Writing 1 as a sum of unit fractions

1. Bounding. $1=\frac1a+\frac1b+\frac1c\le\frac3a$ gives $a\le3$; and $a\ge2$ (else the sum exceeds $1$).

• $a=2$: $\frac1b+\frac1c=\frac12$ with $b\le c$, so $\frac2b\ge\frac12$, $b\le4$; also $b>2$. If $b=3$: $\frac1c=\frac16$, $c=6$. If $b=4$: $\frac1c=\frac14$, $c=4$.
• $a=3$: $\frac1b+\frac1c=\frac23$ with $b\ge3$, so $\frac2b\ge\frac23$ gives $b\le3$; thus $b=3$ and $c=3$.

The solutions are $(2,3,6)$, $(2,4,4)$, $(3,3,3)$.

2. Distinct denominators. For $k=3$: $1=\frac12+\frac13+\frac16$. From a representation with $k$ distinct denominators, take the largest denominator $m$ and replace $\frac1m$ by $\frac1{m+1}+\frac1{m(m+1)}$ (true because $\frac1m-\frac1{m+1}=\frac1{m(m+1)}$). Both new denominators exceed $m$, hence all the others, and they are different from each other. This gives a representation with $k+1$ distinct denominators. Induction finishes.

Example. $1=\frac12+\frac13+\frac17+\frac1{42}$.

Lesson. In equations with a symmetric sum, order the variables and bound the smallest one; the identity $\frac1m=\frac1{m+1}+\frac1{m(m+1)}$ is the basic splitting step for unit fractions.