1 can be Expressed as Sum of 4 Distinct Unit Fractions in 6 Ways/Proof 2

Theorem

The number $1$ can be expressed as the sum of $4$ distinct unit fractions in $6$ different ways:

 $\ds 1$ $=$ $\ds \frac 1 2 + \frac 1 3 + \frac 1 7 + \frac 1 {42}$ $\ds$ $=$ $\ds \frac 1 2 + \frac 1 3 + \frac 1 8 + \frac 1 {24}$ $\ds$ $=$ $\ds \frac 1 2 + \frac 1 3 + \frac 1 9 + \frac 1 {18}$ $\ds$ $=$ $\ds \frac 1 2 + \frac 1 3 + \frac 1 {10} + \frac 1 {15}$ $\ds$ $=$ $\ds \frac 1 2 + \frac 1 4 + \frac 1 5 + \frac 1 {20}$ $\ds$ $=$ $\ds \frac 1 2 + \frac 1 4 + \frac 1 6 + \frac 1 {12}$

Proof

There are $14$ ways to represent $1$ as the sum of exactly $4$ unit fractions.

This includes repeated unit fractions.

The full list is:

 $\text {(1)}: \quad$ $\ds$  $\ds \dfrac 1 2 + \dfrac 1 3 + \dfrac 1 7 + \dfrac 1 {42}$ $\text {(2)}: \quad$ $\ds$  $\ds \dfrac 1 2 + \dfrac 1 3 + \dfrac 1 8 + \dfrac 1 {24}$ $\text {(3)}: \quad$ $\ds$  $\ds \dfrac 1 2 + \dfrac 1 3 + \dfrac 1 9 + \dfrac 1 {18}$ $\text {(4)}: \quad$ $\ds$  $\ds \dfrac 1 2 + \dfrac 1 3 + \dfrac 1 {10} + \dfrac 1 {15}$ $\text {(5)}: \quad$ $\ds$  $\ds \dfrac 1 2 + \dfrac 1 3 + \dfrac 1 {12} + \dfrac 1 {12}$ $\text {(6)}: \quad$ $\ds$  $\ds \dfrac 1 2 + \dfrac 1 4 + \dfrac 1 5 + \dfrac 1 {20}$ $\text {(7)}: \quad$ $\ds$  $\ds \dfrac 1 2 + \dfrac 1 4 + \dfrac 1 6 + \dfrac 1 {12}$ $\text {(8)}: \quad$ $\ds$  $\ds \dfrac 1 2 + \dfrac 1 4 + \dfrac 1 8 + \dfrac 1 8$ $\text {(9)}: \quad$ $\ds$  $\ds \dfrac 1 2 + \dfrac 1 5 + \dfrac 1 5 + \dfrac 1 {10}$ $\text {(10)}: \quad$ $\ds$  $\ds \dfrac 1 2 + \dfrac 1 6 + \dfrac 1 6 + \dfrac 1 6$ $\text {(11)}: \quad$ $\ds$  $\ds \dfrac 1 3 + \dfrac 1 3 + \dfrac 1 4 + \dfrac 1 {12}$ $\text {(12)}: \quad$ $\ds$  $\ds \dfrac 1 3 + \dfrac 1 3 + \dfrac 1 6 + \dfrac 1 {6}$ $\text {(13)}: \quad$ $\ds$  $\ds \dfrac 1 3 + \dfrac 1 4 + \dfrac 1 4 + \dfrac 1 6$ $\text {(14)}: \quad$ $\ds$  $\ds \dfrac 1 4 + \dfrac 1 4 + \dfrac 1 4 + \dfrac 1 4$

The result follows from inspection: solutions $(1)$, $(2)$, $(3)$, $(4)$, $(6)$ and $(7)$ are the only solutions of the above such that the denominators of the summands are distinct.

$\blacksquare$