If $d\in \mathbb{Z}$ is square free (we also consider $d=-1$ to be square free), show that the quadratic ring $\mathbb{Z}[\sqrt{d}]$ is isomorphic to the subring $B \subseteq M_2(\mathbb{Z})$ where $$B=\{\begin{pmatrix} a & b\\ bd & a \end{pmatrix}\ \in M_2(\mathbb{Z})\: : \: a,b\in\mathbb{Z}\}$$
I can see that $\det(A) = a^2-b^2d$ for any matrix $A$ in $B$ and there's probably a relationship between that and elements of the given quadratic ring multiplied by their conjugates but otherwise, I'm not sure how this is isomorphic. Also, why does $d$ have to be square free here?
Every element of the quadratic ring $\Bbb Z[\sqrt{d}]$ is a linear combination $a+b\sqrt{d}$ with integer coefficients. It acts on the $\Bbb Q$-vector space $\operatorname{span}_{\Bbb Q}\{1,\sqrt{b}\}$ by right multiplication. The matrix, corresponding to the number $u=a+b\sqrt{d}$ in the basis $\{1,\sqrt{b}\}$ is $\begin{pmatrix}a & b \\ bd & a\end{pmatrix}$ (it acts on the right, so $1\cdot u$, $\sqrt{d}\cdot u$ give you the rows of the matrix instead of the usual columns). This gives you a homomorphism from the quadratic ring to your ring of matrices. That is injective since no non-zero number goes to $0$ matrix, and surjective because a number can be recovered from its matrix. Thus it is an isomorphism.