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For example in $\mathbb{F}_5$, $2^2=3^2=-1$. However, in $\mathbb{F}_3$, there is no solution to $x^2=-1$.

When do the squareroot(s) exist, and if they do, can we say anything about their multiplicity (or even when they occur)?

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Are you asking about all finite fields or just the residues modulo a prime natural number? – Rory Daulton Jul 28 '14 at 11:00
up vote 8 down vote accepted

Since the unit group of a finite field $\mathbb{F}$ is always cylic of order $|\mathbb{F}|-1=:n$ and $-1$ is the unique element of order $2$ in $\mathbb{F}^ \times$ (if $1 \neq -1$ in $\mathbb{F}$) there is an element $x$ with $x^2=-1$ if and only if $4 \mid n$ so if and only if $|\mathbb{F}| \equiv 1\pmod 4$.

Edit: Just to clarify: If $\mathrm{char}(\mathbb{F})=2$ we have $1=-1$ so $-1$ is obviously a square. The above argument is hence only concerned with odd characteristic.

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For the Jacobi symbol you have $(-1|p) = 1\:$ if $p \equiv 1 \pmod 4.\;$ This answers your question for the prime fields $\mathbb{F}_p.\;$ And obviously if $x^2=-1\;$ then $(-x)^2 = -1.$ Note that $\mathbb{F}_2\;$ is a special case because here you have $1=-1$.

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You forgot to mention that $-1$ also has a square root in $\mathbb{F}_2$. In other words, $\left( {\begin{array}{*{20}{c}} { - 1} \\ 2 \end{array}} \right)=1$ – Rory Daulton Jul 28 '14 at 11:02

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