(13) $a, b$ are two distinct positive numbers, and satisfy $a^{3}-b^{3}=a^{2}-b^{2}$, find all possible integer $c$, such that $c=9ab$.
13 From $a^{3}-b^{3}=a^{2}-b^{2}$, we get
$$
a^{2}+a b+b^{2}=a+b,
$$

so
$$
a b=(a+b)^{2}-(a+b)>0,
$$

from which we obtain $a+b>1$. Also, because
$$
\frac{1}{4}(a+b)^{2}>a b=(a+b)^{2}-(a+b),
$$

thus
$$
1<a+b<\frac{4}{3} .
$$

Furthermore, since
$$
a b=(a+b)^{2}-(a+b),
$$

let $t=a+b$, then $a b=t^{2}-t$. When $t \geqslant 1$, $t^{2}-t$ is monotonically increasing with respect to $t$, so $0<$ $a b<\frac{4}{9}, 0<9 a b<4$. Therefore, $c$ can take $1, 2, 3$.