4. determine all positive integer solutions of the equation

$$
\operator name{kgV}(a, b, c)=a+b+c
$$

## 1st solution:
We can assume $a \leq b \leq c$. Then $a2$, because otherwise the left-hand side is greater than $1 / 2$, on the other hand $y \leq 4$, because otherwise the left-hand side is at most $2 / 5$. This results in the solutions $x=6, y=3$ and $x=y=4$. In the first case, $(a, b, c)=(a, 2 a, 3 a)$, in the second case $(a, b, c)=(a, a, 2 a)$. However, the latter is not possible, because then $\operatorname{kgV}(a, b, c)=c \neq 2 c$. On the other hand, it is easy to calculate that in the first case $\operatorname{kgV}(a, b, c)=6 a=a+b+c$. The triples we are looking for are therefore exactly the permutations of

$$
(a, b, c)=(a, 2 a, 3 a), \quad a \geq 1
$$

## 2nd solution:

We use the following equations:

$$
\operatorname{ggT}(m a, m b, m c)=m \cdot \operatorname{ggT}(a, b, c), \quad \operatorname{kgV}(m a, m b, m c)=m \cdot \operatorname{kgV}(a, b, c)
$$

Let $d=\operator name{ggT}(a, b, c)$ and set $a=x d, b=y d$ and $c=z d$. Then $x, y, z$ are divisors and inserting them into the equation with the above formula yields

$$
\operator name{kgV}(x, y, z)=x+y+z
$$

Now $x, y, z$ are even pairwise divisors, because each divisor of two of the numbers also divides the left side of this equation and thus also the third number, i.e. is equal to 1. From this follows $\operator name{kgV}(x, y, z)=x y z$ and we must find all solutions of the equation
$x y z=x+y+z$. To do this, we assume that $x3 z>x+y+z$, a contradiction. Therefore, $x=1$ and the equation simplifies $\mathrm{to}$

$$
y z=1+y+z \quad \Longleftrightarrow \quad(y-1)(z-1)=2
$$

The only solution is therefore $y=2, z=3$ and the triples $(a, b, c)$ are exactly the permutations of $(d, 2 d, 3 d)$ with $d \geq 1$.

## 3rd solution:

We first introduce some notations. Set $d=\operatorname{ggT}(a, b, c)$ and $x=\operatorname{ggT}(b / d, c / d)$, $y=\operatorname{ggT}(c / d, a / d), z=\operatorname{ggT}(a / d, b / d)$. Then $x, y$ and $z$ are pairwise divisors by definition of $d$. In addition, $a / d$ is divisible by $y$ and $z$, i.e. also by their product, so $A=a /(d x y)$ is an integer. Define $B$ and $C$ analogously, then $\operatorname{ggT}(x, A)=\operatorname{ggT}(y, B)=\operatorname{ggT}(z, C)=1$ and we can write $a=d y z A$, $b=d z x B$ and $c=d x y C$. In addition, $\operatorname{kgV}(a, b, c)=d x y z A B C$. Substituting into the equation and simplifying gives

$$
x y z A B C=y z A+z x B+x y C
$$

Now $x$ divides the left side, therefore also the right side, therefore also $y z A$. However, since $x$ is not related to all three factors, $x=1$ follows. Similarly, $y=z=1$ and after division with $A B C$ the equation now reads

$$
\frac{1}{A B}+\frac{1}{B C}+\frac{1}{C A}=1
$$

We can assume $A \leq B \leq C$. If $A>1$, then the left-hand side is at most equal to $3 / 4$, i.e. $A=1$. If $B=1$, then the LS is greater than 1, if $B \geq 3$, then the LS is at most equal to $1 / 3+1 / 3+1 / 9<1$. Therefore $B=2$ and thus $C=3$. Except for one permutation, the triples we are looking for are all of the form $(a, b, c)=(d, 2 d, 3 d)$, as in the first solution.

## Remarks and scoring scheme: