Find all natural roots of: $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=d$ given that: $a<b<c$

Question

Find all natural roots of: $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=d$ given that: $a<b<c$

Rearranging the equation gives: $$ab+bc+ac=abcd$$
What can we do with this?

Answer 1

First observe that$$d \le \frac{1}{1}+\frac{1}{2}+\frac{1}{3}<2$$Thus $d=1$. And notice that if $a \ge 3$ then$$d \le \frac{1}{3}+\frac{1}{4}+\frac{1}{5}<1$$ So we have $a \le 2$. Its easy to see that $a \neq 1$, so $a=2$. Equation reduces to$$\frac{1}{b}+\frac{1}{c}=\frac{1}{2}$$Again if $b \ge 4$ then$$\frac{1}{b}+\frac{1}{c} \le \frac{1}{4}+\frac{1}{5}<\frac{1}{2}$$Therefore $b \le 3$. $b=2$ doesn't work. So $b=3$ and finally $c=6$.

Thus the only solution is $(a, b, c, d)=(2, 3, 6, 1)$.