Let $p$ be a prime number. If -1 is a square in $\mathbf F_p$, it means that there exists an element of order 4. So $p-1$ is divisible by 4 and therefore $p \equiv 1$ (mod 4). I wonder the converse is also true. Does $p \equiv 1$ (mod 4) mean the existence of an element whose square is $-1$ in $\mathbf F_p$?
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Martin Sleziak
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Jeong
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Yes. The converse is true. If $g \in F_p^{\times}$ is a generator of the multiplicative group of $F_p$, then the square of $g^{\frac{p-1}{4}}$ is $-1$. – Crostul Jun 14 '15 at 14:15
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1Yes, the converse is also true. $$\left(\frac{-1}{p}\right) = (-1)^{\frac{p-1}{2}}$$ grants that $p\equiv 1\pmod{4}$ implies that $-1$ is a quadratic residue in $\mathbb{F}_p$. – Jack D'Aurizio Jun 14 '15 at 14:16
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See $-1$ is a quadratic residue modulo $p$ if and only if $p\equiv 1\pmod{4}$ – Martin Sleziak Jun 18 '15 at 12:16
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Yes, the converse holds. Assume $p = 4k+1$ for some $k$. Every finite field has cyclic multiplicative group, so if we take $g$ to be a generator of this group, $g^{p-1} = g^{4k} = 1$. But then $g^{2k} = -1$ (since $(g^{2k})^2=1$ but $g^{2k}\neq 1$), and so $g^{k}$ is a square root of $-1$.
ptrsinclair
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Assume $p>2$ prime.
$x^2=-1 \iff x^2+1=0 \iff \deg(x)=4$ (That means for is smaller for $n$ such taht $x^n-1$). This means
$x^2+1 | x^{p-1}-1 \iff 4|p-1$.
vudu vucu
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