
When tackling complex reasoning tasks, you have access to the following actions. Use them as needed to progress through your thought process.

[ASSESS]

[ADVANCE]

[VERIFY]

[SIMPLIFY]

[SYNTHESIZE]

[PIVOT]

[OUTPUT]

You should strictly follow the format below:

[ACTION NAME]

# Your action step 1

# Your action step 2

# Your action step 3

...

Next action: [NEXT ACTION NAME]


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Find all mutually coprime triples \( x, y, (x+y) \) of divisors of the number \( a = 1944 = 2^3 \cdot 3^5 \). Note that if a prime \( p > 1 \) divides both \( x \) and \( y \), then \( p \) also divides \( x + y \). Similarly, if \( p > 1 \) divides both \( x \) and \( x + y \), then it also divides \( y \), and any prime \( p > 1 \) that divides both \( y \) and \( x + y \) also divides \( x \). Thus, the numbers \( x, y, (x+y) \) have distinct elementary divisors, except for one. Since there are only two such elementary divisors, namely 2 and 3, and three numbers, one of them must be 1 (note that \( x + y \neq 1 \)).

Case 1: \( x = 1, y = 2m, x + y = y + 1 = 3n, n, m \in \mathbb{Z}, m \geq 1, n \geq 1 \).

Then \( 3n - 2m = 1 \rightarrow \left\{ \begin{array}{l} n = 1 + 2s \\ m = 1 + 3s \end{array}, s \in \mathbb{Z} \rightarrow \left\{ \begin{array}{c} y = 6s + 2 \\ x + y = 6s + 3 \end{array} \right. \right. \). In this case, \( y \) may only contain powers of 2, i.e., \( y = 2, 4, 8 \), and \( x + y \) powers of 3, i.e., \( x + y = 3, 9, 27, 81 \). The latter is possible for \( s = 0 \) and \( s = 1 \).

Present the answer in LaTex format: \boxed{Your answer}
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