Find all triples satisfying an equation

Question

Another question I saw recently:

Find all triples of positive integers $(a,b,c)$ such that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1$.

Can someone help me with it?

Answer 1

As $\frac{1}{n}>0$ for all natural $n$, all of $a,b,c$ have to be at least two. Let $a\leq b\leq c$. Unless $a=b=c=3$, $a=2$. Then we have $\frac{1}{b}+\frac{1}{c}=\frac{1}{2}$, so similarly either $b=c=4$ or $b=3$. In the latter case, we get $c=6$. So, $(a,b,c)=(3,3,3),(2,4,4),(2,3,6)$ are all the solutions.

Answer 2

Hint

Without loss of generality, let $a<b<c$ and set $t=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$.

• If $a=1$ then $t>1$
• If $a\ge 3$ then $t<1$

Thus $a=2$