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:

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


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:

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


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.

$\blacksquare$