Problem 1. Find all triples $(x, y, z)$ of positive integers such that

$$
\sqrt{\frac{2005}{x+y}}+\sqrt{\frac{2005}{x+z}}+\sqrt{\frac{2005}{y+z}}
$$

is a positive integer.

Oleg Mushkarov
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 $p q r \neq 0$ and $s$ is a rational number. Then $\sqrt{p}+\sqrt{q}=s-\sqrt{r}$ and we get by squaring that

$$
p+q+2 \sqrt{p q}=s^{2}+r-2 s \sqrt{r} \Longleftrightarrow 2 \sqrt{p q}=s^{2}+r-p-q-2 s \sqrt{r}
$$

Squaring the last identity gives $4 p q=M^{2}+4 s^{2} r-4 M s \sqrt{r}$, 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 $x=14.2005$, $y=z=2.2005$. Therefore the solutions of the problem are all triples $(x, y, z)$ in which two numbers are equal to 2.2005 and the third one is equal to 14.2005.