Pardon my ignorance, but I've been stuck with this silly problem for almost a whole day while trying to learn some quadratic forms theory. Let $F$ be a field of characteristic $\neq 2$. Szymiczek's book on the algebraic theory of quadratic forms contains an exercise where the equivalence of the following are supposed to be proven:

1. $I^2(F)=0$.
2. $\langle 1,a,b,ab\rangle=0\in W(F)$ for all $a,b\in F^\times$.
3. $\langle 1,a,b,ab\rangle$ is hyperbolic.
4. $\langle 1,a,b,ab\rangle$ is isotropic.
5. $\langle 1,a\rangle$ is universal for all $a\in F^\times$.
6. $\langle a,b\rangle$ is universal for all $a,b\in F^\times$.

For some reason I seem to have got stuck on showing $(4)\Rightarrow (5)\Rightarrow (6)$, everything else being trivial or easy.

EDIT: I actually have $(5)\Rightarrow (6)$ now, so I'm looking for $(4)\Rightarrow (5)$.

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Here is one way to do it. Assume that (4) holds for all $a,b \in F^\times$, and now let $a \in F^\times$ be arbitrary. If $\langle 1,a \rangle$ is isotropic, then it is of course universal, so we may assume it is anisotropic, but then $\langle 1,a \rangle$ is the norm form $N$ of a separable quadratic extension $E/F$. So $\langle 1, a, b, ab \rangle = N \perp bN$ for all $b \in F^\times$.
Now let $t \in F^\times$ be arbitrary; then $N \perp (-t)N$ is isotropic by assumption, so there are elements $x,y \in E$, not both zero, such that $N(x) - tN(y) = 0$. But then $t = N(x y^{-1})$, so it is indeed represented by $\langle 1, a \rangle$. Since $t \in F^\times$ was arbitrary, this means that $\langle 1,a \rangle$ is universal.