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

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).

@@ Line 9: / Line 9: @@
 {{solution}}
+Partial Solution:
+Consider <math>P(x)=(x-a)(x-b)(x-c)</math>.
+Let <math>S_k= a^k+b^k+c^k</math>.
+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>.
+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).
 ==See also==

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

Page

Toolbox

Search

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

Revision as of 08:04, 1 November 2022

Problem

Solution

See also