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Bulgarian Mathematical Olympiad

Number theory Difficulty 5.4 AIME, harder Prove it Bulgaria

Problem:
Find all pairs of positive integers $(m, n)$ , $m > n$ , such that
$\left[m^2 + m n, m n - n^2\right] + [m - n, m n] = 2^{2005}$
where $[a, b]$ denotes the least common multiple of $a$ and $b$ .

Solution

Solution:
The left hand side of the given equation is a multiple of $m$ , $n$ and $m-n$ . Therefore $m = 2^a$ , $n = 2^b$ and $m - n = 2^c$ for some nonnegative integers $a$ , $b$ and $c$ , where $a > b$ . It is obvious that $2^b (2^{a-b} - 1) = 2^c$ , whence $a - b = 1$ .

Plugging $b = a - 1$ in the given equation we obtain
$\begin{aligned} \left[2^{2a} + 2^{2a-1}, 2^{2a-1} - 2^{2a-2}\right] + \left[2^a - 2^{a-1}, 2^{2a-1}\right] &= 2^{2a-1} + 3 \cdot 2^{2a-1} \\ &= 2^{2a+1} = 2^{2005} \end{aligned}$
Hence $a = 1002$ , $m = 2^{1002}$ and $n = 2^{1001}$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.