<s>[INST] The elliptic curve for $x_1^9+x_2^9+\dots+x_6^9 = y_1^9+y_2^9+\dots+y_6^9$ I. Theorem: "If there are $a,b,c,d,e,f$ such that,
$$a+b+c = d+e+f\tag1$$
$$a^2+b^2+c^2 = d^2+e^2+f^2\tag2$$
$$3u^3-3uv+w=-def\tag3$$
where $u=a+b+c,\; v = ab+ac+bc,\;w = abc$, 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 $k=1,2,3,9$." 
A rational point implies that,
$$\small14(a^6+b^6+c^6-d^6-e^6-f^6)^2-9(a^4+b^4+c^4-d^4-e^4-f^4)(a^8+b^8+c^8-d^8-e^8-f^8) = t^2$$
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 $k = 1,2,3,9$, a rational point on a homogeneous cubic, and thus one can find an infinite more. (The first soln had $a,b,c,d,e,f = 9, 14, -19, 17, -18, 5$.)
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 algebra, let $c_1 = \tfrac{1}{8}(n^2+3),\; c_2 = \tfrac{1}{2}(n^3-9n)$, then one is to find rational $x$ such that,
$$Poly_1:= 7c_1x+c_2$$
$$Poly_2:= -7x^3-21c_1x+c_2$$
and,
$$D:=Poly_1 Poly_2 = \text{square}\tag5$$
a situation similar to the linked post for sixth powers. If a rational point $x$ can be found for some constant $n$, it is a simple matter to transform it into an elliptic curve. (The first soln used $n = 1/2$.)
Question: Is it possible to find a non-trivial polynomial solution to $(5)$ as a non-zero square $y$?
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 $v$.
 [/INST] 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
</s>