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Most of this appears in my answers to Isotropic ternary forms

Most of this appears in my answers to Isotropic ternary forms

For all four recipes, $$ x^2 + y^2 + z^2 = 1469 \left( u^2 + uv + v^2 \right)^2 $$ Since $u^2 + uv + v^2 \geq 3 u^2 / 4$ and $u^2 + uv + v^2 \geq 3 v^2 / 4,$ this gives us explicit bounds on the absolute values of $u,v$ that gives us all (primitive) solutions of $ 2(x^2 + y^2 + z^2) = 113 (yz+zx+xy) $ with the absolute values of $x,y,z$ up to a desired bound.

Indeed, we were able to choose all four coefficient matrices with this pattern: $$ R = \left( \begin{array}{ccc} \alpha & \beta & \gamma \\ \gamma & - \beta + 2 \gamma & \alpha - \beta + \gamma \\ \alpha - \beta + \gamma & 2 \alpha - \beta & \alpha \end{array} \right) $$

The rows constitute a cycle of three neighboring, but not reduced, binary quadratic forms under the action of the matrix $$ P = \left( \begin{array}{rr} 0 & 1 \\ -1 & -1 \end{array} \right), $$ where $P^3 = I.$ As soon as we write $X = RU$ we get the identity $$ x^2 + y^2 + z^2 = \left( \alpha^2 + (\alpha - \beta + \gamma)^2 + \gamma^2 \right) \cdot \left( u^2 + uv + v^2 \right)^2 $$