Integer solutions of an algebraic equation I'm trying to find integer solutions $(a,b,c)$ of the following algebraic equation with additional conditions $b>a>0$, $c>0$.
$(-a^2+b^2+c^2)(a^2-b^2+c^2)(a^2+b^2-c^2) + 2 a b (-a^2+b^2+c^2)(a^2-b^2+c^2) - 2 b c (a^2-b^2+c^2)(a^2+b^2-c^2) - 2 a c (-a^2+b^2+c^2) (a^2+b^2-c^2) = 0$
Is it possible to do using some math packages?
 A: There are no points other than the trivial points $(n, \pm n, 0)$, $(n,0,\pm n)$ and $(0,n,\pm n)$, and in particular, no integer points with all coordinates positive. For example, taking the affine part $f(\alpha,\beta, 1)$ of the above curve $f(a,b,c)$ and then setting $x:=\alpha+\beta$, $y:=\alpha\beta$, one verifies that the curve defined by $x$ and $y$ (of which the original (genus 4) curve is a cover) is of genus $1$, i.e., elliptic, and has (finite) Mordell-Weil group $\mathbb{Z}/6\mathbb{Z}$ (I asked Magma for this, returning this curve: http://www.lmfdb.org/EllipticCurve/Q/14/a/5  - although there may well be a more direct argument). These are all either points at infinity or having $y(=\alpha\beta)=0$, i.e., already this subcover of the original curve has no ``good" points.
A: The equation is homogeneous in 3 variables, thus it is associated with a plane curve. First, I would check if the curve has genus less than 2. If the genus is 0 or 1, the curve is parametrizable or elliptic, respectively. In particular, for parametrizable curves, you can generate integer solutions, provided that at least one integer solution exists and you know it.
