proof, suppose $\dim\varphi=5$ and $(C_0(\varphi),\tau_0)$ contains ${(Q,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})}$. The centralizer of ${(Q,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})}$ is a quaternion algebra with orthogonal involution $(Q',\rho)$ such that $$(C_0(\varphi),\tau_0)={(Q,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})}\otimes(Q',\rho).$$ Let $V\subset\operatorname{Sym}(C_0(\varphi),\tau_0)$ be the vector space of $\tau_0$-symmetric elements of trace $0$. The map $x\mapsto x^2$ defines a quadratic form $s\colon V\to F$ that is similar to $\varphi$ by the equivalence $\mathsf{B}_2\equiv\mathsf{C}_2$, see [@KMRT (15.16)]. Let $y\in Q'$ be a $\rho$-skew-symmetric unit and let $Q^0\subset Q$ be the vector space of pure quaternions. The restriction of $s$ to the subspace $Q^0\otimes y\subset V$ is similar to ${{\left\langle{1,-a,-b}\right\rangle}}$, hence the proof is complete.

\[qf.cor\] A $5$-dimensional quadratic form $\varphi$ over $F$ is $F_Q$-minimal if and only if the following conditions hold:

1.  $\varphi$ is similar to a Pfister neighbour of the $3$-fold Pfister form ${{\langle\!\langle{a,b,\lambda}\rangle\!\rangle}}$ for some $\lambda\in F^\times$, and

2.  $C_0(\varphi)\simeq M_2(Q')$ for some quaternion $F$-algebra $Q'$ such that $Q\otimes_FQ'$ is a division algebra.

Proposition \[qf.prop\] shows that $\varphi$ is $F_Q$-minimal if and only if $(C_0(\varphi),\tau_0)$ is hyperbolic but $(C_0(\varphi),\tau_0)$ does not contain ${(Q,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})}$. By Theorem \[deg4symp.thm\], this condition is equivalent to $$\label{eq:qf}
    (C_0(\varphi),\tau_0)\simeq\operatorname{Ad}_{{{\langle\!\langle{\lambda}\rangle\!\rangle}}}\otimes(Q',{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})$$ for some quaternion $F$-algebra $Q'=(a',b')_F$ and some $\lambda\in
  F^\times$ with $Q\otimes Q'$ a division algebra and ${{\langle\!\langle{a,b,\lambda}\rangle\!\rangle}}\simeq{{\langle\!\langle{a',b',\lambda}\rangle\!\rangle}}$. It follows from the isomorphism  that $\varphi$ 