Suppose $F$ is a finite field with $q$ elements, $l$ is a prime and $\tilde{F}$ is the degree $l$ extension of $F$. Given elements $a,b \in F$ with $b \neq 0$, how many elements in $\tilde{F}$ have $a$ as their trace and $b$ as their norm? If an explicit solution in terms of $a$ and $b$ cannot be given, can good lower and upper bounds depending on $l$ and $q$ be given?
We know that $\tilde{F}$ has $q^{l-1}$ elements with any given trace (by the $F$-linearity of the trace map) and $\frac{q^l - 1}{q - 1}$ elements with any given non-zero norm (the norm map is given by raising to the $\frac{q^l - 1}{q - 1}$-th power).
In the case $l = 2$, an arbitrary element of norm b and trace a has characteristic polynomial $X^2 - aX + b$.
- If there exists an $x \in F$ with $b = x^2$ and $a = 2x$, then the polynomial is a square and $x$ is the only element with $b$ as norm and $a$ as trace.
- If the polynomial is irreducible over $F$, then it splits into two different linear factors over $\tilde{F}$, whereby there exist two elements with $b$ as norm and $a$ as trace.
- If the polynomial is reducible but not a square, then $a$ can not be the trace of an element with norm $b$.