Multiplication of quadratic residues

Question

I know that the following is right, where QR is a qiadratic residue and NQR is not a quadratic residue.

QR*QR=QR

NQR*NQR=QR

And why must a QR ratio consist of a QR numerator and QR denomerator or both have to be NQR.

Answer 1

One way to understand this is through group theory. The squaring map \begin{align*} \varphi: \mathbb{F}_p^\times &\to \mathbb{F}_p^\times\\ x &\mapsto x^2 \end{align*} is a group homomorphism. Since $\mathbb{F}_p$ is a field, then $$ x^2 = 1 \iff 0 = x^2 - 1 = (x-1)(x+1) \iff x = \pm 1 $$ so $\ker(\varphi) = \{\pm 1\}$. This shows that the subgroup of squares $\operatorname{img}(\varphi)$ has index $2$ in $\mathbb{F}_p^\times$, so $$ \mathbb{F}_p^\times/\operatorname{img}(\varphi) = \{\operatorname{QR}, \operatorname{NQR}\} \cong C_2 = \{\pm 1\} $$ where $C_2$ is the cyclic group of order $2$ (written multiplicatively). (As Jack D'Aurizio notes, we can also consider this as the map $\mathbb{F}_p^\times \to C_2$, $x \mapsto x^{\frac{p-1}{2}}$.) In this isomorphism, we have \begin{align*} \text{QR} &\leftrightarrow 1\\ \text{NQR} &\leftrightarrow -1 \, . \end{align*} Since this group has order $2$, by Lagrange's Theorem we have \begin{align*} \operatorname{QR} \cdot \operatorname{QR} &= \operatorname{QR}^2 = \operatorname{QR}\\ \operatorname{NQR} \cdot \operatorname{NQR} &= \operatorname{NQR}^2 = \operatorname{QR} \, . \end{align*}

As an analogy, recall a more familiar example of a group with an index $2$ subgroup: the integers $\mathbb{Z}$ and the subgroup $2 \mathbb{Z}$ of even integers. We have similar rules for even and odd numbers, just with $\cdot$ replaced by $+$:

\begin{align*} \operatorname{even} + \operatorname{even} &= \operatorname{even}\\ \operatorname{odd} + \operatorname{odd} &= \operatorname{even} \end{align*}

Answer 2

your tags say abstract algebra. The quadratic residues are a subgroup of the multiplicative group of $\mathbb Z / p \mathbb Z.$ The nonresidues are the only other coset.