Interpretation of the group structure of $x^2-y^2=a$ from Norm of quadratic field We want to know the interpretation of the group structure of $x^2-y^2=a \bmod p$.
For Pell's equation, $x^2-Dy^2=1$ s.t. $D>1$, we can interpret the group structure of the sollution $(x,y)$ as a norm of unit group of $\mathbb{Q}(\sqrt D)$.
However, if $D=1$, $\mathbb{Q}(\sqrt{D})=\mathbb{Q}$. Thus we cannot see $x^2-Dy^2$ as norm of $\mathbb{Q}(\sqrt{D})$.
But this also have a group structure as below.
$$e:=\langle 1,0,1 \rangle$$
$$\langle x,y,a\rangle * \langle z,w,b\rangle:=\langle xz+yw,xw+yz,ab \rangle$$
$$\langle x,y,a\rangle^{-1}:=\langle xa^{-1},-ya^{-1},a^{-1} \rangle$$
How can we interpret this group structure? This group structure is used to prove the number of sollution of $x^2-y^2=a \bmod p$ is $p-1$.
 A: For a field $K$, if $A$ is a commutative ring (not necessarily a field!) containing $K$ that is finite-dimensional as a $K$-vector space, the norm map ${\rm N}_{A/K}  \colon A \rightarrow K$ is defined as ${\rm N}_{A/K}(a) = \det(m_a)$, where $m_a \colon A \rightarrow A$ is multiplication by $a$ (a $K$-linear map). Then multiplicativity of the norm map follows from (i) the equation $m_{ab} = m_a \circ m_b$ and (ii) multiplicativity of the determinant. Strictly speaking, we don't need $A$ to be commutative but just that $K$ is in the center of $A$ (e.g., $K = \mathbf R$ and $A = $ Hamilton quaternions).
For each $d \in K$, apply this definition to the ring $A = K[t]/(t^2-d)$, which is $2$-dimensional as a $K$-vector space no matter what $d$ is (it could be $0$ or $1$). For $x$ and $y$ in $K$, multiplication by $x+yt$ on $K[t]/(t^2-d)$ with respect to the $K$-basis $\{1,t\}$ has matrix representation $(\begin{smallmatrix}x&dy\\y&x\end{smallmatrix})$, which has determinant $x^2- dy^2$. Take $d = 1$ to see that the expression $x^2 - y^2$ is multiplicative by viewing it as a norm from the ring $K[t]/(t^2-1) \cong K \times K$ to $K$.
