1. Find all triples of integers $(a, b, c)$ such that each of the fractions

$$
\frac{a}{b+c}, \quad \frac{b}{c+a}, \quad \frac{c}{a+b}
$$

has an integer value.
1. The considered triplet of fractions is symmetric in the sense that if we replace the triplet of integers $(a, b, c)$ with any permutation of them, we get the same (up to order) triplet of fractions. Similarly, if we replace the numbers $a, b, c$ with their opposites. This fact will facilitate the following case analysis.

Assume, then, that the numbers $a, b, c$ are such that all three considered fractions have integer values. If any of them is zero, it suffices to consider the case $a=0$ without loss of generality. Substituting into the considered fractions, we get that the fractions $b / c$ and $c / b$ have integer values. Hence, $b$ and $c$ are non-zero and $|b| \geq |c|$ and simultaneously $|c| \geq |b|$, so $c = \pm b$. Additionally, the number $b+c$ is the denominator of the first fraction, so $b+c \neq 0$, thus $b=c$. Overall, we get (obviously valid) triplets $(0, c, c)$ and their permutations for any non-zero integer $c$.

It remains to solve the case where $a b c \neq 0$.

Given the observation from the first paragraph, we will assume that at least two of the numbers $a, b, c$ are positive. If all three were positive, the fraction with the smallest of the numbers $a, b, c$ in the numerator would lie between 0 and 1, so it could not have an integer value.

Let, then, $a, b$ be positive numbers and $c=-d$ for a positive $d$. Substituting into the problem, we get that the fractions

$$
\frac{a}{d-b}, \quad \frac{b}{d-a}, \quad \frac{d}{a+b}
$$

have integer values. From the last one, it is clear that $d \geq a+b$. Therefore, the first fraction has a positive denominator, and since its value is an integer, it must be $a \geq d-b$ or $d \leq a+b$. Thus, $d=a+b$ or $c=-a-b$, and we get, in summary, triplets $(a, b, c)$ of non-zero numbers for which $a+b+c=0$. All such triplets satisfy the condition, as the value of all three considered fractions is -1 for them.

Answer. The problem is satisfied by all triplets $(0, c, c), (c, 0, c)$, and $(c, c, 0)$, where $c$ is a non-zero integer, and all triplets $(a, b, c)$ of non-zero integers for which $a+b+c=0$.

For a complete solution, award 6 points. If any of the valid triplets is missing (e.g., another permutation of the triplet $(0, c, c)$ is not mentioned), award no more than 5 points. Incomplete solution: for a complete solution of the case where one of the numbers $a, b, c$ is 0, award 2 points; for solving the case where $a, b, c$ are non-zero, award 4 points, of which 1 point for merely excluding the cases where all numbers $a, b, c$ are positive or negative. For the observation in the first paragraph about permutations of triplets and changes in signs, if it does not lead to solving one of the two mentioned cases, do not award points. For finding all solutions (without justification that no others exist), award 1 point.