Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution)
Line 10: Line 10:
 
{{solution}}
 
{{solution}}
 
Partial Solution:
 
Partial Solution:
 +
 
Consider <math>P(x)=(x-a)(x-b)(x-c)</math>.
 
Consider <math>P(x)=(x-a)(x-b)(x-c)</math>.
 
Let <math>S_k= a^k+b^k+c^k</math>.
 
Let <math>S_k= a^k+b^k+c^k</math>.

Revision as of 03:00, 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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2005_Canadian_MO_Problems/Problem_5&oldid=206216"