Witt index of the sum of 24 squares Consider the quadratic forms $q(x)=x_1^2+\dots +x_8^2$ in 8 variables and $p(x)=x_1^2+\dots+x_{24}^2$ in 24 variables over the field of rational numbers $\mathbb{Q}$. Let $E/\mathbb{Q}$ be a field extension. Is it true that if $q_E$ is anisotropic over $E$, then the Witt inddex of $p_E$ over $E$ is less or equal 8?
 A: The answer is YES. The quadratic form $q$ from the question is a Pfister form, and 
$p=q\oplus q\oplus q$.
Therefore, the affirmative answer follows from the next lemma.


Lemma.
    Let $q$ be an anisotropic Pfister quadratic form  of dimension $m=2^n$ over a field $k$
    of characteristic  different from $2$.
    Let $a,b,c\in k^\times$.
    Consider the quadratic form $p:=aq\oplus bq\oplus cq$ of dimension $3m$.
    Then  the Witt index $i(p)\le m$.


Proof.
We may assume that $a=1$.
Consider the quadratic form $s:=q\oplus bq$ of dimension $2m$, 
it is again a Pfister form,
and by Pfister's theorem it is either anisotropic or  (completely) split.
If $s$ is split, then $p=s\oplus cq$ is a direct sum of the  split form $s$ 
and the anisotropic form $cq$, and by Witt's decomposition theorem $i(p)=i(s)=m$.
If $s$ is anisotropic, then $p\oplus (-cq)=s\oplus cq\oplus(-cq)$ 
is a direct sum of the anisotropic form $s$
and the split form $cq\oplus(-cq)$ of dimension $2m$ and of Witt index $m$. 
By Witt's decomposition theorem  $i(p\oplus(-cq))=i(cq\oplus(-cq))=m$,
and therefore, $i(p)\le m$.
A: The answer seems to be affirmative.
It is well known that if the level of a number field is finite then it belongs to $\{1,2,4\}$.
Now in your situation, as $q_E$ is anisotropic, $E$ should be formally real. So $p_E$ is anisotropic as well, so its Witt index is $0$.
