Difference between revisions of "2012 USAMO Problems/Problem 3"

Revision as of 10:10, 11 July 2020

Problem

Determine which integers $n > 1$ have the property that there exists an infinite sequence $a_1$ , $a_2$ , $a_3$ , $\dots$ of nonzero integers such that the equality $a_k + 2a_{2k} + \dots + na_{nk} = 0$ holds for every positive integer $k$ .

Partial Solution

For $n$ equal to any odd prime $p$ , the sequence $\left\{a_i = \left(\frac{1-n}{2}\right)^{m_p\left(i\right)}\right\}$ , where $p^{m_p\left(i\right)}$ is the greatest power of $p$ that divides $i$ , gives a valid sequence. Therefore, the set of possible values for $n$ 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 $n=2$ , $|a_1| = 2 |a_2| = \cdots = 2^m |a_{2^m}|$ implies that for any positive integer $m$ , $|a_1| \ge 2^m$ , which is impossible.

We proceed to prove that the infinite sequence exists for all $n\ge 3$ .

First, one notices that if we have $a_{xy} = a_x a_y$ for any integers $x$ and $y$ , then it is suffice to define all $a_x$ for $x$ prime, and one only needs to verify the equation (*)

$a_1+2a_2+\cdots+na_n=0$

for the other equations will be automatically true.

To proceed with the construction, I need the following fact: for any positive integer $m>2$ , there exists a prime $p$ such that $\frac{m}{2} <p \le m$ .

To prove this, I am going to use Bertrand's Theorem ([1]) without proof. The Theorem states that, for any integer $n>1$ , there exists a prime $p$ such that $n<p\le 2n-1$ . In other words, for any positive integer $m>2$ , if $m=2n$ with $n>1$ , then there exists a prime $p$ such that $\frac{m}{2} < p < m$ , and if $m=2n-1$ with $n>1$ , then there exists a prime $p$ such that $\frac{m+1}{2} <p\le m$ , both of which guarantees that for any integer $m>2$ , there exists a prime $p$ such that $\frac{m}{2} <p \le m$ .

Go back to the problem. Suppose $n\ge 3$ . Let the largest two primes not larger than $n$ are $P$ and $Q$ , and that $n\ge P > Q$ . By the fact stated above, one can conclude that $2P > n$ , and that $4Q = 2(2Q) \ge 2P > n$ . Let's construct $a_n$ :

Let $a_1=1$ . There will be three cases: (i) $Q>\frac{n}{2}$ , (ii) $\frac{n}{2} \ge Q > \frac{n}{3}$ , and (iii) $\frac{n}{3} \ge Q > \frac{n}{4}$ .

Case (i): $2Q>n$ . Let $a_x = 1$ for all prime numbers $x<Q$ , and $a_{xy}=a_xa_y$ , then (*) becomes:

$Pa_P + Qa_Q = C_1$

Case (ii): $2Q\le n$ but $3Q > n$ . In this case, let $a_2=-1$ , and $a_x = 1$ for all prime numbers $2<x<Q$ , and $a_{xy}=a_xa_y$ , then (*) becomes:

$Pa_P + Qa_Q - Qa_{2Q} = C_2$

or

$Pa_P - Qa_Q = C_2$

Case (iii): $3Q\le n$ . In this case, let $a_2=3$ , $a_3=-2$ , and $a_x = 1$ for all prime numbers $3<x<Q$ , and $a_{xy}=a_xa_y$ , then (*) becomes:

$Pa_P + Qa_Q + 3Qa_{2Q} - 2Qa_{3Q} = C_3$

or $Pa_P + Qa_Q = C_3$

In each case, by Bezout's Theorem, there exists non zero integers $a_P$ and $a_Q$ which satisfy the equation. For all other primes $p > P$ , just let $a_p=1$ (or any other non-zero integer).

This construction is correct because, for any $k> 1$ ,

$a_k + 2 a_{2k} + \cdots n a_{nk} = a_k (1 + 2 a_2 + \cdots n a_n ) = 0$

Since Bertrand's Theorem is not elementary, we still need to wait for a better proof.

--Lightest 21:24, 2 May 2012 (EDT)

@@ Line 9: / Line 9: @@
 ==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.
@@ Line 58: / Line 61: @@
 --[[User:Lightest|Lightest]] 21:24, 2 May 2012 (EDT)
-==Solution 2 (Bezout)==
-Motivation: The condition that it must work for all positive integers <math>k</math> is annoying. Thus, we construct the sequence so that the integers <math>a_k</math> through <math>a_{nk}</math> are all a multiple of the original <math>a_1</math> through <math>a_n</math>.
-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=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> (that is, the maximum number <math>j</math> such that <math>\frac{i}{n^j}</math> is an integer).
-n=even<math>></math>2: Let <math>b=n-1</math>.We have <math>b>n/2</math>, so <math>b</math> has no multiples except for itself in the numbers 1 through n. Also we get <math>\gcd(b,n)=\gcd(1,n)=1</math>.
-By Bezout we have <math>x*n+y*b=1</math> for nonzero integer <math>x, y</math>. Then let <math>(n+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</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 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> for some c by using the construction, for which <math>S=a_k+2a_{2k}+...+na_{nk}</math> adds up to <math>S=cn(n-1)/2-cn(n-1)/2=0</math>.
-Note that for even <math>n</math>, we have <math>(a_k,a_{2k},...a_{nk})</math> is <math>(c, c, c,..., y'c, x'c)</math> for some c by construction. We can check that this adds up to <math>S=c(n(n+1)/2-n-b+nx'+by')=c(n(n+1)/2-n-b+n+b-n(n+1)/2)=0</math>.
--tigershark22
 ==See Also==

2012 USAMO (Problems • Resources)
Preceded by Problem 2	Followed by Problem 4
1 • 2 • 3 • 4 • 5 • 6
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Difference between revisions of "2012 USAMO Problems/Problem 3"

Revision as of 10:10, 11 July 2020

Contents

Problem

Partial Solution

Solution that involves a non-elementary result

See Also