2006 USAMO Problems/Problem 1

From AoPSWiki

Revision as of 01:27, 14 April 2008 by Mathcrazed (Talk | contribs)

(diff) ← Older revision | Current revision (diff) | Newer revision → (diff)

Problem

Let $p$ be a prime number and let $s$ be an integer with $0 < s < p$ . Prove that there exist integers $m$ and $n$ with $0 < m < n < p$ and

$\left\{ \frac{sm}{p} \right\} < \left\{ \frac{sn}{p} \right\} < \frac{s}{p}$

if and only if $s$ is not a divisor of $p-1$ .

Note: For $x$ a real number, let $\lfloor x \rfloor$ denote the greatest integer less than or equal to $x$ , and let $\{x\} = x - \lfloor x \rfloor$ denote the fractional part of $x$ .

Solution

We proceed by contradiction. Assume that $s|(p-1)$ . Then for some positive integer $k$ , $sk=p-1$ . The conditions given are equivalent to stating that $sm \mod p < sn \mod p< s\mod p$ . Now consider the following array modulo p:

$\mbox{row 1}\qquad s\qquad 2s\qquad \hdots\qquad ks\\\mbox{row 2}\qquad s-1\qquad 2s-1\qquad \hdots\qquad ks-1\\\vdots\\\mbox...$

Obviously, there are $s$ rows and $k$ columns. The first entry of each row is simply $((r-1)k+1)s\mod p$ . Since we wish for $sm\mod p$ and $sn\mod p$ to both be in the first column, while also satisfying the given conditions, we can easily see that $sm$ must be in a row $m_r$ with $m_r>n_r$ , where $n_r$ denotes the row $sn \mod p$ is in. However, since the values of each entry decreases while $((r-1)k+1)s$ keeps increasing, we can see that the condition can never be satisfied and thus, $s\not|(p-1).$

To prove the other direction, let $sj+r=p-1$ for positive integers $j$ and $r$ with $j$ being the largest integer such that $sj<p-1$ and $r<s$ . Since $s\not|(p-1)$ , $1<s<p-1$ .