The elliptic curve for $x_1^9+x_2^9+\dots+x_6^9 = y_1^9+y_2^9+\dots+y_6^9$

Question

I. Theorem: "If there are $a,b,c,d,e,f$ such that,

$$a+b+c = d+e+f\tag1$$

$$a^2+b^2+c^2 = d^2+e^2+f^2\tag2$$

$$3u^3-3uv+w=-def\qquad\tag3$$

with $(u,v,w)$ as the symmetric polynomials $u=a+b+c,\; v = ab+ac+bc,\;w = abc$, then,

$$(a + u)^k + (b + u)^k + (c + u)^k + (d - u)^k + (e - u)^k + (f - u)^k = \\ (a - u)^k + (b - u)^k + (c - u)^k + (d + u)^k + (e + u)^k + (f + u)^k\tag4$$

for $k=1,2,3,9$."

The first solution to $\sum\limits^6 x_i^9 = \sum\limits^6 y_i^9$ was found by Lander in 1967. In a 2010 paper, Bremner and Delorme realized that it had the form of $(4)$, was good for $k = 1,2,3,9$, was a rational point on a homogeneous cubic and thus one can find an infinite more. The first solution had $(a,b,c,d,e,f) = (9, 14, -19, 17, -18, 5).$

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II. An alternative method is to directly solve $(1),(2)$ with simple identities such as,

$$(a,b,c) = (3 + 3 m + n - 3 m n + x,\; -6 m - 2 n + x,\; -3 + 3 m + n + 3 m n + x)$$ $$(d,e,f) = (3 - 3 m + n + 3 m n + x,\quad 6 m - 2 n + x,\,\; -3 - 3 m + n - 3 m n + x)$$

which incidentally also obeys,

$$-a+nb+c = -d+ne+f$$

Then substitute it into $(3)$, and end up only with a quadratic in $m$ whose discriminant $D$ must be made a square. After some minor algebra, then one is to find rational $x$ such that,

$$y^2 = (-7x^3-21c_1x+c_2)(7c_1x+c_2)\tag5$$

where $c_1=(n^2+3)/8,$ and $c_2=(n^3-9n)/2.$ The situation is similar to the linked post for sixth powers.

Question: Is it possible to find an $x$ that is a non-trivial polynomial solution to $(5)$?

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P.S. To clarify re comments, a polynomial solution would be $x,y$ as rational functions of $n$. Or $x,y,n$ as rational functions of a variable $v$.

Answer 1

For the equation, \begin{equation*} y^2=(-7x^3-21c_1x+c_2)(7c_1x+c_2) \end{equation*} where $c_1=(n^2+3)/8$ and $c_2=(n^3-9n)/2$, and assuming $n$ is rational, we have a rational point $(0,c_2)$, so the quartic is birationally equivalent to an elliptic curve.

Using an ancient Ms-Dos version of Derive, it is easy to use Mordell's method to find the curve \begin{equation*} v^2=u^3+49((n^2+3)u+4(3n^6-5n^4+145n^2+49))^2 \end{equation*} which has $2$ points of order $3$ at \begin{equation*} ( \, \, 0 \, , \, \pm \, 28(3n^6-5n^4+145n^2+49)\, \, ) \end{equation*} and numerical experimentation suggests that $\mathbb{Z}/3\mathbb{Z}$ is the torsion subgroup.

The reverse transformation is \begin{equation*} x=\frac{8n(n^2-9)u}{(252n^6-420n^4+7n^2(u+1740)+3(7u+v+1372))} \end{equation*}

Further numerical experimentation suggests that, for $1 \le n \le 19$, the curve has rank one for $n=6,7,8,9,12,15,16,17$, rank two for $n=18$ and rank zero otherwise - the curve is singular for $n=3$.

The presence of rank $0$ curves means (I think!!) that there cannot be a polynomial solution in $n$, valid for all $n$. It might be possible to find a parametric subset of n-values by investigating the curves with strictly positive rank. The problem is that the curves do not have a point of order $2$, so finding generators can take time.

Allan MacLeod