<s>[INST] Structure of $x^2 + xy + y^2 = z^2$ integer quadratic form The pythagorean triples $x^2 + y^2 = z^2$ can be solved in integers using rational parameterization of solutions to $x^2 + y^2 = 1$.
It goes through $(1,0)$, then consider the line $y = -k (x - 1)$ so that $x^2 + k^2(x-1)^2 = 1$
We get $(1+ k^2 )x^2 - 2k^2 x + k^2 = 1$ or $x =  \frac{k^2-1}{k^2+1}$ and $y=\frac{-2k}{k^2+1}$ and $z=1$
Then set $k= m/n$, for $(x,y,z) = (k^2-1,2k,k^2+1)=(m^2-n^2,2mn,m^2+n^2)$

What happens for 60-degree angle triangles $x^2+xy+y^2 = z^2$ ?  We could look for rational solutions to
$x^2 + xy + y^2 = 1 $ and $(1,0)$ works again.  Intersect with the line $y= k (x-1)$ ... 
get integer solutions: $(m^2-n^2, -m^2+2mn, m^2 - m n + n^2)$  A similar derivation was obtained earlier on math.StackExchange

In the case of pythagorean triple we can build new solutions $(m^2-n^2,2mn,m^2+n^2)$ from old using the maps
$$ (m,n) \mapsto (2m-n,m) \text{ or } (2m+n,m) \text{ or } (m+2n,n)$$
Can I find something similar to the Pythagorean triple tree for this quadratic form, $x^2 + xy + y^2 = z^2$?

 [/INST] Generally speaking, this equation has a lot of formulas for the solution. Because it is symmetrical.
Write the formula can someone come in handy. the equation:
$Y^2+aXY+X^2=Z^2$
Has a solution:
$X=as^2-2ps$
$Y=p^2-s^2$
$Z=p^2-aps+s^2$
more:
$X=(4a+3a^2)s^2-2(2+a)ps-p^2$
$Y=(a^3-8a-8)s^2+2(a^2-2)ps+ap^2$
$Z=(2a^3+a^2-8a-8)s^2+2(a^2-2)ps-p^2$
more:
$X=(a+4)p^2-2ps$
$Y=3p^2-4ps+s^2$
$Z=(2a+5)p^2-(a+4)ps+s^2$
more:
$X=8s^2-4ps$
$Y=p^2-(4-2a)ps+a(a-4)s^2$
$Z=-p^2+4ps+(a^2-8)s^2$
For the particular case: $Y^2+XY+X^2=Z^2$  You can draw more formulas.
$X=3s^2+2ps$
$Y=p^2+2ps$
$Z=p^2+3ps+3s^2$
more:
$X=3s^2+2ps-p^2$
$Y=p^2+2ps-3s^2$
$Z=p^2+3s^2$
In the equation:  $X^2+aXY+bY^2=Z^2$ there is always a solution and one of them is quite simple.
$X=s^2-bp^2$
$Y=ap^2+2ps$
$Z=bp^2+aps+s^2$
$p,s$ - integers asked us.
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