I am following Theorem 2.3.1 of Maclachlan's and Reid's The Arithmetic of Hyperbolic 3-Manifolds. We define a quaternion algebra $A=\left(\frac{a,b}{F}\right)$ over a field $F$ of characteristic $\neq 2$ by the vector space spanned by $\{1,i,j,k\}$ with $i^2=a,\,j^2=b,$ and $ij=-ji=k$ (of course letting $a,b\in F^\ast$). It is earlier established in this book that $$ \left(\frac{a,b}{F}\right)\cong\left(\frac{ax^2,by^2}{F}\right) $$ for any $x,y\in F^\ast$. Further, it is established that $M_2(F)\cong\left(\frac{1,1}{F}\right)$.
Keep the definition of $A$ from above. What I am attempting to show is that if there exist $x,y\in F^\ast$ such that $ax^2+by^2=1$, then $A\cong M_2(F)$. I suspect this is trivial but I'm a bit stuck. Thanks for your help.
Edit: to add a bit of what I've done, it's clear that we have $$ \frac{1}{a}=\frac{x^2}{1-by^2}. $$ If we can prove the RHS is a square, then we're good.
Edit 2: Equivalently, it could be proven that the existence of those $x,y$ imply that $A$ is not a division algebra, and I actually figured that out. Let $\alpha\in A$ such that $\alpha=1+xi+yj$. Then $\alpha\overline\alpha=1-ax^2-by^2=0$, and clearly $\alpha,\overline\alpha\neq 0$. I'm still interested in how the first assertion might be proven, since it is definitely true.
Using the equation $ax^2+by^2=1$, you can deduce the existence of an orthonormal basis with respect to that bilinear form:
$$ v=\left(\frac{x}{1-by^2},0\right)\\ w=\left(0,\frac{y}{by^2}\right)\\ $$
The transformation of the metric space $(V,B)\to (V,\cdot)$ using the matrix $T=\begin{bmatrix}\frac{1-by^2}{x}&0\\0&\frac{by^2}{y}\end{bmatrix}$ maps $v,w$ onto $(1,0), (0,1)$, and moreover the bilinear forms match:
$$ Tv\cdot Tw=B(v,w) $$
So, they actually match for all pairs of vectors, and the associated Clifford algebras are isomorphic.