb{F}_{q^n}|\mathbb{F}_{q}}(\alpha) = a$ and $N_{\mathbb{F}_{q^n}|\mathbb{F}_{q}}(\alpha) = b$.
The assumption that $a$ is non-zero is because normal elements cannot have zero trace and a primitive element $\alpha \in \mathbb{F}_{q^n}$ must have norm a primitive element of $\mathbb{F}_{q}$.
My point is, the article I'm reading asserts that if $n \leq 2$, such $\alpha$ is already prescribed by its trace and norm, but I cannot see this. Can anyone help me?
The case $\mathbb{F}_{q^2}|\mathbb{F}_{q}$ we then have $Tr(\alpha) = \alpha + \alpha^q$ and $N(\alpha) = \alpha^{q+1}$. I cannot see why all possible values for the norm and trace in $\mathbb{F}_{q}$ are achieved by primitive normal elements of $\mathbb{F}_{q^2}$.
Edit: I'm trying to think about it's minimal polynomial. There is a fact (I will not prove here but it is true): In $\mathbb{F}_{q^2} | \mathbb{F}_{q}$, every primitive element is also normal. So, the minimal polynomial of a primitive normal element of $\mathbb{F}_{q^2}$ must be
$$X^2 -aX + b,$$ where $a = Tr(\alpha)$ and $b = N(\alpha)$. Still cannot see why every possible value for $N(\alpha)$ (any primitive element of $\mathbb{F}_{q}$) and $Tr(\alpha)$ (any non-zero element of $\mathbb{F}_{q}$) can be achieved.
Actually, I misinterpreted what the author said. They said that if $n \leq 2$ the problem is not meaningful. In fact, if you can prescribe the norm and trace of a primitive and normal element $\alpha \in \mathbb{F}_{q^2}$, then it's minimal polynomial must be:
$$X^2 -Tr(\alpha)X + N(\alpha),$$
and this must be irreducible for every possible value of $Tr(\alpha)$ and $N(\alpha)$. This is not true however. In $\mathbb{F}_{17^2} | \mathbb{F}_{17}$, the polynomial
$$X^2 -5X + 3 = (X+2)(X+10),$$ therefore there is no $\alpha \in \mathbb{F}_{17^2}$ such that $Tr(\alpha) = 5$ and $N(\alpha) = 3$ (3 is primitive in $\mathbb{F}_{17}$).