Exercise 1.3.46* (The sum of each triplet is divisible by each member of the triplet.)

Let $k \geq 3$ be a fixed integer. Find all sets $a_{1}, a_{2}, \dots, a_{k}$ of positive integers such that the sum of any triplet is divisible by each member of the triplet.

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Since ${a_{1}, \dots, a_{k}}$ is a set, there is no repetition, and we can assume

a_{1} < a_{2} < \dots < a_{k} .

If I understand the problem correctly, for a fixed $k$ , we must find a set ${a_{1}, \dots, a_{k}}$ , verifying for all $i_{1}, i_{2}, i_{3} \in [[1, k]]$ , such that $i_{1} < i_{2} < i_{3}$ ,

a_{i_{1}} ∣ a_{i_{1}} + a_{i_{2}} + a_{i_{3}}, a_{i_{2}} ∣ a_{i_{1}} + a_{i_{2}} + a_{i_{3}}, a_{i_{3}} ∣ a_{i_{1}} + a_{i_{2}} + a_{i_{3}} .

Proof. If $k = 3$ , we must find all triplets $(a, b, c) \in ℕ *$ , with $0 < a < b < c$ , such that

\begin{align} a ∣ a + b + c, b ∣ a + b + c, c ∣ a + b + c . & (1) \end{align}

There are obvious solutions, such as $(1, 2, 3)$ , or more generally $(a, 2 a, 3 a)$ for $a \in ℕ^{∗}$ . Indeed, $a ∣ 6 a, 2 a ∣ 6 a, 4 a ∣ 6 a$ . We prove first that there is no other solution.

The relations (1) are equivalent to

\begin{align} a ∣ b + c, b ∣ a + c, c ∣ a + b . & (2) \end{align}

Therefore, there exist positive integers $λ, μ, ν$ such that

\begin{align} b + c = λa, a + c = μb, a + b = νc . & (3) \end{align}

Using $0 < a < b < c$ , we obtain $νc = a + b < 2 c$ , thus $ν < 2$ . Moreover $1 \leq ν$ , so $ν = 1$ , and

a + b = c .

Therefore (2) is equivalent to

a + b = c, a ∣ 2 b, b ∣ 2 a .

There are integers $u, v \in ℕ^{∗}$ such that

2 b = ua, 2 a = vb,

thus $4 b = uvb$ , where $b \neq 0$ , so

4 = uv,

where $2 b = ua < ub$ , therefore $u > 2$ . Hence $u = 4, v = 1$ . This gives

b = 2 a .

Then $(a, b, c) = (a, 2 a, 3 a)$ . The only solutions are $(a, 2 a, 3 a)$ , where $a \in ℕ^{∗}$ .

We consider now the case $k = 4$ :

{a_{1}, a_{2}, a_{3}, a_{4}} = {a, b, c, d}, where a < b < c < d .

Then $(a, b, c)$ and $(b, c, d)$ are the triplets studied in the first part, so

(a, b, c) = (a, 2 a, 3 a), (b, c, d) = (b, 2 b, 3 b) (a, b \in ℕ^{∗}) .

Therefore $b = 2 a$ and $2 b = 3 a$ , thus $4 a = 3 a$ , $4 = 3$ : this is a contradiction.

There is no solution for $k = 4$ . A fortiori, there is no solution for $k > 4$ .

The only solutions are ${a, 2 a, 3 a}$ , where $a$ is a positive integer. □

richardganaye

2024-07-15 15:29

Exercise 1.3.46* (The sum of each triplet is divisible by each member of the triplet.)

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