hyperbolic;

2.  $A(t)$ contains an element $y$ satisfying $$\label{y.eq} 
        \sigma(y)=-y\qquad\text{and}\qquad y^2=at^2+b.$$

If conditions *(a)* and *(b)* hold, then $A[t]$ contains an element $y_0$ such that $$\label{y0.eq}
  \sigma(y_0)y_0=-(at^2+b).$$ Moreover, the following conditions are equivalent:

1.  $(A,\sigma)$ contains ${(Q,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})}$;

2.  $A[t]$ contains an element $y$ satisfying .

The equivalence of (a) and (b) readily follows from the description of anisotropic involutions that become hyperbolic over a quadratic extension in [@BST 3.3]. If $y\in A(t)$ satisfies , then the version of the Cassels–Pfister theorem for algebras with involution in [@T:96] yields an element $u\in A(t)$ such that $\sigma(u)u=1$ and $uy\in A[t]$. Then $y_0=uy$ satisfies .

If (a’) holds, then $A$ contains two skew-symmetric elements $i$, $j$ such that $i^2=a$, $j^2=b$, and $ji=-ij$. Then $y=it+j\in A[t]$ satisfies , so (b’) holds. Conversely, suppose (b’) holds and let $y\in A[t]$ satisfy . We then have $\sigma(y)y=-(at^2+b)$, hence the degree of $y$ is $1$ since $\sigma$ is anisotropic. Thus, $y=\lambda t+\mu$ for some $\lambda$, $\mu\in A$. It follows from that $\lambda$ and $\mu$ are skew-symmetric and satisfy $\lambda^2=a$, $\mu^2=b$, and $\mu\lambda=-\lambda\mu$, hence they generate a $\sigma$-stable subalgebra of $(A,\sigma)$ isomorphic to ${(Q,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})}$.

\[division.prop\] Let $(A,\sigma)$ be a central simple $F$-algebra with involution. Assume $A$ is division. 