The Diophantine equation,

$$x_1^k+x_2^k+x_3^k = y_1^k+y_2^k+y_3^k\tag1$$ 

for either $k=5$ or $6$ is quite well explored, and it has long been known that it has an infinite number of **primitive** solutions (as points on an *elliptic curve*, or an infinite family of polynomials,). However,

$$x_1^k+x_2^k+x_3^k+x_4^k = y_1^k+y_2^k+y_3^k+y_4^k\tag2$$

for either $k=7$ or $8$, results seem to be harder to find. Only ***one*** solution found back in 2006 is known for $k=8$. Many solutions exist when $k=7$, but it is not known if there are infinite many. Choudhry [reduced][1] $(2)$ when it is true for $k=1,3,7$ plus a fourth constraint (re Wolfgang's comment below), namely $\sum\limits^4 x_i = 0$ (and since $k=1$, then also $\sum\limits^4 y_i=0$ ), to a multi-variable *cubic equation* by passing through the simultaneous equations, in Choudhry's notation,

$$\begin{aligned}
X_1X_2X_3\,&=Y_1Y_2Y_3\\
2(X_1^4 + X_2^4 + X_3^4) - 5(X_1^2 + X_2^2 + X_3^2)^2\, &= 2(Y_1^4 + Y_2^4 + Y_3^4) - 5(Y_1^2 + Y_2^2 + Y_3^2)^2
\end{aligned}\tag3$$

Using that cubic, Choudhry (3) and Wroblewski (21) found a total of $3+21=24$ solutions to $(2)$ valid for $k=1,3,7$. Results can be found in [euler.free.fr][2]. Using the $x_i,y_i$, I derived their $X_i, Y_i$. Surprisingly, seven had a simple linear constraint, namely $X_3=2Y_3$. Let,

$$X_1,\,X_2,\,X_3 =u_1 u_2,\;u_3 u_4,\;2t$$
$$Y_1,\,Y_2,\,Y_3 =2u_1 u_4,\;u_2 u_3,\;t$$

then solutions are,

$$\begin{array}{|c|c|c|c|c|c|}
\text{#}&u_1&u_2&u_3&u_4&t\\
1&227 &27 &113 &13 &1305\\
2&431 &187 &365 &49 &6929\\
3&127 &139 &313 &1 &23647\\
4&303 &304 &338 &37 &24616\\
5&871 &163 &4364 &127 &254405 \\
6&439 &2459 &247 &1175 &261851 \\
7&14737 &139 &15899 &25 &544069 \\
\end{array}$$

It was hard to find commonalities for the other seventeen solutions.

**Questions:**

1. Can the simultaneous eqns $(3)$, or the cubic described by Choudhry in the paper, be reduced to an *elliptic curve*? 
2. Why does almost $1/3$ of the $24$ known solutions have $X_3=2Y_3$? Translated into the addends $x_i,y_i$ of $(2)$, this is equivalent to the **5th** constraint that $x_1+x_2 = 2y_1+2y_2$. *Is there some identity behind it?* For example, I found this 7th deg multi-grade,$$1 + 5^k + (3+2y)^k + (3-2y)^k + (-3+3y)^k + (-3-3y)^k = \\(-2+x)^k + (-2-x)^k + (5-y)^k + (5+y)^k$$ for $k = 1,3,5,7,\;$ if $x^2-10y^2 = 9$, guaranteeing an infinite supply with the constraint $x_2 = 5x_1$, though I doubt anything for $(2)$ would be as simple as this.

**Note:**

However, the fifth degree version,

$$x_1^5+x_2^5+x_3^5 = y_1^5+y_2^5+y_3^5$$

$$x_1+x_2+x_3 = y_1+y_2+y_3=0$$

***does*** have a polynomial identity behind it, found by Choudhry and Wrobleski. (See *"A quintic Diophantine equation with applications to two Diophantine systems concerning fifth powers"*)

**P.S.** Eq.1 is briefly discussed in this [MO post][3].


  [1]: http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.101.2751
  [2]: http://euler.free.fr/database.txt
  [3]: https://mathoverflow.net/questions/150428/