If $(A,\sigma)_{F_Q}$ is hyperbolic, then $A$ contains $Q$ and $A[t]$ contains an element $y_1$ such that $y_1^2=at^2+b$.

Since $(A,\sigma)_{F_Q}$ is hyperbolic, the algebra $A_{F_Q}$ is not division, hence Merkurjev’s index reduction theorem [@Mer:92 Th. 1] shows that $A$ contains $Q$, hence also two elements $i$, $j$ such that $i^2=a$, $j^2=b$, and $ji=-ij$. Then $y_1=it+b$ satisfies $y_1^2=at^2+b$.

Thus, the algebra with involution $(A,\sigma)$ in Theorem \[totdec.thm\] below is such that $A(t)$ contains an element $y$ satisfying , and $A[t]$ contains elements $y_0$, $y_1$ satisfying $$\sigma(y_0)y_0=-(at^2+b),\qquad y_1^2=at^2+b,$$ but no element satisfying both equations.

For the rest of this section, we focus on the case where $\sigma$ is orthogonal. We then have a discriminant $\operatorname{disc}\sigma\in
F^{\times}/F^{\times2}$ and a Clifford algebra $C(A,\sigma)$, see [@KMRT §§7, 8]. Recall from [@KMRT (8.25)] that the center of $C(A,\sigma)$ is the quadratic étale $F$-algebra obtained by adjoining a square root of $\operatorname{disc}\sigma$. Therefore, if $\operatorname{disc}\sigma=1$ the algebra $C(A,\sigma)$ decomposes into a direct product of two components, which are central simple $F$-algebras, $$C(A,\sigma)\simeq C_+(A,\sigma)\times C_-(A,\sigma).$$ If $\deg A\equiv0\bmod4$, then the canonical involution $\underline{\sigma}$ on $C(A,\sigma)$ restricts to involutions $\sigma_+$ and $\sigma_-$ on $C_+(A,\sigma)$ and $C_-(A,\sigma)$. Moreover, the tensor product $C_+(A,\sigma)\otimes_FC_-(A,\sigma)$ is Brauer-equivalent to $A$, see [@KMRT (8.12), (9.14)].

\[orthog.prop\] 