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I know that in fields of cardinality $p$, $a$ is a quadratic residue if and only if $a^{\frac{p-1}{2}}=1$ (Euler's criterion). Therefore $a^{\frac{p+1}{2}}=a$ and if also $p=3\!\!\!\mod\! 4$ we can exhibit the square roots: $(a^{\frac{p+1}{4}})^2=a$.

Can we do something similar in fields of prime power cardinality (i.e., find square roots of elements with possibly a constraint on the cardinality)? I do not see how trying to do the same thing with the Legendre symbol helps.

This question was marked as a possible duplicate of Quadratic residues in finite field, however, I'm not asking whether an element is a quadratic residue, but assuming it is a quadratic residue, obtain its square roots.

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Yes, we can. That criterion and the formula for the square root work because the multiplicative group of integers mod $p$ is cyclic (of order $p-1$). Well, the multiplicative group of the field of cardinality $p^k$ is cyclic as well (of order $p^k-1$).

Thus, for $p>2$, $a$ is a square in the field of order $p^k$ if and only if $a^{(p^k-1)/2}=1$ or equivalently $a^{(p^k+1)/2}=a$. For $p\equiv 3\mod 4$ and $k$ odd, we also have explicit square roots $\left(a^{(p^k+1)/4}\right)^2=a$.

What about $p=2$? In those fields, everything is a square, and the map $x\to x^2$ is a field automorphism.

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  • $\begingroup$ I see. Does Euler's criterion hold in this case? Because as I see it, that is why the proof works, not the fact that the group is cyclic. Am I missing something? $\endgroup$
    – Tal-Botvinnik
    Feb 9, 2019 at 18:39
  • $\begingroup$ The Euler criterion follows from the group being cyclic. Saying $(b^2)^{|G|/2}=e$ works in any finite group $G$, while the converse that $a$ is a square if $a^{|G|/2}=e$ requires an additional condition on the structure of $G$ - like being cyclic. $\endgroup$
    – jmerry
    Feb 9, 2019 at 21:27
  • $\begingroup$ I see your point but still do not understand the implicance. According to Euler's theorem, we have $$a^{p^{k-1}(p-1)}=1\mod p^k,$$ because of the multiplicativity of the totient function. Doesnt this mean that the roots are given by $$a^{\frac{p^{k-1}(p-1)+1}{2}}?$$ $\endgroup$
    – Tal-Botvinnik
    Feb 11, 2019 at 11:40
  • $\begingroup$ Why are you asking about things mod $p^k$ now? We're working with finite fields of characteristic $p$ - that's not the same structure. $\endgroup$
    – jmerry
    Feb 11, 2019 at 11:43
  • $\begingroup$ Oh of course, the important thing here is the characteristic, not cardinality. $\endgroup$
    – Tal-Botvinnik
    Feb 11, 2019 at 12:01

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