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 "2004 USAMO Problems/Problem 2"

(own solution)
(No difference)

Revision as of 17:36, 26 April 2009

Problem

Suppose $a_1, \dots, a_n$ are integers whose greatest common divisor is 1. Let $S$ be a set of integers with the following properties:

(a) For $i = 1, \dots, n$, $a_i \in S$. (b) For $i,j = 1, \dots, n$ (not necessarily distinct), $a_i - a_j \in S$. (c) For any integers $x,y \in S$, if $x + y \in S$, then $x - y \in S$.

Prove that $S$ must be equal to the set of all integers.

Solution

Suppose $a_i$ has only one element; then for the greatest common divisor to be 1, $1$ has to be the sole element. Then $1$ is in $S$ by (a), $0$ is in $S$ by (b), $0 + 1 = 1\in S\Rightarrow 0 - 1 = - 1\in S$ by (c), and we can apply (c) analogously to get that $n\cdot 1 \in S$ for integers $n$ and hence $S$ is the set of all integers, as desired.

[b]Lemma:[/b] If $x,y\in a_i$, then $ax + by\in S$ for integers $a,b$. [b]Proof:[/b] Assume $a_i$ has at least two elements; $x$ and $y$. By (b), $x - y$ is in $S$, and by the application of (c) above, we get that $n(x - y)$ for integers $n$ is in $S$. Then apply (c) to $n(x - y)$ and $ny$ or $nx$ to get that $ax + by\in S$ for all $a,b\in \mathbb{Z}$.

Now let the terms be $a_1,a_2,\ldots,a_{n}$. By applying our lemma many times, all numbers in the form $\sum c_1a_1$ for a sequence of integers $c_i$ are attainable.

By Schur's theorem (a generalization of the more well-known Chicken McNugget theorem), every integer greater than some integer $n$ is attainable, and hence there are two members of $S$ in the form $\sum c_1a_1$ which are consecutive integers. Furthermore, because such numbers are closed under addition, their sum is in $S$, and hence so is their difference; $1$. Thus, by the argument at the beginning at this proof, $S$ is the set of all integers, as desired.

Resources

2004 USAMO (Problems • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
1 • 2 • 3 • 4 • 5 • 6
All USAMO Problems and Solutions
  • <url>viewtopic.php?t=1478389 Discussion on AoPS/MathLinks</url>
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2004_USAMO_Problems/Problem_2&oldid=31478"