2012 USAMO Problems/Problem 3
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.