(*Much revised for clarity*.) I was considering the system of equations,

$$-a+nb+c = -d+ne+f\tag1$$ $$a+b+c = d+e+f\tag2$$ $$a^2+b^2+c^2 = d^2+e^2+f^2\tag3$$ $$a^6+b^6+c^6 = d^6+e^6+f^6\tag4$$

**Question 1.** Is it true that, for a fixed integer $n$, if the system has an infinite number of co-prime integer solutions, *then $n$ is a multiple of $3$?*

**Method**: Eqs $(2)$ and $(3)$ can easily be given a complete solution. Incorporating $(1)$, I got,

$$(-2 p + \alpha q -\beta u)^k + (\beta p - 2 u)^k+(\beta q + \alpha u)^k =\\ (-2 p + \alpha q +\beta u)^k + (\beta p + 2 u)^k+(\beta q -\alpha u)^k\tag5$$

where $\alpha = n+1,\;\beta = n-1$. It is also true for $k=6$ 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,

$$\color{red}{-}Poly_1 Poly_2 = \text{square}\tag6$$

A trivial solution is $q = \frac{(3-n)p}{2n}$ which yields,

$$\color{red}{-}Poly_1 Poly_2 = \frac{(-9+n^2)^2(-1+n)^4p^4}{4n^2}$$

**Example**: Let $n=12$, 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 for $k=6$ if,

$$225 p^3 + 458 p^2 q + 587 p q^2 + 1392 q^3 = -3 (75 p + 464 q) u^2\tag8$$

An initial point is $p,q = -104,17.$ Hence $(8)$ can be easily turned into an ** elliptic curve**, so there is an infinite number of integer solutions to $(7)$.

**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 $n<150$.)

obey $(1)$, but instead focuses on the constraint, $$a+nb+c=d+ne+f\tag3$$ and for what integer $n$ are permissible. $\endgroup$ – Tito Piezas III Nov 6 '15 at 1:56not