Difference between revisions of "2012 USAMO Problems/Problem 3"
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First, one notices that if we have <math>a_{xy} = a_x a_y</math> for any integers <math>x</math> and <math>y</math>, then it is suffice to define all <math>a_x</math> for <math>x</math> prime, and one only needs to verify the equation (*) | First, one notices that if we have <math>a_{xy} = a_x a_y</math> for any integers <math>x</math> and <math>y</math>, then it is suffice to define all <math>a_x</math> for <math>x</math> prime, and one only needs to verify the equation (*) | ||
− | <cmath>a_1+2a_2+\cdots+na_n</cmath> | + | <cmath>a_1+2a_2+\cdots+na_n=0</cmath> |
for the other equations will be automatically true. | for the other equations will be automatically true. |
Revision as of 21:32, 2 May 2012
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
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.
In the following construction, I am using Bertrand's Theorem ([1]) without proof. The Theorem states that, for any integer , there exists a prime
such that
.
In other words, 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
So, for , let the largest two primes not larger than
are
and
, and that
. By the Theorem stated above, one can conclude that
, and that
. Using this fact, I'm going to construct the sequence
.
Let . There will be three cases:
Case (i). If , then let
for all prime numbers
, and
, then (*) becomes:
Case (ii). If but
, then let
, and
for all prime numbers
, and
, then (*) becomes:
or
Case (iii). If , let
,
, and
for all prime numbers
, and
, then (*) becomes:
or
In each case, there exists non zero integers and
which satisfy the equation. Then for other primes
, just let
(or any number).
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)
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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 |