Solve $\begin{cases} \left|x_1-x_2\right|=\left|x_2-x_3\right|=...=\left|x_{2018}-x_1\right|,\\ x_1+x_2+...+x_{2018}=2018. \end{cases}$

I think there must be such a way to solve systems of equation with the form of rotation and absolute value like this.

I have difficulty in solving the first equation. Since to me, there are quite a lot of cases to consider, for example, $\left|x_1-x_2\right|=\left|x_2-x_3\right|$ leads to $x_1-x_2=x_2-x_3$ and $x_1-x_2=x_3-x_2$ and so on.

  • $\begingroup$ I observe that $x_{i} = 1$ for all $i$ is a solution (not sure if it is unique). Perhaps squaring the absolute sign can help a bit. $\endgroup$ – K_inverse Sep 27 '18 at 1:45
  • $\begingroup$ Like what you've said, it may result in $x_1=x_2=...=x_{2018}=1$. But actually, I think squaring the absolute sign still remains the same. I mean there is no difference between $|x_1-x_2| = |x_2-x_3|$ and $(x_1-x_2)^2=(x_2-x_3)^2$. $\endgroup$ – Martin Tr Sep 27 '18 at 1:58

Here is the answer to verify my comment.

Let $|x_{1} - x_{2}| = C \geq 0$ and note that \begin{align} |x_{1} - x_{2}| = C \implies x_{1} = \pm C + x_{2} \end{align} Similarly, we have \begin{align} x_{1} &= \pm C + x_{2} \\ x_{2} &= \pm C + x_{3} \\ & \; \; \vdots \\ x_{2017} &= \pm C + x_{2018} \\ x_{2018} &= \pm C + x_{1} \\ \end{align} Keep substituting, we obtain \begin{align} x_{1} &= \pm C + x_{2} \\ &= \pm C \pm C + x_{3} \\ & \; \; \vdots \\ &= \pm C \pm C \pm \cdots \pm C + x_{1} \end{align} Thus, we have \begin{align} \pm 2018C = 0 \implies C = 0 \end{align} It means that $x_{1} = x_{2} \cdots = x_{2018}$ and let them be $A$.

So now, we use the second equation \begin{align} x_{1} + x_{2} + \cdots + x_{2018} &= 2018 \\ 2018A &= 2018 \\ A &= 1 \end{align}

Conclusion: $x_{1} = x_{2} \cdots = x_{2018} = 1$

  • 1
    $\begingroup$ There is no reason why all your $\pm$ signs must be the same. So $\pm C\pm C\pm\cdots\pm C$ could be for example $C-C+C-C+\cdots$ which is always zero and does not prove that $C$ is zero. $\endgroup$ – David Sep 27 '18 at 2:49
  • $\begingroup$ Here is another solution: $x_1=0$, $x_2=2$, $x_3=0$, $x_4=2$,... $\endgroup$ – David Sep 27 '18 at 2:53
  • $\begingroup$ Thank you a lot for your detailed solution. The idea $|x_1-x_2|=C$ is great. However, your solution could be true if there were odd amounts of variables $x_i$ (that will definitely lead to $C=0$). $\endgroup$ – Martin Tr Sep 27 '18 at 2:56
  • $\begingroup$ @David, oh yes you are right. So, there should be multiple solutions depending on how $C$'s are cancelled out, right? $\endgroup$ – K_inverse Sep 27 '18 at 2:59

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.