2. Find the smallest natural $k$ such that for some natural number $a$, greater than 500,000, and some natural number $b$, the equality $\frac{1}{a}+\frac{1}{a+k}=\frac{1}{b}$ holds. (I. Bogdanov)
Answer. $k=1001$. Solution. Estimation. Let $a+k=c$ and $\text{GCD}(a, c)=d$. Then $a=d a_{1}, c=d c_{1}$ and $\frac{1}{a}+\frac{1}{c}=\frac{a_{1}+c_{1}}{d a_{1} c_{1}}$. Since the numbers $a_{1} c_{1}$ and $a_{1}+c_{1}$ are coprime, $d$ must divide $a_{1}+c_{1}$. Therefore, $d \geq a_{1}+c_{1}$ and $d^{2} \geq d\left(a_{1}+c_{1}\right)=a+c>10^{6}$, from which $d \geq 1001$ and $k=d\left(c_{1}-a_{1}\right) \geq 1001$. Example. $a=500500, k=1001$ : $\frac{1}{500500}+\frac{1}{501501}=\frac{1}{250500}$.