# First Supplement to Law of Quadratic Reciprocity

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## Contents

## 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

- 1971: George E. Andrews:
*Number Theory*... (previous) ... (next): $\text {4-1}$ Basic Properties of Congruences: Exercise $6$