# 2006 USAMO Problems/Problem 1

## Problem

(Kiran Kedlaya) 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 $x$.

## Solutions

### Solution 1

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:

$$\begin{array}{ccccc} \mbox{Row 1}& s& 2s&\hdots& ks\\ \mbox{Row 2}& s-1& 2s-1& \hdots&ks-1\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ \mbox{Row }s& 1& s+1& \hdots& (k-1)s+1 \end{array}$$

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 and $r. Since $s\not|(p-1)$, $1.

Note that for $0, $ls\ge s\mod p$. Thus, the first integer multiplied by s modulo p that will be less than s is $j+1.$ $(j+1)s\equiv s-r-1\mod p$. Since $\lbrace s,2s,\hdots,js,(j+1)s,\hdots,xs,\hdots,(p-1)s \rbrace$ is a complete residue system mod p with the exception of the 0 term, we can find an $x>j+1$ such that $xs \equiv s-1 \mod p$.

Thus, choose $m=j+1$ and $n=x$ to complete the proof.