12.39 Let $N$ be the set of natural numbers, for $a \in N, b \in N$, find all natural number solutions $(x, y)$ of the equation $x^{a+b}+y=x^{a} y^{b}$.
(Australian Mathematics Competition, 1983)
［Solution］ From the given equation, we have
$$
\begin{aligned}
y & =x^{a} y^{b}-x^{a+b} \\
& =x^{a}\left(y^{b}-x^{b}\right) .
\end{aligned}
$$

Therefore, $x^{a}$ is a divisor of $y$. Let $y=x^{a} z$, where $z \in \mathbb{N}$.
Substituting $y=x^{a} z$ into (1), we get
$$
\begin{aligned}
z & =x^{a b} z^{b}-x^{b} \\
& =x^{b}\left(x^{a b-b} z^{b}-1\right) .
\end{aligned}
$$

Therefore, $x^{b}$ is a divisor of $z$. Let $z=x^{b} u$, where $u \in \mathbb{N}$. Substituting $z=x^{b} u$ into (2), we get
$$
\begin{array}{l}
u=x^{a b-b} x^{b^{2}} u^{b}-1, \\
1=x^{a b+b^{2}-b} u^{b}-u
\end{array}
$$

Therefore, $u$ is a divisor of 1,
so
$$
u=1 .
$$

And
$$
x^{a b+b^{2}-b}=2 \text {, }
$$

Thus
$$
\begin{array}{c}
x=2, \\
a b+b^{2}-b=1, \\
a=1, \quad b=1 .
\end{array}
$$

Therefore
$$
a=1, \quad b=1 .
$$
$$
y \doteq x^{a} x^{b} u=4 \text {. }
$$

Hence, the original equation has natural number solutions only when $a=b=1$:
$$
x=2, \quad y=4 .
$$