- $q(x, y, z) = 5x^2-y^2-11z^2$
- $q(x, y, z) = 3x^2-y^2+22z^2$
So for part (1), I apply Hasse-Minkowski and check over the $\mathbb{Q}_p$. It's trivial to see for $\mathbb{R}$. We only need to check for $\mathbb{Q}_{13}$ and $\mathbb{Q}_5$ right?. Now mod 5, the equation becomes $-y^2-z^2 = 0 \mod 5$ and thus there is a solution since -1 is square mod 5. Checking mod 11, this becomes $5x^2-y^2 = 0 \mod 11$. Now enough to check if 5 is square mod 11. But it's clear that it is by Legendre. Thus there is a solution and the form isotropic over $\mathbb{Q}$.
Part(2) is similar. We check mod 3, $-y^2+z^2 = 0 \mod 3$ and clearly there is a solution. For mod 11, we have that $3x^2-y^2 = 0 \mod 11$. Now 3 is a square mod 11 thus there is a solution. Thus it's isotropic over $\mathbb{Q}$.
Note that for other $p$, the coefficients are units thus must be isotropic. Am I right?
next day: those recipes below give all primitive. Two small solutions to $5x^2 = y^2 + 11 z^2$ are $(2,3,1)$ and $(3,1,2).$ Two small solutions to $y^2 = 3x^2 + 22 z^2$ are $(1,5,1)$ and $(3,7,1).$
We get infinitely many primitive solutions for the first one taking $$ x = 5 u^2 + 6 uv + 4 v^2 \; \; , \; \; y = 9 u^2 + 2 uv -6 v^2 \; \; , \; \; z = 2 u^2 + 6 uv + 2 v^2 \; \; , \; \; $$ and the genuinely different (even considering absolute values) $$ x = 6 u^2 + 2 uv + 2 v^2 \; \; , \; \; y = 13 u^2 + 8 uv -3 v^2 \; \; , \; \; z = u^2 -4 uv - v^2 \; \; . \; \; $$
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Second one
$$ x = 3 u^2 + 8 uv -2 v^2 \; \; , \; \; y = 7 u^2 + 4 uv + 10 v^2 \; \; , \; \; z = u^2 - 2 uv - 2 v^2 \; \; , \; \; $$ and the different $$ x = u^2 + 8 uv -6 v^2 \; \; , \; \; y = 5 u^2 - 4 uv + 14 v^2 \; \; , \; \; z = u^2 - 2 uv - 2 v^2 \; \; . \; \; $$
In both cases you may take absolute values of $x,y,z$ as there are no mixed terms.
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The example I like to show is solving $$ 2(x^2 + y^2 + z^2) - 113(yz + zx + xy)=0, $$ four "recipes," $$ \left( \begin{array}{r} x \\ y \\ z \end{array} \right) = \left( \begin{array}{r} 37 u^2 + 51 uv + 8 v^2 \\ 8 u^2 -35 uv -6 v^2 \\ -6 u^2 + 23 uv + 37 v^2 \end{array} \right) $$
$$ \left( \begin{array}{r} x \\ y \\ z \end{array} \right) = \left( \begin{array}{r} 32 u^2 + 61 uv + 18 v^2 \\ 18 u^2 -25 uv -11 v^2 \\ -11 u^2 + 3 uv + 32 v^2 \end{array} \right) $$
$$ \left( \begin{array}{r} x \\ y \\ z \end{array} \right) = \left( \begin{array}{r} 38 u^2 + 45 uv + 4 v^2 \\ 4 u^2 -37 uv -3 v^2 \\ -3 u^2 + 31 uv + 38 v^2 \end{array} \right) $$
$$ \left( \begin{array}{r} x \\ y \\ z \end{array} \right) = \left( \begin{array}{r} 29 u^2 + 63 uv + 22 v^2 \\ 22 u^2 -19 uv -12 v^2 \\ -12 u^2 -5 uv + 29 v^2 \end{array} \right) $$
For all four recipes, $$ x^2 + y^2 + z^2 = 1469 \left( u^2 + uv + v^2 \right)^2 $$