[color=blue][size=150]PERU TST IMO - 2006[/size]
Saturday, may 20.[/color]

[b]Question 02[/b]
Find all pairs $(a,b)$ real positive numbers $a$ and $b$ such that:

$[a[bn]]= n-1,$

for all $n$ positive integer.
Note: [x] denotes the integer part of $x$.

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[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=88510]Spanish version[/url]

$\text{\LaTeX}{}$ed by carlosbr
To solve the problem, we need to find all pairs \((a, b)\) of real positive numbers such that the equation \([a[bn]] = n - 1\) holds for all positive integers \(n\). Here, \([x]\) denotes the integer part of \(x\).

1. **Express the given equation in terms of inequalities:**
   \[
   [a[bn]] = n - 1
   \]
   This implies:
   \[
   n - 1 \leq a[bn] < n
   \]

2. **Analyze the inner floor function \([bn]\):**
   Since \([bn]\) is the integer part of \(bn\), we have:
   \[
   bn - 1 < [bn] \leq bn
   \]

3. **Substitute \([bn]\) into the inequality:**
   \[
   n - 1 \leq a(bn - 1) < n
   \]
   Simplifying the inequality:
   \[
   n - 1 \leq abn - a < n
   \]

4. **Separate the inequality into two parts:**
   \[
   n - 1 \leq abn - a
   \]
   \[
   abn - a < n
   \]

5. **Solve the first part of the inequality:**
   \[
   n - 1 \leq abn - a
   \]
   Adding \(a\) to both sides:
   \[
   n - 1 + a \leq abn
   \]
   Dividing by \(n\):
   \[
   \frac{n - 1 + a}{n} \leq ab
   \]
   As \(n\) approaches infinity, the left-hand side approaches 1:
   \[
   1 \leq ab
   \]

6. **Solve the second part of the inequality:**
   \[
   abn - a < n
   \]
   Adding \(a\) to both sides:
   \[
   abn < n + a
   \]
   Dividing by \(n\):
   \[
   ab < 1 + \frac{a}{n}
   \]
   As \(n\) approaches infinity, the right-hand side approaches 1:
   \[
   ab \leq 1
   \]

7. **Combine the results:**
   From the two inequalities, we have:
   \[
   1 \leq ab \leq 1
   \]
   Therefore:
   \[
   ab = 1
   \]

8. **Determine the values of \(a\) and \(b\):**
   Since \(ab = 1\), we can express \(b\) in terms of \(a\):
   \[
   b = \frac{1}{a}
   \]

9. **Verify the solution:**
   Substitute \(b = \frac{1}{a}\) into the original equation:
   \[
   [a \left\lfloor \frac{n}{a} \right\rfloor] = n - 1
   \]
   Since \(\left\lfloor \frac{n}{a} \right\rfloor\) is an integer, let \(k = \left\lfloor \frac{n}{a} \right\rfloor\). Then:
   \[
   a k = n - 1
   \]
   This holds true for all positive integers \(n\) if \(a\) is irrational and \(0 < a < 1\).

Therefore, the pairs \((a, b)\) that satisfy the given equation are:
\[
(a, b) = (a, \frac{1}{a}) \text{ where } a \text{ is an irrational number and } 0 < a < 1.
\]

The final answer is \( \boxed{ (a, b) = (a, \frac{1}{a}) \text{ where } a \text{ is an irrational number and } 0 < a < 1. } \)