Natural numbers $a,b,c$ satisfaying $abc=2(a+b+c)$

Question

How can one find all natural numbers such that: $a≤b≤c$

$$abc=2(a+b+c)$$

I tried this :

$abc-2c=2a+2b$ so $c=\frac{2(a+b)}{ab-2}$

Answer 1

The only possible cases are - $(1,3,8) , (1,4,5) , (2,2,4)$

How? I looked at it this way-

$a=1 \implies c=\frac{2b+2}{b-2} $ Check out for $b=3,4$ at $b=4$, $c=5$ ,so you know you have to stop.

$a=2 \implies c=\frac{b+2}{b-1} $ Check out for $b=2,3$ at $b=3$, $c<3$ ,so you know that's enough.

$a=3 \implies c=\frac{2b+6}{3b-2} $ Check out for $b=3$ $c<3$, That's done.

A very tiresome way to do , but it works.

Just a means of simplification, put $a=b=c$ as the boundary condition. You get $a=\sqrt{6}$ Which means you can't go beyond $a=2$

Also, in your equation, put $b=c$ to get the boundary condition for $b$ . you get the boundary to be $\frac{2+\sqrt{4+2a}}{a}$ for each $a$

Answer 2

We need $abc = 2(a+b+c)$ and we've got $a \le b \le c$.

Hence $2(a+b+c) \le 6c$

That means we need $ab \le 6$. This gives you finitely many values of $a,b$ to test ($a=1,2,3$, with $ab=6$) and for each of these it is easy to find a $c$ or show no $c$ exists so I leave that to you.