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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 $v$.

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 $(a,b,c,d,e,f) = (9, 14, -19, 17, -18, 5).$

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 $(5)$?

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 $v$.

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 $v$.

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 $v$.

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 this 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 $v$.

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 $v$.