Difference between revisions of "2012 USAMO Problems/Problem 3"
m (→See Also) |
m (Changed to "Bertrand's Postulate", as "Bertrand's Theorem" refers to something unrelated) |
||
(7 intermediate revisions by 2 users not shown) | |||
Line 9: | Line 9: | ||
==Solution that involves a non-elementary result== | ==Solution that involves a non-elementary result== | ||
+ | |||
+ | (Since Bertrand's is well known and provable using elementary techniques, I see nothing wrong with this-tigershark22) | ||
+ | |||
For <math>n=2</math>, <math> |a_1| = 2 |a_2| = \cdots = 2^m |a_{2^m}| </math> implies that for any positive integer <math>m</math>, <math>|a_1| \ge 2^m</math>, which is impossible. | For <math>n=2</math>, <math> |a_1| = 2 |a_2| = \cdots = 2^m |a_{2^m}| </math> implies that for any positive integer <math>m</math>, <math>|a_1| \ge 2^m</math>, which is impossible. | ||
Line 21: | Line 24: | ||
To proceed with the construction, I need the following fact: for any positive integer <math>m>2</math>, there exists a prime <math>p</math> such that <math>\frac{m}{2} <p \le m</math>. | To proceed with the construction, I need the following fact: for any positive integer <math>m>2</math>, there exists a prime <math>p</math> such that <math>\frac{m}{2} <p \le m</math>. | ||
− | To prove this, I am going to use '''Bertrand's | + | To prove this, I am going to use '''Bertrand's Postulate''' ([http://en.wikipedia.org/wiki/Bertrand's_postulate]) without proof. The Theorem states that, for any integer <math>n>1</math>, there exists a prime <math>p</math> such that <math>n<p\le 2n-1</math>. In other words, for any positive integer <math>m>2</math>, if <math>m=2n</math> with <math>n>1</math>, then there exists a prime <math>p</math> such that <math>\frac{m}{2} < p < m</math>, and if <math>m=2n-1</math> with <math>n>1</math>, then there exists a prime <math>p</math> such that <math>\frac{m+1}{2} <p\le m</math>, both of which guarantees that for any integer <math>m>2</math>, there exists a prime <math>p</math> such that <math>\frac{m}{2} <p \le m</math>. |
Line 58: | Line 61: | ||
--[[User:Lightest|Lightest]] 21:24, 2 May 2012 (EDT) | --[[User:Lightest|Lightest]] 21:24, 2 May 2012 (EDT) | ||
− | |||
− | |||
==See Also== | ==See Also== | ||
Line 65: | Line 66: | ||
{{USAMO newbox|year=2012|num-b=2|num-a=4}} | {{USAMO newbox|year=2012|num-b=2|num-a=4}} | ||
+ | {{MAA Notice}} | ||
[[Category:Olympiad Number Theory Problems]] | [[Category:Olympiad Number Theory Problems]] |
Latest revision as of 06:38, 27 October 2022
Problem
Determine which integers have the property that there exists an infinite sequence
,
,
,
of nonzero integers such that the equality
holds for every positive integer
.
Partial Solution
For equal to any odd prime
, the sequence
, where
is the greatest power of
that divides
, gives a valid sequence. Therefore, the set of possible values for
is at least the set of odd primes.
Solution that involves a non-elementary result
(Since Bertrand's is well known and provable using elementary techniques, I see nothing wrong with this-tigershark22)
For ,
implies that for any positive integer
,
, which is impossible.
We proceed to prove that the infinite sequence exists for all .
First, one notices that if we have for any integers
and
, then it is suffice to define all
for
prime, and one only needs to verify the equation (*)
for the other equations will be automatically true.
To proceed with the construction, I need the following fact: for any positive integer , there exists a prime
such that
.
To prove this, I am going to use Bertrand's Postulate ([1]) without proof. The Theorem states that, for any integer , there exists a prime
such that
. In other words, for any positive integer
, if
with
, then there exists a prime
such that
, and if
with
, then there exists a prime
such that
, both of which guarantees that for any integer
, there exists a prime
such that
.
Go back to the problem. Suppose . Let the largest two primes not larger than
are
and
, and that
. By the fact stated above, one can conclude that
, and that
. Let's construct
:
Let . There will be three cases: (i)
, (ii)
, and (iii)
.
Case (i): . Let
for all prime numbers
, and
, then (*) becomes:
Case (ii): but
. In this case, let
, and
for all prime numbers
, and
, then (*) becomes:
or
Case (iii): . In this case, let
,
, and
for all prime numbers
, and
, then (*) becomes:
or
In each case, by Bezout's Theorem, there exists non zero integers and
which satisfy the equation. For all other primes
, just let
(or any other non-zero integer).
This construction is correct because, for any ,
Since Bertrand's Theorem is not elementary, we still need to wait for a better proof.
--Lightest 21:24, 2 May 2012 (EDT)
See Also
2012 USAMO (Problems • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.