Equivalent conditions on quadratic forms over a field This is a question (Q.10 in chapter one) from Lam's "Introduction to quadratic forms over fields"
Question: Let $\mathbb F$ be a field, where $\operatorname{char}\mathbb F\neq 2$. Show that the following are equivalent: 


*

*Every four dimensional quadratic space $(V,q)$ with $\det(V)=-1$ is isotropic.

*Every even dimensional quadratic space $(V,q)$ with $\det(V)=-1$ is isotropic.

*Every three dimensional quadratic space $(V, q)$ represents its determinant. 

*Every odd dimensional quadratic space $(V, q)$ represents its determinant. 


(We say that $(V, q)$ represents it's determinant if there is $v\in V$ so that $q(v) = \det V$)
Obviously $(4)\Rightarrow (3)$ and $(2)\Rightarrow (1)$ and I've shown $(1)\Leftrightarrow (3)$ and $(2)\Leftrightarrow (4)$. Thus it suffices to show $(1)\Rightarrow (2)$ or $(3) \Rightarrow (4)$. But no progress so far. I have the feeling that it might be easier to do the induction 
$$m \to m+1 \to m+2,$$
that is, do not show $(1)\Rightarrow (2)$ or $(3) \Rightarrow (4)$ directly and instead showing it in two steps. To this end I showed the following equivalence (for each fixed $m$): 


*Every $2m$-dimensional quadratic space $(V, q)$ with $\det V = -1$ are isotrpoic,

*Every $2m-1$-dimensional quadratic space $(V, q)$ represents its determinant. 


But this is not good enough since I still need the implication from even dimensional $2m$ to odd dimensional $2m+1$ (jumping up one dimension, instead of lowering)
Any hint are welcome. 
 A: I figured out the answer. One can prove $(1)\Rightarrow (2)$ by induction on $r = \dim(V)/2$. The base case is $r=2$ and is just $(1)$. For the induction step, let $(V, q)$ be a quadratic space with $\dim(V) = 2r+2$ and $d(V)  =-1$. Then we have $q = \langle a_1, \cdots, a_{2r}\rangle$ for some $a_i \in \mathbb F$. Let $A = a_1 \cdots a_{2r-1}$ and $h = \langle A, -A\rangle$. Then 
$$\begin{split} 
q\oplus h &= \langle a_1, \cdots a_{2r+2}\rangle \oplus \langle A, -A\rangle \\
&\cong \langle a_1, \cdots, a_{2r-1}, -A\rangle \oplus \langle A, a_{2r}, a_{2r+1}, a_{2r+2}\rangle \\
&=: q_1 \oplus q_2. 
\end{split}$$ 
Note that both $q_1, q_2$ have $d(q_i) = -1$ (note that determinant is defined only mod $(\mathbb F^*)^2$). Thus first by $(1)$, In the quadratic space $(V_2, q_2)$ there is $v_2\in V_2$ so that $q_2(v_2)=0$. Then there is a hyperbolic space $h$ in $V_2$. Since $h$ is nondegenerate, $V_2 \cong h \oplus Z$, where $Z = h^\perp$. Thus 
$$ q\oplus h \cong q_1 \oplus h \oplus Z.$$
By Witt's Cancellation Theorem we have 
$$q \cong q_1 \oplus Z.$$
Second, by the induction hypothesis, since $d(q_1)=-1$, then $q_1$ is isotropic. Thus $q \cong q_1 \oplus Z$ is also isotropic. That finishes the proof.  
