Difference between revisions of "2004 USAMO Problems/Problem 2"
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[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>. | [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. | 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. |
Revision as of 19:12, 26 April 2009
Problem
Suppose are integers whose greatest common divisor is 1. Let be a set of integers with the following properties:
(a) For , . (b) For (not necessarily distinct), . (c) For any integers , if , then .
Prove that must be equal to the set of all integers.
Solution
Suppose has only one element; then for the greatest common divisor to be 1, has to be the sole element. Then is in by (a), is in by (b), by (c), and we can apply (c) analogously to get that for integers and hence is the set of all integers, as desired.
[b]Lemma:[/b] If , then for integers . [b]Proof:[/b] Assume has at least two elements; and . By (b), is in , and by the application of (c) above, we get that for integers is in . Then apply (c) to and or to get that for all .
Now let the terms be . By applying our lemma many times, all numbers in the form for a sequence of integers 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 .
By Schur's theorem (a generalization of the more well-known Chicken McNugget theorem), every integer greater than some integer is attainable, and hence there are two members of in the form which are consecutive integers. Furthermore, because such numbers are closed under addition, their sum is in , and hence so is their difference; . Thus, by the argument at the beginning at this proof, 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>