Hi,

for which primes $\displaystyle p$ are the quadratic forms

$\displaystyle 7x^2-y^2-5z^2$

and

$\displaystyle w^2+x^2+3y^2+11z^2$

isotropic over the field over $\displaystyle p$-adic numbers $\displaystyle \mathbb Q_p$ ?

A abbreviated form for the two quadratic forms above is $\displaystyle \left \langle 7,-1,5 \right \rangle$ and $\displaystyle \left \langle 1,1,3,11 \right \rangle$.

Define $\displaystyle \varepsilon (\left \langle a_1,...,a_n \right \rangle):=\prod_{1\leq i< j\leq n} (a_i,a_j) \in \left \{ \pm 1 \right \}$.

I can use the fact that a regular quadratic form $\displaystyle \phi$ over a field $\displaystyle F$ is isotropic if $\displaystyle dim(\phi)=3$ and

$\displaystyle \varepsilon(\phi)=(-1,-d)$.

[$\displaystyle (. , .)$ denotes the Hilbert symbol and $\displaystyle d:=disc(\phi)$ the discriminant of $\displaystyle \phi$.]

Moreover, a regular quadratic form $\displaystyle \phi$ over a field $\displaystyle F$ is isotropic if $\displaystyle dim(\phi)=4$ and

$\displaystyle d \neq 1$ or {$\displaystyle d=1$ and $\displaystyle \varepsilon(\phi)=(-1,-1)$}.

$\displaystyle d \neq 1$ means actually that the residue classes of $\displaystyle d$ and $\displaystyle -1$ are not equal modulo $\displaystyle F^*^2$.

Let's come to the first form:

$\displaystyle \varepsilon (\left \langle 7,-1,-5 \right \rangle)=(7,-1)(7,-5)(-1,-5)$

$\displaystyle d=disc(\left \langle 7,-1,-5 \right \rangle)=35 $, therefore $\displaystyle -d=-35$.

Now I have to check for which primes $\displaystyle p$

$\displaystyle (7,-1)(7,-5)(-1,-5) = (-1,-35)$ in $\displaystyle \mathbb Q_p$.

But I don't know how to solve it.

Please can you help me?

Thanks!

Regards

Alexander