# 2. CONDITION

Solve the equation in natural numbers

$$
\text { LCM }(a ; b)+\text { GCD }(a ; b)=a b .
$$

(GCD - greatest common divisor, LCM - least common multiple).

#
# Solution.

Let $\quad$ GCD $(a ; b)=d$. Then $a=a_{1} d, b=b_{1} d$, where GCD $\left(a_{1} ; b_{1}\right)=1$. Then LCM $(a ; b)=a_{1} b_{1} d$. From this, $a_{1} b_{1} d+d=a_{1} d b_{1} d$, or $a_{1} b_{1}+1=a_{1} b_{1} d$. Hence, $a_{1} b_{1}(d-1)=1$. That is, $a_{1}=b_{1}=1$ and $d=2$, so $a=b=2$.

Comment on the solution. Another solution can be obtained by using the equality $LCM(a ; b) \cdot GCD(a ; b)=a b$.

Answer: $a=b=2$.