number of quadratic residues in finite field Is there a way to determine how many quadratic residues are there in the finite field $F_q$ for $q = p^k$?
It seems if $q=p$ exactly $(p-1)/2$ are residues and the same amount are not. Does analogy holds for $q = p^k$?
Thanks for any input.
 A: $s:k^{\times}\to k^{\times}, s(a)=a^2$ is a (multiplicative) group homomorphism.  In characteristic $\neq 2$, the kernel of $s$ has order two and if $k$ is finite, the set of quadratic residues, $s(k^{\times})$, has size $|k^{\times}/\{\pm1\}|=(q-1)/2$.
A: In response to the comment saying 'mimic the proof of $(p-1)/2$', here is that proof, taken from this document from ETH Zurich.
The theorem is:

Let $q$ be an odd prime power, and $a \in \mathbb{F}_q$,
then:

*

*$a \in QR(q), \iff a^{(q-1)/2} = 1$

*$a \in QNR(q), \iff a^{(q-1)/2} = -1$

*$| QNR(q) | = \frac{q-1}{2} = | QR(q) |$

And the essence of the proof is:

1:  A polynomial of degree $d$, with coefficients from a field $R$, can
have at most $d$ roots in $R$.
2: Lagrange's Theorem: In a finite group $G$, $x^{|G|} = 1$ for any $x \in G$.
On the one hand, by Fact 1, the polynomial $x^{q-1} -1 = 0$ cannot have more than $q - 1$ roots by Legrange's theorem, because all the elements in $\mathbb{F}_q^*$ are roots.
Consequently, since $x^{q-1} - 1 = ( x^{(q-1)/2} + 1 ) ( x^{(q - 1)/2} - 1 )$ and the ring of polynomials over $\mathbb{F}_q$ has no (non-trivial) zero divisors, again Fact 1 implies that both factors $( x^{(q-1)/2} + 1 )$ and $( x^{(q-1)/2} - 1 )$ must have exactly $\frac{q-1}{2}$ roots.

