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

[b]Question 01[/b]

Find all $(x,y,z)$ positive integers, such that:

$\sqrt{\frac{2006}{x+y}} + \sqrt{\frac{2006}{y+z}} + \sqrt{\frac{2006}{z+x}},$

is an integer.

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[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=88509]Spanish version[/url]

$\text{\LaTeX}{}$ed by carlosbr
To solve the given problem, we need to find all positive integer solutions \((x, y, z)\) such that the expression 

\[
\sqrt{\frac{2006}{x+y}} + \sqrt{\frac{2006}{y+z}} + \sqrt{\frac{2006}{z+x}}
\]

is an integer. Let's denote this integer by \(k\). Therefore, we have:

\[
\sqrt{\frac{2006}{x+y}} + \sqrt{\frac{2006}{y+z}} + \sqrt{\frac{2006}{z+x}} = k
\]

1. **Expressing the terms as perfect squares:**
   For the sum of square roots to be an integer, each term \(\sqrt{\frac{2006}{x+y}}\), \(\sqrt{\frac{2006}{y+z}}\), and \(\sqrt{\frac{2006}{z+x}}\) must be rational. This implies that \(\frac{2006}{x+y}\), \(\frac{2006}{y+z}\), and \(\frac{2006}{z+x}\) must be perfect squares of rational numbers. Let:

   \[
   \frac{2006}{x+y} = r_1^2, \quad \frac{2006}{y+z} = r_2^2, \quad \frac{2006}{z+x} = r_3^2
   \]

   where \(r_1, r_2, r_3\) are rational numbers. This gives us:

   \[
   x + y = \frac{2006}{r_1^2}, \quad y + z = \frac{2006}{r_2^2}, \quad z + x = \frac{2006}{r_3^2}
   \]

2. **Solving for \(x, y, z\):**
   We can solve these equations for \(x, y, z\) in terms of \(r_1, r_2, r_3\):

   \[
   x = \frac{2006}{r_3^2} + \frac{2006}{r_1^2} - \frac{2006}{r_2^2}
   \]
   \[
   y = \frac{2006}{r_1^2} + \frac{2006}{r_2^2} - \frac{2006}{r_3^2}
   \]
   \[
   z = \frac{2006}{r_2^2} + \frac{2006}{r_3^2} - \frac{2006}{r_1^2}
   \]

3. **Ensuring \(x, y, z\) are positive integers:**
   Since \(x, y, z\) must be positive integers, the expressions \(\frac{2006}{r_1^2}\), \(\frac{2006}{r_2^2}\), and \(\frac{2006}{r_3^2}\) must be integers. This implies that \(r_1, r_2, r_3\) must be such that \(r_1^2, r_2^2, r_3^2\) are divisors of 2006. The prime factorization of 2006 is:

   \[
   2006 = 2 \times 17 \times 59
   \]

   Therefore, the possible values for \(r_1, r_2, r_3\) are such that \(r_1^2, r_2^2, r_3^2\) are divisors of 2006. The possible values for \(r_1, r_2, r_3\) are 1, \(\sqrt{2}\), \(\sqrt{17}\), \(\sqrt{59}\), \(\sqrt{34}\), \(\sqrt{118}\), \(\sqrt{1003}\), and \(\sqrt{2006}\).

4. **Checking integer solutions:**
   We need to check which combinations of \(r_1, r_2, r_3\) yield integer values for \(x, y, z\). After checking, we find the following valid combinations:

   - \(r_1 = r_2 = r_3 = 1\):
     \[
     x = y = z = \frac{2006}{1^2} = 2006
     \]

   - \(r_1 = 1, r_2 = r_3 = 2\):
     \[
     x = \frac{2006}{2^2} + \frac{2006}{1^2} - \frac{2006}{2^2} = 1003
     \]
     \[
     y = \frac{2006}{1^2} + \frac{2006}{2^2} - \frac{2006}{2^2} = 1003
     \]
     \[
     z = \frac{2006}{2^2} + \frac{2006}{2^2} - \frac{2006}{1^2} = 7021
     \]

   - \(r_1 = r_2 = r_3 = 3\):
     \[
     x = y = z = \frac{2006}{3^2} = 9027
     \]

Thus, the solutions are:

\[
(x, y, z) = (2006, 2006, 2006), (1003, 1003, 7021), (9027, 9027, 9027)
\]

The final answer is \( \boxed{ (2006, 2006, 2006), (1003, 1003, 7021), (9027, 9027, 9027) } \)