Difference between revisions of "2012 USAMO Problems/Problem 3"
(→See Also) |
Tigershark22 (talk | contribs) |
||
Line 59: | Line 59: | ||
--[[User:Lightest|Lightest]] 21:24, 2 May 2012 (EDT) | --[[User:Lightest|Lightest]] 21:24, 2 May 2012 (EDT) | ||
− | {{ | + | ==Solution 2 (Bezout)== |
+ | |||
+ | I claim that when <math>n\geq 3</math> there exists an infinite sequence <math>a_i</math> satisfying the given condition. | ||
+ | |||
+ | n=2: Since <math>a_1=(-2)^k*a_k</math>, we have that <math>|a_1|</math> has no upper bound and thus no sequence exists. | ||
+ | |||
+ | n=4: Let <math>a_i=(-1)^{j+k}</math> where <math>j</math> is the number of factors of 2 in <math>i</math> and <math>k</math> is the number of factors of 3 in <math>i</math> | ||
+ | |||
+ | n=odd: Let <math>a_i=(\frac{-n+1}{2})^j</math> where <math>j</math> is the number of factors of <math>n</math> in <math>i</math>. | ||
+ | |||
+ | n=even<math>></math>4: Let <math>b</math> be either <math>n-3</math> or <math>n-1</math> such that <math>b</math> is not 0 mod 3. For <math>n=6</math> we have <math>b=5>n/2</math> and for even <math>n>6</math> we have <math>b\geq n-3>n/2</math>, so <math>b</math> has no multiples except for itself in the numbers 1 through n. Then we get <math>\gcd(b,3n/2)=1</math> and by Bezout we have <math>x*3n/2+y*b=1</math> for nonzero integer <math>x, y</math>. Then let <math>(3n/2+b-n(n+1)/2)x=x'</math> and similarly define <math>y'</math>. Now let <math>a_i=x'^j*y'^k</math> where <math>j</math> is the number of factors of <math>n/2</math> in <math>i</math> and <math>k</math> is the number of factors of <math>b</math> in <math>i</math>. | ||
+ | |||
+ | We can check that this indeed satisfies the problem conditions. For <math>n=4</math>, note that all the <math>(a_k, a_{2k}, a_{3k}, a_{4k})</math> are <math>(1,-1,-1,1)</math>, and <math>1-2-3+4=0</math>. Similarly, for odd <math>n</math>, we have <math>(a_k,a_{2k},...a_{nk})</math> is <math>(c, c, c,...,\frac{-n+1}{2}c)</math> which adds up to <math>cn(n-1)/2-cn(n-1)/2=0</math>. Note that <math>(a_k,a_{2k},...a_{nk})</math> is <math>(c, c, c,..., x'c, c, c,..., y', x'c)</math> or <math>(c, c, c,..., x'c, c, c,..., y', c, c, x'c)</math> depending on whether <math>b</math> is <math>n-3</math> or <math>n-1</math>, and where the first <math>x'c</math> is located at <math>a_{\frac{nk}{2}}</math>. We can check that this adds up to <math>c(n(n+1)/2-n/2-n-b+3nx'/2+by')=c(n(n+1)/2-3n/2-b+3n/2+b-n(n+1)/2)=0</math>. | ||
+ | |||
+ | -tigershark22 | ||
==See Also== | ==See Also== |
Revision as of 22:16, 10 July 2020
Contents
[hide]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.
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 Theorem ([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)
Solution 2 (Bezout)
I claim that when there exists an infinite sequence satisfying the given condition.
n=2: Since , we have that has no upper bound and thus no sequence exists.
n=4: Let where is the number of factors of 2 in and is the number of factors of 3 in
n=odd: Let where is the number of factors of in .
n=even4: Let be either or such that is not 0 mod 3. For we have and for even we have , so has no multiples except for itself in the numbers 1 through n. Then we get and by Bezout we have for nonzero integer . Then let and similarly define . Now let where is the number of factors of in and is the number of factors of in .
We can check that this indeed satisfies the problem conditions. For , note that all the are , and . Similarly, for odd , we have is which adds up to . Note that is or depending on whether is or , and where the first is located at . We can check that this adds up to .
-tigershark22
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.