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I. The first solution to,

$$\sum^6_{n=1} a_n^9 =\sum^6_{n=1} b_n^9$$

$$13^9+18^9+23^9-5^9-10^9-15^9 = 9^9+21^9+22^9-1^9-13^9-14^9$$

was found in 1967 by computer search by Lander et al. It stood for 40 years as a "numerical curiosity" until Bremner and Delorme discovered it had the highly structured form,

$$\small(u + 9)^k + (u + 14)^k + (u + 19)^k + (u - 9)^k + (u - 14)^k + (u - 19)^k = \\ \small(u + 5)^k + (u + 17)^k + (u + 18)^k + (u - 5)^k + (u - 17)^k + (u - 18)^k$$

with $\small u = 9+14+(-19)=5+17+(-18)=4$, was in fact good for $\small k=1,2,3,9$ and, using an elliptic curve, there were an infinite more.

II. However, this can be generalized to more terms and powers. Define,

$$F_n = x_1^n+x_2^n+x_3^n+x_4^n-(y_1^n+y_2^n+y_3^n+y_4^n)$$

$$P_k =\small(u+x_1)^k+(u+x_2)^k+(u+x_3)^k+(u+x_4)^k+(u-x_1)^k+(u-x_2)^k+(u-x_3)^k+(u-x_4)^k$$

$$Q_k =\small(u+y_1)^k+(u+y_2)^k+(u+y_3)^k+(u+y_4)^k+(u-y_1)^k+(u-y_2)^k+(u-y_3)^k+(u-y_4)^k$$

Theorem 1: If $F_2 = 0$, and $42F_4u^4+28F_6u^2+3F_8 = 0$, then $P_k = Q_k$ for $k=1,2,3,9$.

  • For $x_4=y_4=0$, $F_1 = F_2 = 0$, and the special case $u = x_1+x_2+x_3$, this reduces to the family studied by Bremner and Delorme, hence there are an infinite number of rational solutions. See this post.

Theorem 2: If $F_2 = F_4 = 0$, and $28F_6u^2+F_8 = 0$, then $P_k = Q_k$ for $k=1,2,3,4,5,\color{blue}8$.

  • There are 2 known solutions, one of which is $x_i = 115, 309, 559, 653,\;$ $y_i = 15, 347, 541, 659$, and $u=285$.

Theorem 3: If $F_2 = F_4 = 0$, and $28F_6u^2+3F_8 = 0$, then $P_k = Q_k$ for $k=1,2,3,4,5,\color{blue}9$.

  • There are 11 known solutions, one of which is $x_i = 63, 122, 197, 240,\;$ $y_i =10, 167, 168, 243$, and $u=52$.

Question: Is it possible to find an elliptic curve also for Theorem 2 and 3 and prove it has an infinite number of rational solutions?

For Theorem 1, there was the fortunate condition that the $u$ was a linear function of the $x_i$. I tried to find something similar for the other two theorems using Mathematica's Integer Relations but there was nothing common. One thing I noticed was if the terms are signed, then 5 of the 11 known solutions had the form,

$$(a+p)^k+(a-p)^k+(b+q)^k+(b-q)^k =(a+r)^k+(a-r)^k+(b+s)^k+(b-s)^k$$

hence were in fact $F_1 = F_2 = F_4 = 0$.

P.S. One solution was found by Jarek Wroblewski, and all the rest by Roger Glendenning.

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