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

Latest revision as of 18:15, 2 June 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 solutions are either $x_i=0$ for all $i$ or $x_i=(n-1)^{-\frac{2018}{2017}}$ for all $i$ . However, the second solution is not an integer unless $n=2$ , so for $n>2$ , the only solution is $x_i=0$ . For $n=2$ , the two solutions are $x_1=x_2=0$ or $x_1=x_2=1$ .

~ eevee9406

@@ Line 10: / Line 10: @@
 == Solution ==
-{{solution}}
+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 solutions 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>. However, the second solution is not an integer unless <math>n=2</math>, so for <math>n>2</math>, the only solution is <math>x_i=0</math>. For <math>n=2</math>, the two solutions are <math>x_1=x_2=0</math> or <math>x_1=x_2=1</math>.
+~ [https://artofproblemsolving.com/wiki/index.php/User:Eevee9406 eevee9406]
 == See also ==
 [[OIM Problems and Solutions]]

Page

Toolbox

Search

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

Latest revision as of 18:15, 2 June 2025

Problem

Solution

See also