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

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Partial Solution:
 
Partial Solution:
Consider P(x)=(x-a)(x-b)(x-c).\\
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Consider <math>P(x)=(x-a)(x-b)(x-c)</math>.
Let <math>S_k= a^k+b^k+c^k</math>. \\
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Let <math>S_k= a^k+b^k+c^k</math>.
Since a ,b ,c are roots of P(x), P(x)=0 is the characteristic equation of <math>s_k</math>. \\
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So : \\
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According to Newton’s Sum:
<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}, a+b+c \vert s_{k+3}</math>. \\
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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>.  
 
This proves (b).
 
This proves (b).
  

Revision as of 08:04, 1 November 2022

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