First Supplement to Law of Quadratic Reciprocity

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Theorem

$\paren {\dfrac {-1} p} = \paren {-1}^{\paren {p - 1} / 2} = \begin{cases} +1 & : p \equiv 1 \pmod 4 \\ -1 & : p \equiv 3 \pmod 4 \end{cases}$

where $\paren {\dfrac {-1} p}$ is defined as the Legendre symbol.


Proof

From Euler's Criterion, and the definition of the Legendre symbol, we have that:

$\paren {\dfrac a p} \equiv a^{\paren {p - 1} / 2} \pmod p$

The result follows by putting $a = -1$.

$\blacksquare$


Examples

$-1$ is not Quadratic Residue of $3$

$-1$ is a quadratic non-residue of $3$.


$-1$ is Quadratic Residue of $5$

$-1$ is a quadratic residue of $5$.


$-1$ is not Quadratic Residue of $7$

$-1$ is a quadratic non-residue of $7$.


$-1$ is not Quadratic Residue of $11$

$-1$ is a quadratic non-residue of $11$.


$-1$ is Quadratic Residue of $13$

$-1$ is a quadratic residue of $13$.


$-1$ is Quadratic Residue of $17$

$-1$ is a quadratic residue of $17$.


$-1$ is not Quadratic Residue of $19$

$-1$ is a quadratic non-residue of $19$.


Also see


Sources