Difference between revisions of "2004 USAMO Problems/Problem 2"
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Revision as of 18:36, 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.
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 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>