Let $k$ be a field, and for any $a,b\in k^\times$ let $(a,b)$ be the quaternion algebra over $k$ with parameters $a,b$. Is the following correct? I haven't been able to find a place where this is spelled out, so I have doubts about its validity.
For any $a,b,c,d\in k^\times$, the quaternion algebras $(a,b),(c,d)$ are Brauer equivalent (i.e. $(a,b)\otimes M_n(k)\simeq(c,d)\otimes M_n(k)$ for some $n$) if and only if $(a,b)\simeq(c,d)$.
Proof. Each Brauer class contains only one division algebra. This is because by Wedderburn's theorem, any finite-dimensional simple algebra over a field $k$ is isomorphic to $M_n(D)$ for some division algebra $D$, with $n$ unique and $D$ unique up to isomorphism. Moreover, for any division algebra $D$ over $k$, $D\otimes M_n(k)\simeq M_n(D)$.
We conclude by using the fact that each quaternion algebra over $k$ is either split (isomorphic to $M_2(k)$) or a division algebra.