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I have been working on this problem for quite a while and frankly, I ran out of ideas quite a while ago and hence I decided to ask it here(my ideas are below). I will greatly appreciate any help.

Given $(x_0,x_1,x_2,x_3)\in \mathbb{Q}$ where all $(x_0,x_1,x_2,x_3)$ are larger than one and they are all unique and given $$f(x) = \dfrac{x(x^2 - 1)}{(x^2 + 1)^2}$$ I was trying to determine a valid $(x_0,x_1,x_2,x_3)$ such that $$f(x_2) - f(x_0) - f(x_3) = 0$$ and $$2f(x_2) - f(x_1) - f(x_3) = 0$$

My approaches

  1. Trigonometry

As $\cos \theta$ and $\sin \theta$ can be parametarized by $\dfrac{2x}{x^2 + 1}$ and $\dfrac{x^2 - 1}{x^2 + 1}$, $f(x)$ can be thought to be $\dfrac{\sin 2\theta}{4}$. Hence, the first condition can be thought as $$\sin 2\theta _2 - sin 2\theta _0 - \sin 2\theta _3 = 0$$ Yet I failed to reach any ideas from here.

  1. Groebner basis

I am quite new to this idea of Groebner basis yet using Sage, I found 9 Groebner bases of an ideal defined by the variance given by the two conditions above. Yet I found that two of those 9 were the two conditions above while for the others to be 0, $(x_0,x_1,x_2,x_3)$ cannot be all unique or larger than one. Yet as I said, I am quite new to this concept as I am quite new to algebraic geometry, I may be wrong. If this is the case I will greatly appreciate any corrections

  1. Brute Force

I defined two terms $m_i$ and $n_i$ where they are both integers which gives $$f(x_i) = \dfrac{n_im_i(n_i^2 - m_i^2)}{(n_i^2 + m_i^2)^2}$$ where $x_i = \dfrac{n_i}{m_i}$. Here, I ran a octa-loop(horrendous amounts of loops), making sure n_i and m_i are mutually prime, with hope that some $n_0, m_0$.....$n_3, m_3$ might be found.

I was unsuccessful.

I'm fairly confident that there is a better way to go about doing this and I am just unfamiliar with it. Any help is appreciated. Thank you in advance!

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  • $\begingroup$ Are those the right equations? It's immediate that $f(x_2)=0.$ Also, $x_1$ never appears. $\endgroup$ – saulspatz Mar 20 '18 at 1:29
  • $\begingroup$ Oh, sorry about that. I made a mistake. The second one is wrong $\endgroup$ – Isamu Isozaki Mar 20 '18 at 1:31
  • $\begingroup$ The second equation was supposed to be $2f(x_2) - f(x_1) - f(x_3) = 0$. Thanks for the comment! I corrected it now $\endgroup$ – Isamu Isozaki Mar 20 '18 at 1:32
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HINT

The issue conditions can be presented in the form of $$f(x_1) - f(x_2) = f(x_2) - f(x_3) = f(x_0).\tag1$$ Thus, $f(x_1),\ f(x_2)$ and $f(x_3)$ form an arithmetic progression with a difference $-f(x_0).$

Let $$x_i = \dfrac{s_i}{t_i},\quad y_i = x_i - \dfrac1{x_i},\quad z_i = x_1 + \dfrac1{x_i},\tag2$$ then $$ \begin{cases} f(x_2) - f(x_0) = f(x_3)\\ f(x_2) + f(x_0) = f(x_1) \end{cases}\rightarrow \begin{cases} 2f(x_2) = f(x_1) + f(x_3)\\ f(x_2) = f(x_0) + f(x_3) \end{cases}\rightarrow \begin{cases} \dfrac{2y_2}{y_2^2+4} = \dfrac{y_1}{y_1^2+4} + \dfrac{y_3}{y_3^2+4}\\ \dfrac{y_2}{y_2^2+4} = \dfrac{y_0}{y_0^2+4} + \dfrac{y_3}{y_3^2+4}, \end{cases}$$ $$ \begin{cases} \dfrac{8y_2}{y_2^2+4}+2 = \dfrac{4y_1}{y_1^2+4}+1 + \dfrac{4y_3}{y_3^2+4} + 1\\ \dfrac{y_2}{y_2^2+4} = \dfrac{y_0}{y_0^2+4} + \dfrac{y_3}{y_3^2+4} \end{cases}\rightarrow \begin{cases} \dfrac{2(y_2+2)^2}{z_2^2} = \dfrac{(y_1+2)^2}{z_1^2} + \dfrac{(y_3+2)^2}{z_3^2}\\ \dfrac{y_0}{z_0^2} = \dfrac{y_2}{z_2^2} - \dfrac{y_3}{z_3^2} \end{cases} $$ \begin{cases} 2((y_2+2)z_1z_3)^2 = ((y_1+2)z_2z_3)^2 + ((y_3+2)z_2z_1)^2\\[4pt] y_0(z_2z_3)^2 = z_0^2(y_2z_3^2 - y_3z_2^2), \end{cases} $$\begin{cases} 2((s_2+t_2)(s_1^2+t_1^2)(s_3^2+t_3^2))^2 = ((s_1+t_1)(s_2^2+t_2^2)(s_3^2+t_3^2))^2 + ((s_3+t_3)(s_2^2+t_2^2)(s_1^2+t_1^2))^2\\[4pt] (s_0^2-t_0^2)((s_2^2+t_2^2)(s_3^2+t_3^2))^2 = (s_0^2+t_0^2)^2((s_2^2-t_2^2)(s_3^2+t_3^2)^2 - (s_3^2-t_3^2)(s_2^2+t_2^2)^2). \end{cases}\tag3$$ Both of the terms in the RHS of the first equation of $(3)$ are even.

