**I. Fifth Powers** The Diophantine equation, $$x_1^k+x_2^k+x_3^k = y_1^k+y_2^k+y_3^k\tag1$$ for $k=5$ is quite well-explored. It has an infinite number of **primitive** solutions (as points on an *elliptic curve*, or an infinite family of polynomials). They can be simultaneously true for $k=1,5$ and in [a 2013 paper][1], Choudhry and Wroblewki found an infinite subset that satisfy the side condition, $$\sum\limits^3 x_i = \sum\limits^3 y_i = 0$$ --- **II. Seventh Powers** Since the first solution to, $$x_1^k+x_2^k+x_3^k+x_4^k = y_1^k+y_2^k+y_3^k+y_4^k\tag2$$ for $k=7$ was found almost 30 years ago by Randy Ekl in 1996, only $113$ primitive solutions (in positive and negative integers) have been found so far. In [a 2000 paper][2], Choudhry found solutions **simultaneously** true for $k = 1,3,7$ and satisfy the analogous side condition, $$\sum\limits^4 x_i = \sum\limits^4 y_i = 0$$ Wrobrewski would later find more solutions, for a total of 24. They used the form, $$(X_1-X_2-X_3)^k+(-X_1+X_2-X_3)^k+(-X_1-X_2+X_3)^k+(X_1+X_2+X_3)^k = (Y_1-Y_2-Y_3)^k+(-Y_1+Y_2-Y_3)^k+(-Y_1-Y_2+Y_3)^k+(Y_1+Y_2+Y_3)^k$$ which is identically true for $k=1$ and is true for $k=3,7$ if the two conditions, $$X_1X_2X_3=Y_1Y_2Y_3$$ $$2(X_1^4 + X_2^4 + X_3^4) - 5(X_1^2 + X_2^2 + X_3^2)^2= 2(Y_1^4 + Y_2^4 + Y_3^4) - 5(Y_1^2 + Y_2^2 + Y_3^2)^2$$ are met. Using the $x_i,y_i$ (see [euler.free.fr][3]), I derived their $X_i, Y_i$. Surprisingly, seven had a simple linear constraint, namely $X_3=2Y_3$. Let, $$X_1,\,X_2,\,X_3 =u_1 u_2,\;u_3 u_4,\;2u_5$$ $$Y_1,\,Y_2,\,Y_3 =2u_1 u_4,\;u_2 u_3,\;u_5$$ then solutions are, $$\begin{array}{|c|c|c|c|c|c|} \text{#}&u_1&u_2&u_3&u_4&u_5\\ 1&127 &139 &313 &1 &23647\\ 2&227 &27 &113 &13 &1305\\ 3&14737 &139 &15899 &25 &544069 \\ 4&303 &304 &338 &37 &24616\\ 5&431 &187 &365 &49 &6929\\ 6&871 &163 &4364 &127 &254405 \\ 7&439 &2459 &247 &1175 &261851 \\ \end{array}$$ Looking at $u_4$, it is tempting to speculate there is a pattern in the $u_i$ and that there is infinitely many $u_i$. --- **Question:** 1. The first condition $X_1X_2X_3=Y_1Y_2Y_3$ is easily met. After doing so, can the quartic second condition be split as an intersection of two quadric surfaces [as Elkies did][4] for $x_1^4+x_2^4+x_3^4= 1$ or [as Bremner and Ulas did][5] for $x_1^6+x_2^6+x_3^6= y^2$? **P.S.** Eq.1 is briefly discussed in this [MO post][6]. [1]: https://www.jstor.org/stable/44240281 [2]: https://www.jstor.org/stable/44238512 [3]: http://euler.free.fr/database.txt [4]: https://math.stackexchange.com/questions/509526/ [5]: https://math.stackexchange.com/a/4633598/4781 [6]: https://mathoverflow.net/questions/150428/