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

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)

@@ Line 4: / Line 4: @@
 holds for every positive integer <math>k</math>.
-==Solution==
+==Partial Solution==
 For <math>n</math> equal to any odd prime <math>p</math>, the sequence <math>\left\{a_i = \left(\frac{1-n}{2}\right)^{m_p\left(i\right)}\right\}</math>, where <math>p^{m_p\left(i\right)}</math> is the greatest power of <math>p</math> that divides <math>i</math>, gives a valid sequence. Therefore, the set of possible values for <math>n</math> is at least the set of odd primes.
-{{solution}}
+==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.
+We proceed to prove that the infinite sequence exists for all <math>n\ge 3</math>.
+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=0</cmath>
+for the other equations will be automatically true.
+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 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>.
+Go back to the problem. Suppose <math>n\ge 3</math>. Let the largest two primes not larger than <math>n</math> are <math>P</math> and <math>Q</math>, and that <math>n\ge P > Q</math>. By the fact stated above, one can conclude that <math>2P > n</math>, and that <math>4Q = 2(2Q) \ge 2P > n</math>. Let's construct <math>a_n</math>:
+Let <math>a_1=1</math>. There will be three cases: (i) <math>Q>\frac{n}{2}</math>,  (ii) <math>\frac{n}{2} \ge Q > \frac{n}{3}</math>, and (iii) <math>\frac{n}{3} \ge Q > \frac{n}{4}</math>.
+Case (i): <math>2Q>n</math>. Let <math>a_x = 1</math> for all prime numbers <math>x<Q</math>, and <math>a_{xy}=a_xa_y</math>, then (*) becomes:
+<cmath> Pa_P + Qa_Q = C_1</cmath>
+Case (ii): <math>2Q\le n</math> but <math>3Q > n</math>. In this case, let <math>a_2=-1</math>, and <math>a_x = 1</math> for all prime numbers <math>2<x<Q</math>, and <math>a_{xy}=a_xa_y</math>, then (*) becomes:
+<cmath> Pa_P + Qa_Q - Qa_{2Q} = C_2 </cmath>
+or
+<cmath> Pa_P - Qa_Q = C_2</cmath>
+Case (iii): <math>3Q\le n</math>. In this case, let <math>a_2=3</math>, <math>a_3=-2</math>, and <math>a_x = 1</math> for all prime numbers <math>3<x<Q</math>, and <math>a_{xy}=a_xa_y</math>, then (*) becomes:
+<cmath> Pa_P + Qa_Q + 3Qa_{2Q} - 2Qa_{3Q} = C_3 </cmath>
+or
+<cmath> Pa_P + Qa_Q = C_3</cmath>
+In each case, by Bezout's Theorem, there exists non zero integers <math>a_P</math> and <math>a_Q</math> which satisfy the equation. For all other primes <math>p > P</math>, just let <math>a_p=1</math> (or any other non-zero integer).
+This construction is correct because, for any <math>k> 1</math>,
+<cmath>a_k + 2 a_{2k} + \cdots n a_{nk} =  a_k (1 + 2 a_2 + \cdots n a_n ) = 0 </cmath>
+Since Bertrand's Theorem is not elementary, we still need to wait for a better proof.
+--[[User:Lightest|Lightest]] 21:24, 2 May 2012 (EDT)
 ==See Also==
@@ Line 13: / Line 66: @@
 {{USAMO newbox|year=2012|num-b=2|num-a=4}}
+{{MAA Notice}}
+[[Category:Olympiad Number Theory Problems]]

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

Latest revision as of 05:38, 27 October 2022

Contents

Problem

Partial Solution

Solution that involves a non-elementary result

See Also