If $F$ is a finite field of order $q$, where $q$ is an odd prime power, then when is $2$ a quadratic residue in $F$?

I know the result for when $q$ is prime. I also know a theorem which says that:

If $p$ is an odd prime and $gcd(a,p)=1$, then for any $k\geq 1$, $a$ is a quadratic residue modulo $p^{k}$ if and only if $a$ is a quadratic residue modulo $p$. So this tells me that if $q=p^{k}$ then $2$ is a quadratic residue modulo in $F$ iff $p\equiv 1,7 \mod 8$. But is there no equivalent result which only involves congruences of $q$?

Many thanks for any help!


1 Answer 1


You correctly recalled the extension to the law of of quadratic reciprocity. So in a prime field $\Bbb{F}_p=\Bbb{Z}/p\Bbb{Z}$ $2$ is a square if and only if $p\equiv\pm1\pmod8$.

In an extension field $\Bbb{F}_q$, $q=p^n$, $p$ an odd prime, you can think of it as follows. If $p\equiv\pm1\pmod8$, then $2$ is a square already in the prime field $\Bbb{F}_p\subseteq\Bbb{F}_q$. But if $p\equiv\pm3\pmod8$, then the polynomial $x^2-2$ is irreducible over $\Bbb{F}_p$. Therefore $K=\Bbb{F}_p[x]/\langle x^2-2\rangle$ is a field, and the zeros of $x^2-2$ both (they are negatives of each other) exist in $K$. We see that $[K:\Bbb{F}_p]=2$, so $K$ has $p^2$ elements. Because to each prime power $q$ there is a unique field of cardinality $q$ we see that $2$ is a square in the field $\Bbb{F}_{p^2}$ when $p\equiv\pm3\pmod8$.

So in the case $p\equiv\pm3\pmod8$ we see that $2$ is a square in the field $\Bbb{F}_{p^n}$ if and only if $K\subseteq\Bbb{F}_{p^n}$. By one of the other basic results about finite fields we know that $\Bbb{F}_{p^2}\subseteq\Bbb{F}_{p^n}$ if and only if $2\mid n$.

Summary: $2$ is a square in the field $\Bbb{F}_{p^n}$, $p$ odd, if and only if either $p\equiv\pm1\pmod8$ or $n$ is even.

  • 3
    $\begingroup$ Equivalently, if and only if $p^n\equiv\pm1\pmod8$. Although that's only shorter to state, not easier to see why it's true. $\endgroup$ Feb 25, 2016 at 7:10

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