2025 - Turkmenistan National Math Olympiad

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2025 - Turkmenistan National Math Olympiad

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infinitely many solutions

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Source: Turkmenistan Math Olympiad - 2025

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A_E_R

#1 h Apr 6, 2025, 9:48 AM

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Let k,m,n>=2 positive integers and GCD(m,n)=1, Prove that the equation has infinitely many solutions in distict positive integers: x_1^m+x_2^m+⋯x_k^m=x_(k+1)^n

This post has been edited 1 time. Last edited by A_E_R, Apr 6, 2025, 9:54 AM
Reason: More understandable

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ND_

#2 h Apr 6, 2025, 10:05 AM

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A_E_R wrote:

Let $k,m,n \geq 2$ positive integers and $GCD(m,n)=1$ , Prove that the equation has infinitely many solutions in distinct positive integers: ${x_1}^m+{x_2}^m+\ldots {x_k}^m=x_{k+1}^n$

What are distinct? If it is $(x_1, x_2, \ldots x_n)$ , then:
$x_1=a(a^2+b^2), x_2=b(a^2+b^2), x_3=(a^2+b^2)$ with $(k,m,n)=(2,2,3)$ works

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5 posts

A_E_R

#3 h Apr 6, 2025, 10:21 AM

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I think,it is a special case. We need a general version

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44 posts

ND_

#4 h Apr 6, 2025, 10:34 AM

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$x_{k+1}=\sum_{p=1}^{k} {a_p}^m$ , $x_{p}=a_p \cdot x_{k+1}$ for $1\leq p \leq k$

$\sum_{p=1}^{k} {x_p}^m ={x_{k+1}}^m \sum_{p=1}^{k} {a_p}^m={x_{k+1}}^{m+1}$

This post has been edited 1 time. Last edited by ND_, Apr 6, 2025, 10:52 AM
Reason: Explanation

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#5 h Apr 6, 2025, 10:46 AM

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ND_ wrote:

$x_{k+1}=\sum_{p=1}^{k} {a_p}^m$ , $x_{p}=a_p \cdot x_{k+1}$ for $1\leq p \leq k$

This is not a solution! Or I can not see it.

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#6 h Apr 6, 2025, 12:18 PM

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It can be made into a solution with some adjustments.

Since $m$ and $n$ are coprime, there are positive integers $u$ and $v$ such that $un=vm+1.$ By choosing $x_{k+1}=y_{k+1}^u$ and $x_i=y_i^v$ for $i= \overline{1,k},$ the equation becomes $y_1^{vm}+y_2^{vm}+\ldots+y_k^{vm}=y_{k+1}^{vm+1}.$
As ND_ indicated above, a solution to this is given by $y_{k+1}=\sum_{i=1}^k a_i^{vm}$ and $y_i=a_iy_{k+1}$ for $i=\overline{1,k},$ so we found an infinite family of solutions:

$x_{k+1}= \left( \sum_{i=1}^k a_i^{vm} \right)^u$ and $x_i= \left(a_i \sum_{i=1}^k a_i^{vm} \right)^v$ for $i=\overline{1,k}.$

We just have to make sure that these numbers are distinct. For this we impose $a_i >1$ for $i=\overline{1,k},$ which ensures that the numbers $x_i$ with $i=\overline{1,k}$ are pairwise distinct.

If $u \le v$ then clearly $x_{k+1}<x_i$ for $i=\overline{1,k}.$

If $u>v$ then

$x_{k+1} \ge \left( \sum_{i=1}^k a_i^{vm} \right)^{v+1}= \left( \sum_{i=1}^k a_i^{vm} \right)^v \left( \sum_{i=1}^k a_i^{vm} \right) > \left( \sum_{i=1}^k a_i^{vm} \right)^v \cdot a_i^v=x_i,$ for $i=\overline{1,k}.$

This post has been edited 1 time. Last edited by Filipjack, Apr 6, 2025, 12:20 PM
Reason: comma

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