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  1. $q(x, y, z) = 5x^2-y^2-11z^2$
  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?

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  • $\begingroup$ if you are correct, there are triples of integer $x,y,z$ (not all zero) for each one, that cause each form to become zero. There is a theorem that says that these integers can be found fairly small, if there are any $\endgroup$ – Will Jagy Dec 12 '18 at 1:21
  • $\begingroup$ I don't want to find them, I just want to show they are isotropic. $\endgroup$ – kmini Dec 12 '18 at 1:28
  • $\begingroup$ What book is this from? The method, I mean. $\endgroup$ – Will Jagy Dec 12 '18 at 1:31
  • $\begingroup$ Alright, this is Theorem 4.1, page 80 in Cassels, just part (i). On page 82 he points out that conditions (ii) and (iii) can be ignored if we know the form is indefinite $\endgroup$ – Will Jagy Dec 12 '18 at 1:34
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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, $$ x^2 + y^2 + z^2 = 1469 \left( u^2 + uv + v^2 \right)^2 $$

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