Equation $(a-3)cb=a(c+b)$ for natural numbers. Let $a$, $b$, and $c$ be positive integers. Suppose that $c \leq  b \leq a$ and that they satisfy the relation
$$
(a-3)cb=a(c+b).
$$
What can be said about the solutions?
 A: This equation can be rewritten as
$$\frac{3}{a}+\frac{1}{b}+\frac{1}{c}=1.$$
Now 


*

*If $c>5$, then there is no solutions (the lhs $<1$).

*If $c=5$, then the only solution is $a=b=c=5$.

*If $c=4$ and $b>5$, then there is no solutions (the lhs $<1$).

*If $c=4$ and $b=4$, then $a=6$.

*If $c=3$ and $b>6$, then there is no solutions (the lhs $<1$).

*If $c=3$ and $b=6$, then $a=6$.

*If $c=3$ and $b=4,5$, then there is no solutions (direct verification).

*If $c=3$ and $b=3$, then $a=9$.

*If $c=2$ and $b>8$, then there is no solutions (the lhs $<1$).

*If $c=2$ and $b=8$, then $a=8$.

*If $c=2$ and $b=7$, then there is no solutions.

*And finally, we have solutions $(a,b,c)=(9,6,2), (10,5,2), (12,4,2), (18,3,2)$.
Hence the complete list of triples $(a,b,c)$ is:
$$(5,5,5),(6,4,4),(9,3,3),(8,8,2),(9,6,2), (10,5,2), (12,4,2), (18,3,2).$$
A: Note that $c \leq 5$. Now consider $c=1,2,3,4,5$. Check for each case! You can easily find $(5,5,5),(6,4,4),(9,3,3),(8,8,2),(9,6,2), (10,5,2), (12,4,2), (18,3,2)$ are the only solutions.
