3. Find all pairs of natural numbers $m$ and $n$ that satisfy the system of equations

$$
\left\{\begin{array}{l}
\operatorname{LCM}(m, n)+\operatorname{GCD}(m, n)=m+n \\
m-n=2015
\end{array}\right.
$$

Answer. Possible options for the pair $(m, n):(2016,1),(2020,5),(2028,13),(2046,31)$, $(2080,65),(2170,155),(2418,403),(4030,2015)$.
Solution. Method I. The product of the LCM and GCD of any two natural numbers is equal to the product of these numbers. From this fact, the first equation, and Vieta's theorem, it follows that the LCM and GCD of the numbers $m$ and $n$ are the roots of the same quadratic trinomial as $m$ and $n$ themselves, which means they are equal to $m$ and $n$ (in some order). Since $m>n$ (from the second equation), $m$ is the LCM, and $n$ is the GCD. Therefore, $m=k n$ for some natural number $k>1$. Then from the first equation,

$$
(k-1) n=2015=5 \cdot 13 \cdot 31 .
$$

Thus, $n$ can be any natural divisor of the number 2015, and $m$ is found from the second equation, which gives the answer mentioned above.

Method II. Let $n=a d, m=b d$, where $d$ is their GCD, and the numbers $a$ and $b$ are coprime. Then the LCM is $a b d$; from the first equation, we get that $a b d+d=a d+b d$, that is, $a b+1=a+b$, from which $(a-1)(b-1)=0$. Therefore, at least one of the numbers $a$ and $b$ is equal to 1. The second equation shows that $m>n$, so $a=1, b>1$, which means $m$ is divisible by $n$. The rest follows as in the first method.