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Difference between revisions of "2018 OIM Problems/Problem 1"

(Created page with "== Problem == For each natural number <math>n \ge 2</math>, find the integer solutions to the following system of equations: <cmath>x_1 = (x_2 + x_3 + x_4 + \cdots + x_n)^{20...")
 
(solution)
Line 10: Line 10:
  
 
== Solution ==
 
== Solution ==
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{{solution}}
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Since the right-hand side (RHS) of each equation is nonnegative, then the left-hand side (LHS) is too; this implies that each <math>x_i</math> is nonnegative. As a result, we can take the <math>2018^\text{th}</math> root of both sides of the first and second equations, resulting in:
 +
<cmath>\sqrt[2018]{x_1} = x_2 + x_3 + x_4 + \cdots + x_n</cmath>
 +
<cmath>\sqrt[2018]{x_2} = x_1 + x_3 + x_4 + \cdots + x_n</cmath>
 +
Subtracting yields:
 +
<cmath>\sqrt[2018]{x_1}-\sqrt[2018]{x_2}=x_2-x_1</cmath>
 +
If <math>x_1>x_2</math>, then the LHS is positive but the RHS is negative, which is not possible, and the same goes for <math>x_1<x_2</math>. Thus <math>x_1=x_2</math>, and if we repeat this process for each consecutive pair of equations we eventually find that all <math>x_i</math> are equal.
 +
Using this property on the first equation results in:
 +
<cmath>x_1 = ((n-1)x_1)^{2018}</cmath>
 +
<cmath>\Rightarrow x_1=(n-1)^{2018}x_1^{2018}</cmath>
 +
<cmath>\Rightarrow (n-1)^{2018}x_1^{2018}-x_1=0</cmath>
 +
<cmath>\Rightarrow x_1((n-1)^{2018}x_1^{2017}-1)=0</cmath>
 +
<cmath>\Rightarrow x_1=0\text{ or }x_1=(n-1)^{-\frac{2018}{2017}}</cmath>
 +
Both solutions work. Thus the solution sets are either <math>x_i=0</math> for all <math>i</math> or <math>x_i=(n-1)^{-\frac{2018}{2017}}</math> for all <math>i</math>.
 +
 
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~ Edited by [https://artofproblemsolving.com/wiki/index.php/User:Aoum aoum]
  
 
== See also ==
 
== See also ==
 
[[OIM Problems and Solutions]]
 
[[OIM Problems and Solutions]]

Revision as of 17:39, 23 March 2025

Problem

For each natural number $n \ge 2$, find the integer solutions to the following system of equations:

\[x_1 = (x_2 + x_3 + x_4 + \cdots + x_n)^{2018}\] \[x_2 = (x_1 + x_3 + x_4 + \cdots + x_n)^{2018}\] \[\cdots\] \[x_n = (x_1 + x_2 + x_3 + \cdots + x_{n-1})^{2018}\]

~translated into English by Tomas Diaz. ~orders@tomasdiaz.com

Solution

Since the right-hand side (RHS) of each equation is nonnegative, then the left-hand side (LHS) is too; this implies that each $x_i$ is nonnegative. As a result, we can take the $2018^\text{th}$ root of both sides of the first and second equations, resulting in: \[\sqrt[2018]{x_1} = x_2 + x_3 + x_4 + \cdots + x_n\] \[\sqrt[2018]{x_2} = x_1 + x_3 + x_4 + \cdots + x_n\] Subtracting yields: \[\sqrt[2018]{x_1}-\sqrt[2018]{x_2}=x_2-x_1\] If $x_1>x_2$, then the LHS is positive but the RHS is negative, which is not possible, and the same goes for $x_1<x_2$. Thus $x_1=x_2$, and if we repeat this process for each consecutive pair of equations we eventually find that all $x_i$ are equal. Using this property on the first equation results in: \[x_1 = ((n-1)x_1)^{2018}\] \[\Rightarrow x_1=(n-1)^{2018}x_1^{2018}\] \[\Rightarrow (n-1)^{2018}x_1^{2018}-x_1=0\] \[\Rightarrow x_1((n-1)^{2018}x_1^{2017}-1)=0\] \[\Rightarrow x_1=0\text{ or }x_1=(n-1)^{-\frac{2018}{2017}}\] Both solutions work. Thus the solution sets are either $x_i=0$ for all $i$ or $x_i=(n-1)^{-\frac{2018}{2017}}$ for all $i$.

~ Edited by aoum

See also

OIM Problems and Solutions

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