## Task B-1.2.

If $a \neq 0$ and $b \neq 0$ are two distinct real numbers for which

$$
\frac{a}{2017}+\frac{2017}{a}=\frac{b}{2017}+\frac{2017}{b}
$$

calculate $\sqrt{a \cdot b}$.
## Solution.

Let's write the given equality in the following form:

$$
\begin{aligned}
& \frac{a}{2017}+\frac{2017}{a}=\frac{b}{2017}+\frac{2017}{b} \\
& \Leftrightarrow \frac{1}{2017}(a-b)=2017\left(\frac{1}{b}-\frac{1}{a}\right) \\
& \Leftrightarrow \frac{1}{2017}(a-b)=2017\left(\frac{a-b}{a b}\right) \\
& \Leftrightarrow \frac{1}{2017}(a-b)-\frac{2017(a-b)}{a b}=0 \\
& \Leftrightarrow(a-b)\left(\frac{1}{2017}-\frac{2017}{a b}\right)=0
\end{aligned}
$$

Since $a \neq 0, b \neq 0$ and $a \neq b$, it follows that $\frac{1}{2017}-\frac{2017}{a b}=0$. 1 point

Then $\frac{1}{2017}=\frac{2017}{a b}$, which means $\sqrt{a b}=2017$.