Definition:Quadratic Residue

Definition

Let $p$ be an odd prime.

Let $a \not \equiv 0 \pmod p$.

Then $a$ is a quadratic residue of $p$ iff $x^2 \equiv a \pmod p$ has a solution.

That is, iff $\exists x \in \Z: x^2 \equiv a \pmod p$.

Quadratic Non-Residue

If there is no such integer $x$ such that $x^2 \equiv a \pmod p$, then $a$ is a quadratic non-residue of $p$.

Quadratic Character

Any integer $a$ is either a quadratic residue or a quadratic non-residue of $p$.

Whether it is or not is known as the quadratic character of $a$ modulo $p$.

Quadratic Character of Congruent Integers

Note that if $a \equiv b \pmod p$ then $x^2 \equiv a \pmod p$ has a solution iff $x^2 \equiv b \pmod p$.

So congruent integers are of the same quadratic character.

Therefore it is sufficient to consider the quadratic character of the non-zero least positive residues modulo $p$.

Example

Take $p = 11$.

To list the quadratic residues of $11$ it is enough to work out the squares $1^2, 2^2, \ldots, 10^2$ modulo $11$.

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle 1^2\)	\(\equiv\)	\(\displaystyle 1\)	\(\displaystyle \)	\(\displaystyle \pmod {11}\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle 2^2\)	\(\equiv\)	\(\displaystyle 4\)	\(\displaystyle \)	\(\displaystyle \pmod {11}\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle 3^2\)	\(\equiv\)	\(\displaystyle 9\)	\(\displaystyle \)	\(\displaystyle \pmod {11}\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle 4^2\)	\(\equiv\)	\(\displaystyle 5\)	\(\displaystyle \)	\(\displaystyle \pmod {11}\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle 5^2\)	\(\equiv\)	\(\displaystyle 3\)	\(\displaystyle \)	\(\displaystyle \pmod {11}\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle 6^2\)	\(\equiv\)	\(\displaystyle 3\)	\(\displaystyle \)	\(\displaystyle \pmod {11}\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle 7^2\)	\(\equiv\)	\(\displaystyle 5\)	\(\displaystyle \)	\(\displaystyle \pmod {11}\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle 8^2\)	\(\equiv\)	\(\displaystyle 9\)	\(\displaystyle \)	\(\displaystyle \pmod {11}\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle 9^2\)	\(\equiv\)	\(\displaystyle 4\)	\(\displaystyle \)	\(\displaystyle \pmod {11}\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle 10^2\)	\(\equiv\)	\(\displaystyle 1\)	\(\displaystyle \)	\(\displaystyle \pmod {11}\)	\(\displaystyle \)

So the quadratic residues of $11$ are $1, 3, 4, 5, 9$.

The quadratic non-residues of $11$ are therefore all the other non-zero least positive residues, that is, $2, 6, 7, 8, 10$.

Also see

Definition of the Legendre symbol.

Note

The case where $a = 0$ has been excluded from the definition, despite the fact that $0 = 0^2$ and so is definitely a square.

The case where $p = 2$ is also excluded, where the only non-zero residue $1$ is also a square.

The main reason for this is so that some useful results can be expressed in a convenient form.

For example, this means that the Number of Quadratic Residues of a Prime $p$ is always equal to $\dfrac {p-1} 2$, which is the same as the number of quadratic non-residues.