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

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## 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

From Sum of 4 Unit Fractions that equals 1:

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.

$\blacksquare$