TITLE: Equivalence of quadratic forms over p-adic fields.
QUESTION [5 upvotes]: There is a theorem that states that two quadratic forms over $\mathbb{Q}_p$ are equivalent iff they have the same rank, discriminant and the same $\epsilon$ invariant.
(The last is defined as follows: if $f = \sum_{i=1}^n a_ix_i^2$ then $\epsilon(f) = \prod_{1\le i < j \le n} (a_i,a_j)$, where $(\cdot,\cdot)$ is the Hilbert symbol over $\mathbb{Q}_p$.)
The proof (from Serre's Arithmetic) is based on a previous theorem, which relates these invariants to the elements the forms represent. 
However, I do not know how I can show that two particular forms are equivalent (with the same invariants of course). So fix some prime $p$ and consider the following two forms over $\mathbb{Q}_p$:
\begin{equation}
f(x) = x_1^2 + x_2^2 + x_3^2 + x_4^2 - (x_5^2 + x_6^2 + x_7^2) \nonumber 
\end{equation}
and
\begin{equation}
g(x) = -(x_1^2 + x_2^2 + x_3^2 + x_4^2 + x_5^2 + x_6^2 + x_7^2) \nonumber 
\end{equation}
How can I show in particular that these two are equivalent (without using the classification theorem above)? Thanks.

REPLY [4 votes]: Since this is only about $p$-adics, an elementary argument that comes to mind would be as follows (unless I totally missed something).
If $p\equiv 1\pmod4$, then $-1=a^2$ for some $a\in\mathbf{Z}_p^*$. Therefore $x^2$ and $(ax)^2=-x^2$ are equivalent.
If $p\equiv -1\pmod4$, then $-1=a^2+b^2$ for some integers $a,b\in\mathbf{Z}_p^*$. In this case $x^2+y^2$ is equivalent to
$$
(ax+by)^2+(bx-ay)^2=(a^2+b^2)(x^2+y^2)=-(x^2+y^2).
$$
Two or four such equivalences convert your $f$ to $g$.
And let's not forget the prime $p=2$. Taking norms of the product of quaternions
$$
(2+i+j+k)(x_1+x_2i+x_3j+x_4k)
$$
tells us that
$$
\begin{aligned}
7(x_1^2+x_2^2+x_3^2+x_4^2)&=(2x_1-x_2-x_3-x_4)^2+(x_1+2x_2-x_3+x_4)^2\\
&+(x_1+x_2+2x_3-x_4)^2+(x_1-x_2+x_3+2x_4)^2.
\end{aligned}
$$
Combining this with the fact that $\sqrt{-7}\in\mathbf{Q}_2$ allows you to turn the sum of four squares to its negative by an equivalence transformation.