NMO Selection Tests for the Junior Balkan Mathematical Olympiad - Second Selection Test

Call good a number $n \in \mathbb{N}$ for which there exist $a, b, c \in \mathbb{N}_{\ge 1}$ such that
$$n = (a, b) \cdot (b, c) + (b, c) \cdot (c, a) + (c, a) \cdot (a, b).$$
We will prove that, regardless of the value of $M$, the good numbers are those of the form $n = 2^{2t}(2k+1)$, where $t, k \in \mathbb{N}$, $k \ge 1$.
Let $k \ge 1$ and $p, q, r > M$ three distinct primes such that $r > \max\{kp, kq\}$. Considering $a = kp, b = kq$ and $c = r$, we have $(a, b) = k$ and $(b, c) = (c, a) = 1$, which leads to $(a, b) \cdot (b, c) + (b, c) \cdot (c, a) + (c, a) \cdot (a, b) = 2k+1$, so $2k+1$ is good.
Let $t \in \mathbb{N}$ and $n$ be a natural number for which there exist $a, b, c > M$, pairwise distinct, such that $n = (a, b) \cdot (b, c) + (b, c) \cdot (c, a) + (c, a) \cdot (a, b)$. For every $t \in \mathbb{N}$, we have $2^t a, 2^t b, 2^t c > M$ and $2^{2t} \cdot n = (2^t a, 2^t b) \cdot (2^t b, 2^t c) + (2^t b, 2^t c) \cdot (2^t c, 2^t a) + (2^t c, 2^t a) \cdot (2^t a, 2^t b)$, so all the numbers of the form $n = 2^{2t}(2k+1)$ are good.

Further, we prove that the numbers of the form $n = 2^t$ or $n = 2^{2t+1}(2k+1)$, where $t, k \in \mathbb{N}$, $k \ge 1$, are not good, so they don't satisfy the conditions of the problem either.
First, we prove that any good number that is even is actually divisible by 4.
Consider $n_0$ an even number. If $n_0 = (a, b) \cdot (b, c) + (b, c) \cdot (c, a) + (c, a) \cdot (a, b)$, then $(a, b)$, $(b, c)$ and $(c, a)$ can't be simultaneously odd. For example, if $2 \mid (a, b)$, then $2 \mid (b, c) \cdot (c, a)$, so $a, b$ and $c$ are even (the other cases are completely similar). It follows that $4 \mid n_0$. Also, the number $\frac{n_0}{4}$ is good too, because $\frac{n_0}{4} = (a', b') \cdot (b', c') + (b', c') \cdot (c', a') + (c', a') \cdot (a', b')$, where $a' = \frac{a}{2}, b' = \frac{b}{2}, c' = \frac{c}{2}$.
Suppose, for the sake of contradiction, that there exists $t \ge 2$ such that $n = 2^t$ is good. Consequently, $2^{t-2}, 2^{t-4}, 2^{t-6}, \dots$ etc. are also good, so either 1 or 2 should be also good. This is a contradiction, since if $a, b, c \ge 1$, then $(a, b) \cdot (b, c) + (b, c) \cdot (c, a) + (c, a) \cdot (a, b) \ge 3$.
Similarly, if $n = 2^{2t+1}(2k+1)$, with $t, k \in \mathbb{N}$, $k \ge 1$ is good, then $2^{2t-1}(2k+1)$, $2^{2t-3}(2k+1), \dots, 2(2k+1)$ are also good, impossible, since $2(2k+1)$ is not divisible by 4.