1. In the set of natural numbers, solve the system of equations

$$
\left\{\begin{array}{l}
a b+c=13 \\
a+b c=23
\end{array}\right.
$$
Solution. If we add the equations, after simplification we get:

$$
(a+c)(b+1)=36
$$

If we subtract the first equation from the second, after simplification we get:

$$
(c-a)(b-1)=10
$$

According to this, $(b+1) \mid 36$ and $(b-1) \mid 10$. But,

$$
36=2 \cdot 18=3 \cdot 12=4 \cdot 9=6 \cdot 6 \text { and } 10=1 \cdot 10=2 \cdot 5 \text {. }
$$

By direct verification, the only possible values for $b$ are 2, 3, and 11. For $b=2$, we get the system

$$
\left\{\begin{array}{l}
2 a+c=13 \\
a+2 c=23
\end{array}\right.
$$

whose solution is $a=1, c=11$. For $b=3$, we get the system

$$
\left\{\begin{array}{l}
3 a+c=13 \\
a+3 c=23
\end{array}\right.
$$

whose solution is $a=2, c=7$. For $b=11$, we get the system

$$
\left\{\begin{array}{l}
11 a+c=13 \\
a+11 c=23
\end{array}\right.
$$

whose solution is $a=1, c=2$.