Every element of $K$ is a square if and only if every 2-dimensional form over $K$ is isotropic. In fact, if $\langle -1, d \rangle$ is isotropic, then $\langle - 1,d \rangle \cong \langle -1, 1\rangle$ and d is a square.
I know that if every element of $K$ is square Then two regular quadratic spaces over K are isometric if and only if they have the same dimension hence $\langle- 1,d\rangle \cong \langle-1, 1\rangle$ and d is square since it is discriminant of quadratic space which is define as $(-1)^n$ det($\phi$) where $\dim(\phi)=2n$ or $\dim(\phi)=2n+1$.
How to prove the above cases?