To solve $ \frac{1}{a}+\frac{1}{a+b}+\frac{1}{a+b+c}=1$ in natural numbers.

Question

I really stuck on the following 7th grade problem (shame on me).

The problem asks to solve an equation in natural numbers (that is to find all possible such natural $a,b,c~$ that) $$ \frac{1}{a}+\frac{1}{a+b}+\frac{1}{a+b+c}=1$$ I have found solutions: $$a=3, ~b=0, ~c=0;$$ $$a=2, ~b=1, ~c=3$$ and $$a=2, ~b=2, ~c=0$$ but there must be a general trick to find all possible solutions, that I missed.

Can anyone suggest some way to solve the equation?

Answer 1

Note that $\frac 1a$ is the largest of the three fractions so must be at least $\frac 13$.

The admissible values for $a$ are then $1, 2, 3$.

If $a=3$ you have $b=c=0$ or the total will be less than $1$.

If $a=2$ then you have $\cfrac 1{2+b}+\cfrac 1{2+b+c}=\cfrac 12$ and the largest fraction must be at least $\frac 14$.

Then consider what happens when $a=1$

Answer 2

$$1=\frac{1}{a}+\frac{1}{a+b}+\frac{1}{a+b+c}\geq\frac{3}{a+b+c},$$ which gives $$a+b+c\leq3$$ and since $a\geq2$, we obtain two easy cases.