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

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).

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 {{solution}}
 Partial Solution:
-Consider P(x)=(x-a)(x-b)(x-c).
-Let <math>S_k= a^k+b^k+c^k</math>.
+Consider <math>P(x)=(x-a)(x-b)(x-c)</math>.
-Since a ,b ,c are roots of P(x), P(x)=0 is the characteristic equation of <math>s_k</math>.
+Let <math>S_k= a^k+b^k+c^k</math>.
-So :
-<math>s_{k+3}-(a+b+c)s_{k+2}+(ab+bc+ca)s_{k+1}-(abc)s_k=0</math>.
+According to Newton’s Sum:
-So clearly if <math>a+b+c \vert s_k, s_{k+1}, a+b+c \vert s_{k+3}</math>.
+<math>S_{k+3}-(a+b+c)S_{k+2}+(ab+bc+ca)S_{k+1}-(abc)S_k=0</math>.
+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).

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

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

Latest revision as of 17:21, 28 November 2023

Problem

Solution

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