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

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Partial Solution:
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Consider P(x)=(x-a)(x-b)(x-c).
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Let <math>S_k= a^k+b^k+c^k</math>.
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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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<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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This proves (b).
  
 
==See also==
 
==See also==

Latest revision as of 08:13, 16 January 2020

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 $S_k= a^k+b^k+c^k$. Since a ,b ,c are roots of P(x), P(x)=0 is the characteristic equation of $s_k$. So : $s_{k+3}-(a+b+c)s_{k+2}+(ab+bc+ca)s_{k+1}-(abc)s_k=0$. So clearly if $a+b+c \vert s_k, s_{k+1}, a+b+c \vert s_{k+3}$. This proves (b).

See also

2005 Canadian MO (Problems)
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Problem 4
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