An elementary question regarding a multiplicative character over finite fields Reading Chapter 2 of Koblitz's Introduction to Elliptic Curves and Modular Forms, I got stuck on the following question. I would like to proceed my reading, so I would appreciate any hint to this. I have no background in Algebraic Number Theory, but if someone explains in that language, I would like to see it and learn it as much as I can at this moment. References will also be appreciated!
Let $q$ be a power of prime $p$ such that $q \equiv 1 \mod 4$ and $p \nmid 2n$.
Question. Let $\chi_{4} : \mathbb{F}_{q}^{*} \rightarrow S^{1}$ be the character of order $4$, show that $\chi_{4}(-4) = 1$.
The hint in the book says we may show that $\chi_{4}(4) = \chi_{4}(-1) = 1$ for $q \equiv 1 \mod 8$ and $\chi_{4}(4) = \chi_{4}(-1) = -1$ for $q \equiv 5 \mod 8$.
If $q \equiv 1 \mod 8$, then $\chi_{4}(4) = \chi_{4}(2)^{2} = \chi_{2}(2) = \pm 1$, where $\chi_{j}$ denotes the character of order $j$. So I think $2$ must be quadratic residue in this case, but I cannot figure out why.
 A: Here $q=p^n$ and $\Bbb{F}_q^*$ is cyclic of order $q-1$ by the basic properties of finite fields. Let $g$ be a generator of that group. The element $-1$ is the only element of order two in that group, so $-1=g^{(q-1)/2}$. We have $\chi_4(g)=\pm i$. We are given that $\ell=(q-1)/4$ is an integer.
$$
\chi_4(-1)=\chi_4(g^{(q-1)/2})=(\pm i)^{(q-1)/2}=\left((\pm i)^2\right)^{\ell}=(-1)^{\ell},
$$
so $\chi_4(-1)$ is either $+1$ or $-1$ according to whether $\ell$ is even or odd, or equivalently, according to whether $q\equiv1\pmod8$ or $q\equiv5\pmod8$.
As you observed, the calculation of $\chi_4(4)$ depends on whether $2$ is a square in $\Bbb{F}_q$ or not. First recall that all the odd numbers $\not\equiv1\pmod8$ are of order two modulo $8$. 
Assume first that we are in the case $q\equiv1\pmod8$. This means that either A) $p\equiv1\pmod8$ or that B) $n$ is even. In case A), by a well known addendum to the law of quadratic reciprocity, $2$ is quadratic residue modulo $p$, and $\chi_4(2)=\pm1$. In case B we have that $\Bbb{F}_p[\sqrt2]\subseteq\Bbb{F}_{p^2}\subseteq\Bbb{F}_{p^n}$, so $2$ is a square of $\Bbb{F}_q$ in this case, too.
The other case is that of $q\equiv5\pmod8$. As $p^2\equiv1\pmod8$ anyway this implies that we must have $p\equiv5\pmod8$ and $n$ must be odd. This time we know that $2$ is a quadratic non-residue modulo $p$. If we had $\Bbb{F}_p[\sqrt2]\subseteq\Bbb{F}_q$ we would also get, by multiplicativity of extension degree in a tower of fields,
$$
2=[\Bbb{F}_p[\sqrt2]:\Bbb{F}_p]\mid [\Bbb{F}_q:\Bbb{F}_p]=n
$$
contradicting the assumption that $n$ is odd. Therefore $2$ is not a square in $\Bbb{F}_q$. Hence $\chi_4(2)=\pm i$, and consequently $\chi_4(4)=-1$.
