Problem 10.3. 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 $b$.

Ivan Landjev
10.3. 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 $2^{b}\left(2^{a-b}-1\right)=2^{c}$, whence $a-b=1$.

Plugging $b=a-1$ in the given equation we obtain

$$
\begin{aligned}
{\left[2^{2 a}+2^{2 a-1}, 2^{2 a-1}-2^{2 a-2}\right]+\left[2^{a}-2^{a-1}, 2^{2 a-1}\right] } & =2^{2 a-1}+3 \cdot 2^{2 a-1} \\
& =2^{2 a+1}=2^{2005}
\end{aligned}
$$

Hence $a=1002, m=2^{1002}$ and $n=2^{1001}$.