3
$\begingroup$

Given rationals $R = a,b,c,d,e,f$. Define,

$$F_n = a^n+b^n+c^n-(d^n+e^n+f^n)$$

  1. If $F_\color{red}1=0$, is there a rational solution to $7F_3x^4+7F_5x^2+F_7 = 0$? Then for $k=1,2,8$,

$$\small(x + a)^k + (x + b)^k + (x + c)^k + (x - d)^k + (x - e)^k + (x - f)^k = \\ \small(x - a)^k + (x - b)^k + (x - c)^k + (x + d)^k + (x + e)^k + (x + f)^k\tag1$$

  1. If $F_2=0$, is there a rational solution to $70F_4x^4+28F_6x^2+F_8 = 0$? Then for $k=1,2,3,8$,

$$\small(x + a)^k + (x + b)^k + (x + c)^k + (x - a)^k + (x - b)^k + (x - c)^k = \\ \small(x + d)^k + (x + e)^k + (x + f)^k + (x - d)^k + (x - e)^k + (x - f)^k\tag2$$

  1. If $F_2=0$, is there a rational solution to $42F_4x^4+28F_6x^2+3F_8 = 0$? Then for $k=1,2,3,9$,

$$\small(x + a)^k + (x + b)^k + (x + c)^k + (x - a)^k + (x - b)^k + (x - c)^k = \\ \small(x + d)^k + (x + e)^k + (x + f)^k + (x - d)^k + (x - e)^k + (x - f)^k\tag3$$

For the third, one solution is $R = 9, 14, 19;\, 17, 18, 5$, and $x = 4$. In fact, using an elliptic curve, one can show there are an infinite more. So this is a settled question.

However, I'm having trouble with the first and second which at first glance seems to be the easier problem. For the first, the best I could find is $R = 5, 19, 19;\, 11, 11, 21$ and $x = \sqrt{13}$. Using some basic Mathematica code, I searched $0<R<24$ as it took too long to search higher. Maybe someone here knows of a more efficient way.

P.S. See also this related post.

$\endgroup$
  • $\begingroup$ $k \equiv n$, isn't it? $\endgroup$ – Alex Silva Jan 21 '15 at 13:57
  • $\begingroup$ @AlexSilva: No, $n$ and $k$ are separate. For example, using the six terms of $(1)$ as $F_2 = 9^2+14^2+19^2-(17^2+18^2+5^2)=0$, if you substitute it into $(2)$ with $x=4$, then you will find that it is good for $k=1,2,3,9$. There is no known $F_2$ yet such that it is good for $k=1,2,3,\color{blue}8$. $\endgroup$ – Tito Piezas III Jan 22 '15 at 0:08
  • $\begingroup$ It looks to me like condition 1. should be $70F_4x^4+28F_6x^2+F_8=0$. $\endgroup$ – Zander Mar 4 '15 at 23:55
  • $\begingroup$ @Zander: I've clarified the questions. Sorry for the confusion. P.S. Note the difference between $(1)$ and $(2)$. $\endgroup$ – Tito Piezas III Mar 5 '15 at 3:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.