number of quadratic residues in finite field

Question

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.

Answer 1

$s:k^{\times}\to k^{\times}, s(a)=a^2$ 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 $|k^{\times}/\{\pm1\}|=(q-1)/2$.

Answer 2

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.