## Task 4 - 070934

Determine all ordered triples $(a, b, c)$ of natural numbers $a, b$, and $c$ for which

$$
\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1
$$

holds. Two triples $\left(a_{1}, b_{1}, c_{1}\right)$ and $\left(a_{2}, b_{2}, c_{2}\right)$ are called equal if and only if $a_{1}=a_{2}, b_{1}=b_{2}$, and $c_{1}=c_{2}$.
This task can be solved using case analysis.

Let $0 < a \leq b \leq c$ and $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 1$.

1. Case: $a = 1$

If $a = 1$, then $\frac{1}{a} = 1$, which contradicts (1). In this case, there is no solution.

2. Case: $a \geq 3$

If $c > 3$, then

$$
\frac{1}{a} + \frac{1}{b} + \frac{1}{c} \leq \frac{1}{3} + \frac{1}{3} + \frac{1}{c} < 1
$$

This is a contradiction. Therefore, $c \leq 3$ is not possible, and we must have $a = 2$.

If $a = 2$, then

$$
\frac{1}{2} + \frac{1}{b} + \frac{1}{c} = 1 \Leftrightarrow \frac{1}{b} + \frac{1}{c} = \frac{1}{2} \Leftrightarrow 2(b + c) = bc \Leftrightarrow 4 = bc - 2(b + c) + 4 = (b - 2)(c - 2)
$$

Since $b \geq a$, $b \geq 2$ must hold, and $b = 2$ is not possible because $4 \neq (2 - 2)(c - 2) = 0$.

$5 \leq b \leq c$ is also not possible, as then $(b - 2)(c - 2) \geq (5 - 2)(5 - 2) = 9 > 4$, which contradicts (2).

Therefore, only the cases $b = 3$ and $b = 4$ are possible.

3.1. Case: $b = 3$

(2) $\Leftrightarrow 4 = c - 2$, so $c = 6$. Indeed, $\frac{1}{2} + \frac{1}{3} + \frac{1}{6} = 1$.

3.2. Case: $b = 4$

(2) $\Leftrightarrow 4 = 2(c - 2)$, so $c = 4$. Indeed, $\frac{1}{2} + \frac{1}{4} + \frac{1}{4} = 1$.

Thus, there are exactly ten solution triples: $(3,3,3)$, $(2,3,6)$, and $(2,4,4)$, as well as the permutations of the 2nd and 3rd triples.

## Solution by Kitaktus