First Supplement to Law of Quadratic Reciprocity

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

• Law of Quadratic Reciprocity
• Second Supplement to Law of Quadratic Reciprocity

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