The equation,
$$((2 p - 13 q + 11 u)^k+(11 p - 2 u)^k+(11 q + 13 u)^k =\\( 2 p - 13 q - 11 u)^k+( 11 p + 2 u)^k+( 11 q - 13 u)^k\tag1$$
for $k=1$ is(Much revised for clarity.) I was considering the system of equations,
$$x_1+12\,x_2+x_3 = y_1+12\,y_2+y_3$$$$-a+nb+c = -d+ne+f\tag1$$ $$a+b+c = d+e+f\tag2$$ $$a^2+b^2+c^2 = d^2+e^2+f^2\tag3$$ $$a^6+b^6+c^6 = d^6+e^6+f^6\tag4$$
It is alsoQuestion 1. Is it true that, for a fixed integer $k=2,6$$n$, if the system has an infinite number of co-prime integer solutions,
$$225 p^3 + 458 p^2 q + 587 p q^2 + 1392 q^3 = -3 (75 p + 464 q) u^2\tag2$$ then $n$ is a multiple of $3$?
An initial point is $p,q = -104,17.$ HenceMethod: Eqs $(2)$ and $(3)$ can be easily turned into an elliptic curve, so there is an infinite number of integer solutions tobe given a complete solution. Incorporating $(1)$. In general, let,
$$\alpha = n+1\\ \beta = n-1$$
then I got,
$$(2 p - \alpha q +\beta u)^k + (\beta p - 2 u)^k+(\beta q + \alpha u)^k =\\ (2 p - \alpha q -\beta u)^k + (\beta p + 2 u)^k+(\beta q -\alpha u)^k\tag3$$$$(-2 p + \alpha q -\beta u)^k + (\beta p - 2 u)^k+(\beta q + \alpha u)^k =\\ (-2 p + \alpha q +\beta u)^k + (\beta p + 2 u)^k+(\beta q -\alpha u)^k\tag5$$
where $\alpha = n+1,\;\beta = n-1$. It is also true for $k=2,6$$k=6$ if there is $p,q,n$ such that,
$$Poly_1:= (-3+n)(5-2n+n^2)p + 4n(1+n^2)q$$
$$Poly_2:= (-3+n)(5-2n+n^2)p^3 + 2(5+11n-5n^2+n^3)p^2q - (5+7n+15n^2-3n^3)pq^2 + 4n(1+n^2)q^3$$
and,
$$\color{red}{-}Poly_1 Poly_2 = \text{square}\tag4$$$$\color{red}{-}Poly_1 Poly_2 = \text{square}\tag6$$
A trivial solution is $q = \frac{(3-n)p}{2n}$ which yields,
$$\color{red}{-}Poly_1 Poly_2 = \frac{(-9+n^2)^2(-1+n)^4p^4}{4n^2}\tag5$$$$\color{red}{-}Poly_1 Poly_2 = \frac{(-9+n^2)^2(-1+n)^4p^4}{4n^2}$$
P.S.Example Note that eqn: Let $(3)$$n=12$, then,
$$((-2 p + 13 q - 11 u)^k+(11 p - 2 u)^k+(11 q + 13 u)^k =\\(-2 p + 13 q + 11 u)^k+( 11 p + 2 u)^k+( 11 q - 13 u)^k\tag7$$
which is already true for $k=1,2$. But it is also obeysfor $k=6$ if,
$$x_1+nx_2+x_3 = y_1+ny_2+y_3$$$$225 p^3 + 458 p^2 q + 587 p q^2 + 1392 q^3 = -3 (75 p + 464 q) u^2\tag8$$
where the example was just the caseAn initial point is $n=12$$p,q = -104,17.$ Hence $(8)$ can be easily turned into an elliptic curve, so there is an infinite number of integer solutions to $(7)$.
Questions:Question 2. For what other positive integer $n$ below a bound can we find solutions to non-zero $(6)$ or with $(3-n)p-2nq \neq 0$? (The constraint is to prevent trivial solutions. I have found $n=12, 15, 21, 30, 33, 135$ but I am not sure if this is exhaustive for $n<150$.)
- For what other positive integer $n$ below a bound can we find solutions to non-zero $(4)$ or with $(3-n)p-2nq \neq 0$? (The constraint is to prevent trivial solutions. I have found $n=12, 15, 21, 30, 33, 135$ but I am not sure if this is exhaustive for $n<150$.)
- Excepting $n=1$, is it true that all other $n$ are multiples of $3$?