Substitutions $$\begin{cases} 2u = (s_1+t_1)(s_2^2+t_2^2)(s_3^2+t_3^2) + (s_3+t_3)(s_2^2+t_2^2)(s_1^2+t_1^2)\\ 2v = (s_1+t_1)(s_2^2+t_2^2)(s_3^2+t_3^2) - (s_3+t_3)(s_2^2+t_2^2)(s_1^2+t_1^2) \end{cases}$$ lead to equation $$u^2 + v^2 = ((s_2+t_2)(s_1^2+t_1^2)(s_3^2+t_3^2))^2,$$ so $$ \begin{cases} (s_2+t_2)(s_1^2+t_1^2)(s_3^2+t_3^2) = (m^2+n^2)l\\ \left[\begin{align} \begin{cases} (s_1+t_1)(s_2^2+t_2^2)(s_3^2+t_3^2) + (s_3+t_3)(s_2^2+t_2^2)(s_1^2+t_1^2) = 2(m^2-n^2)l\\ (s_1+t_1)(s_2^2+t_2^2)(s_3^2+t_3^2) - (s_3+t_3)(s_2^2+t_2^2)(s_1^2+t_1^2) = 4mnl\\ \end{cases}\\ \begin{cases} (s_1+t_1)(s_2^2+t_2^2)(s_3^2+t_3^2) + (s_3+t_3)(s_2^2+t_2^2)(s_1^2+t_1^2) = 4mnl\\ (s_1+t_1)(s_2^2+t_2^2)(s_3^2+t_3^2) - (s_3+t_3)(s_2^2+t_2^2)(s_1^2+t_1^2) = 2(m^2-n^2)l,\\ \end{cases}\\ \end{align} \right. \end{cases} $$ $$\begin{cases} (s_2+t_2)(s_1^2+t_1^2)(s_3^2+t_3^2) = (m^2+n^2)l\\ (s_1+t_1)(s_2^2+t_2^2)(s_3^2+t_3^2) = (m^2-n^2+2mn)l\\ (s_3+t_3)(s_2^2+t_2^2)(s_1^2+t_1^2) = |m^2-n^2-2mn|l\\ m>n\\ (m, n, l) \in\mathbb N^3. \end{cases}\tag4$$ The second equation of $(3)$ can be presented in the form of $$(s_0^2-t_0^2)((s_2s_3-t_2t_3)^2+(s_2t_3+t_2s_3)^2)^2 = (s_0^2+t_0^2)^2((s_2^2-t_2^2)(s_3^2+t_3^2)^2 - (s_3^2-t_3^2)(s_2^2+t_2^2)^2),\tag5$$ so the system of the additional conditions $$\begin{cases} s_0 = k|s_2s_3-t_2t_3|\\ t_0 = k(s_2s_3+t_2s_3)\\ s_0^2-t_0^2 = k^2((s_2^2-t_2^2)(s_3^2+t_3^2)^2 - (s_3^2-t_3^2)(s_2^2+t_2^2)^2)\\ s_2 = p_2^2+q_2^2, \quad t_2 = 2p_2q_2, (p_2,q_2)\in\mathbb N^2,\\ s_3 = p_3^2+q_3^2, \quad t_3 = 2p_3q_3, (p_3,q_3)\in\mathbb N^2, \end{cases}\tag6$$ leads to the rational solutions.

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  • $\begingroup$ Hi! Thanks for the answer! At system of equation 6, $t_0$ > $s_0$ right? Then, as $x_0 = \dfrac{s_0}{t_0}$, this violates the condition that all $x_i$s are above 1. Does this mean that there is no solution to the system of equations given in my question or am I missing something here? Again, thanks for the answer! I found it to be quite ingenious. $\endgroup$ – Isamu Isozaki Apr 3 '18 at 4:44
  • $\begingroup$ I am sorry for not being able to give you the bounty. The time ran out. If you do want it, I will donate 50 of my reputation to you directly. $\endgroup$ – Isamu Isozaki Apr 3 '18 at 5:25
  • $\begingroup$ May I ask how you computed the last step? $\endgroup$ – Isamu Isozaki Apr 3 '18 at 5:29
  • $\begingroup$ I've found the simple way. But I need some time. $\endgroup$ – Yuri Negometyanov Apr 4 '18 at 0:25
  • $\begingroup$ Got it! Thanks! $\endgroup$ – Isamu Isozaki Apr 4 '18 at 12:17
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Fixing or choosing any two points on the curve, with $(x_3,f(x_3))$ and $(x_2,f(x_2))$, one could then find the other two points $x_0,x_1$ by using the two given equations and so $f(x_0)=f(x_2)-f(x_3)$ and $f(x_1)=2f(x_2)-f(x_3)$. To find $x_0$ and $x_1$ we can then use the function $f(x)$.

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  • $\begingroup$ Thanks for the answer but how can I know that those other two are rational too? $\endgroup$ – Isamu Isozaki Mar 20 '18 at 2:18
  • $\begingroup$ I mean how can $x_1$ and $x_0$ guaranteed to be rational in that situation? $\endgroup$ – Isamu Isozaki Mar 20 '18 at 2:19
  • $\begingroup$ I don't remember seeing the condition that $x_i $s are rational but I guess this makes sense (explains your difficulty).@Isamu Isozaki $\endgroup$ – AnyAD Mar 20 '18 at 9:12
  • $\begingroup$ Thanks for the comment! It is written on the third line of the question. $\endgroup$ – Isamu Isozaki Mar 20 '18 at 11:28

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