1. Determine all ordered pairs $(a, b)$ of integers for which $\frac{1}{a}-\frac{1}{b}=\frac{1}{5}$.
First solution.

The numbers $a$ and $b$ are denominators of fractions, so it must be $(a, b) \in \mathbb{Z} \times \mathbb{Z}$ and $a \neq 0, b \neq 0$. We have

$$
\begin{gathered}
\frac{1}{a}-\frac{1}{b}=\frac{1}{5} \\
\frac{b-a}{a b}=\frac{1}{5} \\
a b=5 b-5 a \\
a b+5 a=5 b \\
a(b+5)=5 b \\
a=\frac{5 b+25-25}{b+5} \\
a=5-\frac{25}{b+5}
\end{gathered}
$$

For $a$ to be an integer, the denominator $b+5$ of the fraction must be a divisor of the numerator 25. Therefore,

$$
\begin{aligned}
& b+5 \in\{-1,1,-5,5,-25,25\} \\
& b \in\{-6,-4,-10,0,-30,20\}
\end{aligned}
$$

Since $b \neq 0$,

$$
b \in\{-6,-4,-10,-30,20\}
$$

| $b$ | -6 | -4 | -10 | -30 | 20 |
| :---: | :---: | :---: | :---: | :---: | :---: |
| $a$ | 30 | -20 | 10 | 6 | 4 |

The desired ordered pairs are $(30,-6),(-20,-4),(10,-10),(6,-30)$, and $(4,20)$.

Note. Similarly, we can express

$$
b=-5-\frac{25}{a-5}
$$

Since $a \neq 0$, we get

$$
a \in\{4,6,10,-20,30\}
$$

We obtain the same pairs as the solution.