Difference between revisions of "2004 USAMO Problems/Problem 2"

Revision as of 19:12, 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 if the sequence is of a length which is a power of 2. If not, we "pad" the sequence with many copies of an existing element of the sequence until it does have a length which is a power of 2 - it is apparent that this will not change $S$ .

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 1	Followed by Problem 3
1 • 2 • 3 • 4 • 5 • 6
All USAMO Problems and Solutions

<url>viewtopic.php?p=17440 Discussion on AoPS/MathLinks</url>

@@ Line 15: / Line 15: @@
 [b]Proof:[/b] Assume <math>a_i</math> has at least two elements; <math>x</math> and <math>y</math>. By (b), <math>x - y</math> is in <math>S</math>, and by the application of (c) above, we get that <math>n(x - y)</math> for integers <math>n</math> is in <math>S</math>. Then apply (c) to <math>n(x - y)</math> and <math>ny</math> or <math>nx</math> to get that <math>ax + by\in S</math> for all <math>a,b\in \mathbb{Z}</math>.
-Now let the terms be <math>a_1,a_2,\ldots,a_{n}</math>. By applying our lemma many times, all numbers in  the form <math>\sum c_1a_1</math> for a sequence of integers <math>c_i</math> are attainable.
+Now let the terms be <math>a_1,a_2,\ldots,a_{n}</math>. By applying our lemma many times, all numbers in  the form <math>\sum c_1a_1</math> for a sequence of integers <math>c_i</math> are attainable if the sequence is of a length which is a power of 2. If not, we "pad" the sequence with many copies of an existing element of the sequence until it does have a length which is a power of 2 - it is apparent that this will not change <math>S</math>.
 By Schur's theorem (a generalization of the more well-known Chicken McNugget theorem), every integer greater than some integer <math>n</math> is attainable, and hence there are two members of <math>S</math> in the form <math>\sum c_1a_1</math> which are consecutive integers. Furthermore, because such numbers are closed under addition, their sum is in <math>S</math>, and hence so is their difference; <math>1</math>. Thus, by the argument at the beginning at this proof, <math>S</math> is the set of all integers, as desired.

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Difference between revisions of "2004 USAMO Problems/Problem 2"

Revision as of 19:12, 26 April 2009

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