1 can be Expressed as Sum of 4 Distinct Unit Fractions in 6 Ways
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 1
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 $w$:
\(\ds \dfrac 1 2\) | \(=\) | \(\ds \dfrac 1 w + \dfrac 1 x + \dfrac 1 y\) | ||||||||||||
\(\ds \dfrac 1 2\) | \(<\) | \(\ds \dfrac 1 3 + \dfrac 1 4 + \dfrac 1 5\) | ||||||||||||
\(\ds \dfrac 1 2\) | \(<\) | \(\ds \dfrac 1 4 + \dfrac 1 5 + \dfrac 1 6\) | ||||||||||||
\(\ds \dfrac 1 2\) | \(<\) | \(\ds \dfrac 1 5 + \dfrac 1 6 + \dfrac 1 7\) | ||||||||||||
\(\ds \dfrac 1 2\) | \(>\) | \(\ds \dfrac 1 6 + \dfrac 1 7 + \dfrac 1 8\) |
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:
\(\ds \dfrac 1 x\) | \(<\) | \(\ds \dfrac {w - 2} {2 w}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds x\) | \(>\) | \(\ds \dfrac {2 w} {w - 2}\) |
As $x < y$:
\(\ds x\) | \(<\) | \(\ds \dfrac 1 {\dfrac {w - 2} {2 w} - \dfrac 1 x}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \dfrac 1 x\) | \(<\) | \(\ds \dfrac {w - 2} {2 w} - \dfrac 1 x\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \dfrac 2 x\) | \(<\) | \(\ds \dfrac {w - 2} {2 w}\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds x\) | \(<\) | \(\ds \dfrac {4 w} {w - 2}\) |
Therefore solutions exist only in the domain:
- $\dfrac {2 w} {w - 2} < x < \dfrac {4 w} {w - 2}$
and:
- $w < x$
Case $w = 3$:
\(\ds 6\) | \(<\) | \(\ds x < 12\) | \(\ds \text {and } 3 < x\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds 6\) | \(<\) | \(\ds x < 12\) |
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 $w = 4$:
\(\ds 4\) | \(<\) | \(\ds x < 8\) | \(\ds \text {and } 4 < x\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds 4\) | \(<\) | \(\ds x < 8\) |
Integer solutions in the above domain can again be found by inspection:
- $\tuple {5, 20}, \tuple {6, 12}$
Case $w = 5$:
\(\ds \dfrac {10} 3\) | \(<\) | \(\ds \dfrac {20} 3\) | \(\ds \text {and } 5 < x\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds 5\) | \(<\) | \(\ds \dfrac {20} 3\) |
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:
\(\ds 1\) | \(=\) | \(\ds \frac 1 2 + \frac 1 3 + \frac 1 7 + \frac 1 {42}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 2 + \frac 1 3 + \frac 1 8 + \frac 1 {24}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 2 + \frac 1 3 + \frac 1 9 + \frac 1 {18}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 2 + \frac 1 3 + \frac 1 {10} + \frac 1 {15}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 2 + \frac 1 4 + \frac 1 5 + \frac 1 {20}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 2 + \frac 1 4 + \frac 1 6 + \frac 1 {12}\) |
$\blacksquare$
Proof 2
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.
$\blacksquare$