Find all integers satisfying $m^2=n_1^2+n_1n_2+n_2^2$ I need to solve for $(n_1, n_2, m)$ (where $m, n_1, n_2$ are all integers - may be +ve or -ve) satisfying $m^2=n_1^2+n_1n_2+n_2^2$.
I have found 2 solutions so far - $(3, 5, 7)$ and $(7, 8, 13)$. What is a general solution?
UPDATE: For my purpose, it is further needed that $\gcd(n_1, m) = \gcd(n_2, m) = 1 $.

This equation is symmetric so many solutions.
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.
