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Difference between revisions of "2005 Canadian MO Problems/Problem 5"

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==Solution==
 
==Solution==
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{{solution}}
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Partial Solution:
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Consider <math>P(x)=(x-a)(x-b)(x-c)</math>.
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Let <math>S_k= a^k+b^k+c^k</math>.
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According to Newton’s Sum:
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<math>S_{k+3}-(a+b+c)S_{k+2}+(ab+bc+ca)S_{k+1}-(abc)S_k=0</math>.
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So clearly if <math>a+b+c \vert S_k, S_{k+1},</math> then <math>a+b+c \vert S_{k+3}</math>.
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This proves (b).
  
 
==See also==
 
==See also==
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*[[2005 Canadian MO]]
 
*[[2005 Canadian MO]]
  
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{{CanadaMO box|year=2005|num-b=4|after=Last Question}}
  
 
[[Category:Olympiad Number Theory Problems]]
 
[[Category:Olympiad Number Theory Problems]]

Latest revision as of 17:21, 28 November 2023

Problem

Let's say that an ordered triple of positive integers $(a,b,c)$ is $n$-powerful if $a \le b \le c$, $\gcd(a,b,c) = 1$, and $a^n + b^n + c^n$ is divisible by $a+b+c$. For example, $(1,2,2)$ is 5-powerful.

  • Determine all ordered triples (if any) which are $n$-powerful for all $n \ge 1$.
  • Determine all ordered triples (if any) which are 2004-powerful and 2005-powerful, but not 2007-powerful.

Solution

This problem needs a solution. If you have a solution for it, please help us out by adding it.

Partial Solution:

Consider $P(x)=(x-a)(x-b)(x-c)$. Let $S_k= a^k+b^k+c^k$.

According to Newton’s Sum:

$S_{k+3}-(a+b+c)S_{k+2}+(ab+bc+ca)S_{k+1}-(abc)S_k=0$. So clearly if $a+b+c \vert S_k, S_{k+1},$ then $a+b+c \vert S_{k+3}$. This proves (b).

See also

2005 Canadian MO (Problems)
Preceded by
Problem 4 		1 • 2 • 3 • 4 • 5 Followed by
Last Question
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