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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?

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  • $\begingroup$ Are the signs correct? Just wondering since the same with $-2ab$ instead of $+2ab$ would look temptingly symmetric (and also happen to be reducible). $\endgroup$
    – Joachim König
    Commented Jan 15, 2023 at 10:37
  • $\begingroup$ @JoachimKönig I guess the signs are correct (only symmetry in $a,b$, whence the restriction $a\le b$) $\endgroup$
    – YCor
    Commented Jan 15, 2023 at 10:44

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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.

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  • $\begingroup$ At first, you take c=1. Then you consider substitution x=a+b, y=ab and obtain a curve of degree 6. How do you obtain the curve of the weierstrass form here lmfdb.org/EllipticCurve/Q/14/a/5 ? $\endgroup$
    – Fedor Nilov
    Commented Jan 16, 2023 at 11:35
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    $\begingroup$ @FedorNilov Magma has functions to compute Weierstrass forms of genus 1 curves (when given one rational point), and also for returning the "Cremona label" of the given elliptic curve; I was lazy and just followed the machine here - although at least a Weierstrass model could have been obtained by hand given enough time. $\endgroup$
    – Joachim König
    Commented Jan 16, 2023 at 14:08
  • $\begingroup$ Does Magma give an explicit change of variables formula (converting a curve to Weierstrass form)? $\endgroup$
    – Fedor Nilov
    Commented Jan 16, 2023 at 19:54
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    $\begingroup$ @FedorNilov The equation fulfilled by $x$ and $y$ already happens to be quadratic in $y$, namely $y^2+\frac{1}{4}(x-3)(x-1)(x+1)y-\frac{1}{16}(x-1)^4(x+1)^2/x = 0$. From here, it's really an easy exercise to get a Weierstrass model: complete the square and scale $y$ by some suitable expression in $x$ to get (with the thus shifted variable called $y'$): $(y')^2 = x(x+1)(x^2-3x+4)$; if you want to get to $Y^2=f(X)$ with $f$ cubic, replacing $x$ by $1/x$ (to switch $0$ and $\infty$), and doing some further ``cosmetics" gives, e.g., $Y^2=X(X^2-11X+32)$, with all coordinate changes easy to follow. $\endgroup$
    – Joachim König
    Commented Jan 17, 2023 at 1:53
  • $\begingroup$ By the way, I didn't see the reason for this during the computation, but after all is done, LMFDB database says that this elliptic curve is the modular curve $X_1(14)$, which (away from cusps) parameterizes elliptic curves with a $14$-torsion point. Since Mazur's theorem tells us there are no elliptic curves with rational $14$-torsion point, that's one reason why the rank turns out to be $0$. $\endgroup$
    – Joachim König
    Commented Jan 17, 2023 at 2:11
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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.

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    $\begingroup$ There are obvious integer solutions $(n,\pm n,0)$, $(n,0,\pm n)$, $(0,n,\pm n)$. $\endgroup$
    – YCor
    Commented Jan 15, 2023 at 10:31
  • $\begingroup$ Thus, if the curve is parametrizable, one can generate the solutions under the given inequality constraints $\endgroup$
    – Alm
    Commented Jan 15, 2023 at 10:38
  • $\begingroup$ It is a curve of genus 4, so will have only finitely many rational points. Very likely, $(0:\pm 1;1)$, $(\pm 1:0:1)$ and $(\pm 1:1:0)$ are the only ones (this would imply that the integer solutions listed by @YCor are indeed all). To show this may be fairly hard, however, but perhaps not impossible. $\endgroup$
    – Michael Stoll
    Commented Jan 15, 2023 at 11:50
  • $\begingroup$ Dividing by the involution swapping $a$ and $b$ gives an elliptic curve, which unfortunately has Mordell-Weil rank 1... $\endgroup$
    – Michael Stoll
    Commented Jan 15, 2023 at 11:53
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    $\begingroup$ The six rational points on the original curve are exactly its singularities. $\endgroup$
    – Michael Stoll
    Commented Jan 15, 2023 at 12:56

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