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2012 USAMO Problems/Problem 3

Revision as of 09:54, 11 July 2020 by Tigershark22 (talk | contribs) (→‎Solution 2 (Bezout))

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

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)

Solution 2 (Bezout)

Motivation: The condition that it must work for all positive integers $k$ is annoying. Thus, we construct the sequence so that the integers $a_k$ through $a_{nk}$ are all a multiple of the original $a_1$ through $a_n$.

I claim that when $n\geq 3$ there exists an infinite sequence $a_i$ satisfying the given condition.

n=2: Since $a_1=(-2)^k*a_k$, we have that $|a_1|$ has no upper bound and thus no sequence exists.

n=4: Let $a_i=(-1)^{j+k}$ where $j$ is the number of factors of 2 in $i$ and $k$ is the number of factors of 3 in $i$

n=odd: Let $a_i=(\frac{-n+1}{2})^j$ where $j$ is the number of factors of $n$ in $i$ (that is, the maximum number $j$ such that $\frac{i}{n^j}$ is an integer).

n=even$>$4: Let $b=n-1$.We have $b>n/2$, so $b$ has no multiples except for itself in the numbers 1 through n. Also we get $\gcd(b,n)=\gcd(1,n)=1$.

By Bezout we have $x*n+y*b=1$ for nonzero integer $x, y$. Then let $(n+b-n(n+1)/2)x=x'$ and similarly define $y'$. Now let $a_i=x'^j*y'^k$ where $j$ is the number of factors of $n$ in $i$ and $k$ is the number of factors of $b$ in $i$.

We can check that this indeed satisfies the problem conditions. For $n=4$, note that all the $(a_k, a_{2k}, a_{3k}, a_{4k})$ are $(1,-1,-1,1)$, and $1-2-3+4=0$.

Similarly, for odd $n$, we have $(a_k,a_{2k},...a_{nk})$ is $(c, c, c,...,\frac{-n+1}{2}c)$ for some c by using the construction, for which $S=a_k+2a_{2k}+...+na_{nk}$ adds up to $S=cn(n-1)/2-cn(n-1)/2=0$.

Note that for even $n$, we have $(a_k,a_{2k},...a_{nk})$ is $(c, c, c,..., y'c, x'c)$ or $(c, c, c,..., y'c, c, c, x'c)$ for some c by construction. We can check that this adds up to $S=c(n(n+1)/2-n-b+nx'+by')=c(n(n+1)/2-n-b+n+b-n(n+1)/2)=0$.

-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

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