**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](http://www.ams.org/journals/mcom/2010-79-269/S0025-5718-09-02288-1/S0025-5718-09-02288-1.pdf), 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](https://mathoverflow.net/q/222719/12905),


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


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 $v$.



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


