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

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m (Changed to "Bertrand's Postulate", as "Bertrand's Theorem" refers to something unrelated)
 
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==Solution that involves a non-elementary result==
 
==Solution that involves a non-elementary result==
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(Since Bertrand's is well known and provable using elementary techniques, I see nothing wrong with this-tigershark22)
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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.
  
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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>.
  
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To prove this, I am going to use '''Bertrand's Theorem''' ([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>.
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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>.
  
  
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--[[User:Lightest|Lightest]] 21:24, 2 May 2012 (EDT)
 
--[[User:Lightest|Lightest]] 21:24, 2 May 2012 (EDT)
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==Solution 2 (Bezout)==
 
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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>.
 
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I claim that when <math>n\geq 3</math> there exists an infinite sequence <math>a_i</math> satisfying the given condition.
 
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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.
 
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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).
 
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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>.
 
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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>.
 
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We can check that this indeed satisfies the problem conditions.
 
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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>.
 
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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>.
 
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-tigershark22
 
  
 
==See Also==
 
==See Also==

Latest revision as of 05:38, 27 October 2022

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 Postulate ([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)

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

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