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 for Quadratic Residue, 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
- 1971: George E. Andrews: Number Theory ... (previous) ... (next): $\text {4-1}$ Basic Properties of Congruences: Exercise $6$
- 1998: David Nelson: The Penguin Dictionary of Mathematics (2nd ed.) ... (previous) ... (next): quadratic reciprocity, law of
- 2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): quadratic reciprocity, law of