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

## 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

Let:

- $1 = \dfrac 1 v + \dfrac 1 w + \dfrac 1 x + \dfrac 1 y$

where $ 1 < v < w < x < y$

Suppose $v = 3$ and take the largest potential solution that can be generated:

- $1 \stackrel {?} {=} \dfrac 1 3 + \dfrac 1 4 + \dfrac 1 5 + \dfrac 1 6$

But we find:

- $1 > \dfrac 1 3 + \dfrac 1 4 + \dfrac 1 5 + \dfrac 1 6$

Therefore, there can be no solutions where $v \ge 3$, as that solution was the largest possible.

Hence, $v = 2$ if there are any solutions.

Repeating the above anaylsis on $w$:

\(\displaystyle \dfrac 1 2\) | \(=\) | \(\displaystyle \dfrac 1 w + \dfrac 1 x + \dfrac 1 y\) | |||||||||||

\(\displaystyle \dfrac 1 2\) | \(<\) | \(\displaystyle \dfrac 1 3 + \dfrac 1 4 + \dfrac 1 5\) | |||||||||||

\(\displaystyle \dfrac 1 2\) | \(<\) | \(\displaystyle \dfrac 1 4 + \dfrac 1 5 + \dfrac 1 6\) | |||||||||||

\(\displaystyle \dfrac 1 2\) | \(<\) | \(\displaystyle \dfrac 1 5 + \dfrac 1 6 + \dfrac 1 7\) | |||||||||||

\(\displaystyle \dfrac 1 2\) | \(>\) | \(\displaystyle \dfrac 1 6 + \dfrac 1 7 + \dfrac 1 8\) |

Potential solutions are located where $w = 3, 4, 5$.

Now that $v$ and $w$ are known, the variable $y$ can be written in terms of $x$:

- $y = \dfrac 1 {\dfrac {w - 2} {2 w} - \dfrac 1 x}$

Solutions are only positive when:

\(\displaystyle \dfrac 1 x\) | \(<\) | \(\displaystyle \dfrac {w - 2} {2 w}\) | |||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle x\) | \(>\) | \(\displaystyle \dfrac {2 w} {w - 2}\) |

As $x < y$:

\(\displaystyle x\) | \(<\) | \(\displaystyle \dfrac 1 {\dfrac {w - 2} {2 w} - \dfrac 1 x}\) | |||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \dfrac 1 x\) | \(<\) | \(\displaystyle \dfrac {w - 2} {2 w} - \dfrac 1 x\) | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \dfrac 2 x\) | \(<\) | \(\displaystyle \dfrac {w - 2} {2 w}\) | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle x\) | \(<\) | \(\displaystyle \dfrac {4 w} {w - 2}\) |

Therefore solutions exist only in the domain:

- $\dfrac {2 w} {w - 2} < x < \dfrac {4 w} {w - 2}$

and:

- $w < x$

Case $w = 3$:

\(\displaystyle 6\) | \(<\) | \(\displaystyle x < 12\) | \(\displaystyle \text {and } 3 < x\) | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle 6\) | \(<\) | \(\displaystyle x < 12\) |

Integer solutions in the above domain can then be found by inspection:

- $\tuple {7, 42}, \tuple {8, 24}, \tuple {9, 18}, \tuple {10, 15}$

Case $w = 4$:

\(\displaystyle 4\) | \(<\) | \(\displaystyle x < 8\) | \(\displaystyle \text {and } 4 < x\) | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle 4\) | \(<\) | \(\displaystyle x < 8\) |

Integer solutions in the above domain can again be found by inspection:

- $\tuple {5, 20}, \tuple {6, 12}$

Case $w = 5$:

\(\displaystyle \dfrac {10} 3\) | \(<\) | \(\displaystyle \dfrac {20} 3\) | \(\displaystyle \text {and } 5 < x\) | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle 5\) | \(<\) | \(\displaystyle \dfrac {20} 3\) |

and it is immediately seen that there are no integer solutions in this domain.

All solutions have therefore been found:

- $\tuple {2, 3, 7, 42}, \tuple {2, 3, 8, 24}, \tuple {2, 3, 9, 18}, \tuple {2, 3, 10, 15}, \tuple {2, 4, 5, 20}, \tuple {2, 4, 6, 12}$

Hence:

\(\displaystyle 1\) | \(=\) | \(\displaystyle \frac 1 2 + \frac 1 3 + \frac 1 7 + \frac 1 {42}\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \frac 1 2 + \frac 1 3 + \frac 1 8 + \frac 1 {24}\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \frac 1 2 + \frac 1 3 + \frac 1 9 + \frac 1 {18}\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \frac 1 2 + \frac 1 3 + \frac 1 {10} + \frac 1 {15}\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \frac 1 2 + \frac 1 4 + \frac 1 5 + \frac 1 {20}\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \frac 1 2 + \frac 1 4 + \frac 1 6 + \frac 1 {12}\) |

$\blacksquare$