Can I show the diagonal matrix (1,1,1) and (1,-1,-1) are equivalent over the finite field $\mathbb{F}_3$
Can I show the quadratic forms $x^2 + y^2 + z^2$ and $x^2 - y^2 - z^2$ are equivalent over the finite field $\mathbb{F}_3$
One way to phrase this is that $2\times 2$ matrices and Hamilton quaternions over $\mathbb{F}_3$ are the same as "quadratic spaces"
$$ \left[ \begin{array}{cc} a + b & c + d \\ -c +d & a-b \end{array} \right] \leftrightarrow \left[ \begin{array}{cc} a + b i& c + d i\\ -c +d i & a-bi \end{array}\right] $$
I am trying to do this as elementary algebra and write down the change of basis that turns $x^2 + y^2 + z^2 + w^2$ into $x^2 + y^2 - z^2 - w^2$
These two are certainly equivalent over $\mathbb{C}$ since $\sqrt{-1} \in \mathbb{C}$ and $\sqrt{-1} \in \mathbb{Q}_p$ where $p = 4k+1$, e.g. $\sqrt{-1} \in \mathbb{Q}_5$ but not $\mathbb{Q}_7$
It's not the case that (1,1) is similar to (-1,-1) I don't think
Serre "Course on Arithmetic" says only two quadratic forms up to equivalence. So $x^2+y^2$ and $-x^2-y^2$ are similar.
After a while I came up with:
$$ (x +y)^2+(x -y)^2= 2x^2+2y^2=-x^2-y^2$$
So it is possible to turn this quadratic form into it's negative using invertible linear transformation. Am I mistaken to phrase this as, the two diagonal matrices (1,1) and (1,-1) are similar over the finite field with three elements $\mathbb{F}_3$?
If $p =4k +1$ then $\sqrt{-1}\in \mathbb{F}_p $ such as $2^2=-1$.
However I could not solve the next cases $p =7, 11,19,23,31,...$
Every nonsingular quadratic form $Q$ of dimension $\geq2$ over a finite field $\mathbf F_q$ of characteristic $\neq2$ is equivalent to a diagonal quadratic form of the form $$ \langle 1,\ldots,1,d\rangle, $$ where $d\in \mathbf F_q^\star/\mathbf F_q^{\star2}$ is the discriminant of $Q$, i.e., the determinant of its associated bilinear form (cf. O'Meara: Introduction to quadratic forms, p. 157). This is in fact not so difficult to prove. It follows from the statement that any nonsingular quadratic form of dimension $\geq2$ represents all elements of $\mathbf F_q^\star$, by splitting off quadratic forms of the form $\langle 1\rangle$ until you are left with $\langle d\rangle$. Anyway, since both of your quadratic forms are regular, of dimension $3$ and have discriminant equal to $1$, they are equivalent over any finite field $\mathbf F_q$ of characteristic different from $2$.