Bulgarian Mathematical Olympiad

Solution:
1. We first prove the following lemma.

LEMMA. If $p, q, r$ and $\sqrt{p} + \sqrt{q} + \sqrt{r}$ are rational numbers then $\sqrt{p}$, $\sqrt{q}$ and $\sqrt{r}$ are also rational numbers.

Proof of the lemma. Let $\sqrt{p} + \sqrt{q} + \sqrt{r} = s$, where $pqr \neq 0$ and $s$ is a rational number. Then $\sqrt{p} + \sqrt{q} = s - \sqrt{r}$ and we get by squaring that
$$p + q + 2\sqrt{pq} = s^2 + r - 2s\sqrt{r} \Longleftrightarrow 2\sqrt{pq} = s^2 + r - p - q - 2s\sqrt{r}$$
Squaring the last identity gives $4pq = M^2 + 4s^2 r - 4Ms\sqrt{r}$, where $M = s^2 + r - p - q > 0$. Therefore $\sqrt{r}$ is rational and we see in the same way that $\sqrt{p}$ and $\sqrt{q}$ are rational. This completes the proof of the lemma.

Let the positive integers $x, y$ and $z$ have the required property. Then the lemma implies that $\sqrt{\frac{2005}{x+y}}$, $\sqrt{\frac{2005}{x+z}}$ and $\sqrt{\frac{2005}{y+z}}$ are rational numbers. Set $\sqrt{\frac{2005}{x+y}} = \frac{a}{b}$, where $a$ and $b$ are coprime positive integers. Then $2005 b^2 = (x+y) a^2$ and it follows that $a^2$ divides $2005$. Hence $a = 1$ and therefore $x + y = 2005 b^2$. In the same way we obtain $x + z = 2005 c^2$ and $y + z = 2005 d^2$, where $c$ and $d$ are positive integers. Then
$$\sqrt{\frac{2005}{x+y}} + \sqrt{\frac{2005}{x+z}} + \sqrt{\frac{2005}{y+z}} = \frac{1}{b} + \frac{1}{c} + \frac{1}{d}$$
is a positive integer. Since $b, c$ and $d$ are positive integers, we have
$$1 \leq \frac{1}{b} + \frac{1}{c} + \frac{1}{d} \leq 3$$
If $\frac{1}{b} + \frac{1}{c} + \frac{1}{d} = 3$, then $b = c = d = 1$ and the system $x + y = x + z = y + z = 2005$ has no solutions in positive integers.

If $\frac{1}{b} + \frac{1}{c} + \frac{1}{d} = 2$, then one of the numbers $b, c$ and $d$ is equal to $1$ and the other two are equal to $2$. Again, the system for $x, y$ and $z$ has no solutions in positive integers.

It remains to consider the case $\frac{1}{b} + \frac{1}{c} + \frac{1}{d} = 1$. Suppose that $b \geq c \geq d > 1$. Then $\frac{3}{d} \geq 1$ and therefore $d = 2$ or $d = 3$. If $d = 3$ we get $b = c = 3$, and if $d = 2$ we have $\frac{1}{b} + \frac{1}{c} = \frac{1}{2}$. This equation has two solutions: $b = 3, c = 6$ and $b = c = 4$.

The inspection shows that the system for $x, y$ and $z$ has a solution in positive integers only when $d = 2, b = c = 4$ and in this case $x = 14 \cdot 2005$, $y = z = 2 \cdot 2005$. Therefore the solutions of the problem are all triples $(x, y, z)$ in which two numbers are equal to $2 \cdot 2005$ and the third one is equal to $14 \cdot 2005$.