1. Does there exist a pair of unequal integers $a, b$, for which the equality

$$
\frac{a}{2015}+\frac{b}{2016}=\frac{2015+2016}{2015 \cdot 2016}
$$

holds? If such a pair does not exist, justify it. If such a pair does exist, provide an example.
# Solution.

Multiply the equation by the number $2015 \cdot 2016$. We get the following equation

$$
2016 a + 2015 b = 2015 + 2016
$$

Rewrite the equation in the following form

$$
2016(a-1) = 2015(1-b)
$$

Since the numbers 2016 and 2015 are coprime, then $(a-1)$ is divisible by 2015, i.e., there exists an integer $k$ for which the equality $a-1 = 2015 k \Rightarrow 1-b = 2016 k \Rightarrow a = 2015 k + 1; b = 1 - 2016 k$ holds.

Substituting the values $a = 2015 k + 1; b = 1 - 2016 k$ for any integer $k$ into the original equation, we get a true identity.

For $k = 1, a = 2016, b = -2015$.

For $k = -1, a = -2014, b = 2017$.

For $k = 2, a = 4031, b = -4031$.

For $k = -2, a = -4029, b = 4033$.

Other various examples can also be represented

One could have guessed

$$
\begin{aligned}
& \frac{2016}{2015} + \frac{-2015}{2016} = \frac{2015 + 2016}{2015 \cdot 2016} \\
& \frac{2016^2 - 2015^2}{2015 \cdot 2016} = \frac{2015 + 2016}{2015 \cdot 2016}
\end{aligned}
$$

## Recommendations for checking.

A correct example is provided - 7 points