Problem 1010

I. solution. The product of the greatest common divisor and the least common multiple of two numbers is equal to the product of the two numbers:

$$(a ; b) \cdot[a ; b]=a \cdot b$$

Let's denote the greatest common divisor of $a$ and $b$ by $d$. According to the problem, $d+[a ; b]=a+b$. Using the mentioned relationship, we get the following equation:

$$d+\frac{a \cdot b}{d}=a+b$$

After rearranging, we can factorize it as follows:

$$d^{2}-(a+b) \cdot d+a \cdot b=(d-a)(d-b)=0$$

If $d=a$, that is, the greatest common divisor is equal to $a$, then $b$ is a multiple of $a$. If $d=b$, then $a$ is a multiple of $b$. This proves the statement of the problem.

II. solution. Let $a=d \cdot \ell$ and $b=d \cdot k$, where $k$ and $\ell$ are coprime. Thus, $(a ; b)=d$ and $[a ; b]=d \cdot k \cdot \ell$. The equation is: $d+d \cdot k \cdot \ell=d \cdot \ell+d \cdot k$. Clearly, $d \neq 0$, so we can divide by it:

$$1+k \cdot \ell=\ell+k, \quad k \cdot \ell-\ell-k+1=0, \quad(k-1)(\ell-1)=0$$

from which $k=1$ or $\ell=1$ follows. In both cases, the greatest common divisor is equal to one of the given numbers, and in this case, this number is a divisor of the other.