3. find all triples $(a, b, c)$ of natural numbers such that

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

are also natural numbers.

## Solution 1:
We distinguish three cases, namely that the three numbers are equal, that two of the three numbers are equal and that the three numbers are different.

Case 1: $a=b=c$

This results in the solution $(a, a, a)$.

Case 2: We have two identical and one different number.

Assume that $a=b \neq c$. Substituting leads to $a \mid a+c$. We get $a \mid c$. $c$ is therefore a multiple of $a$. Inserting also leads to $c \mid a+a$, i.e. $c \mid 2 a$. Thus $c=a$ or $c=2 a$ must apply. We looked at the first option in case 1, the second option provides the solution $(a, a, 2 a)$.

Case 3: We have three different numbers.

Assume that $a<b<c$ applies. In particular, then $a+b<2 c$ holds. We also know that $c \mid a+b$. This is only possible if $c=a+b$ holds. We can therefore rewrite the condition $b \mid c+a$ to $b \mid a+b+a$, from which $b \mid 2 a$ follows. Since the assumption $a<b$ applies, this is only possible if we have $b=2 a$. From this follows $c=3 a$ and we get the solution $(a, 2 a, 3 a)$.

In total, we therefore obtain the solutions $(a, a, a),(a, a, 2 a),(a, 2 a, 3 a)$ and the symmetrical permutations thereof, where $a$ may be any natural number. Substitution ensures that each of these triples is actually a solution.

## Solution 2:

Suppose that $c \geq b \geq a$. We now know that $c \mid a+b$ and $a+b \leq 2 c$. It follows that either $c=a+b$ or $2 c=a+b$. We consider these two cases individually.

Case 1: $2 c=a+b$

Since $a \leq b \leq c$ applies, this is only possible for $a=b=c$. We therefore get the solution $(a, a, a)$.

Case 2: $c=a+b$

$b \mid a+c$ applies, i.e. $b \mid a+a+b$ and therefore $b \mid 2 a$. Since $b \geq a$, $b=a$ or $b=2 a$ must apply. This leads to the solutions $(a, a, 2 a)$ and $(a, 2 a, 3 a)$.

In total, we therefore obtain the solutions $(a, a, a),(a, a, 2 a),(a, 2 a, 3 a)$ and the symmetrical permutations thereof, where $a$ may be any natural number. Substitution ensures that each of these triples is actually a solution.