Sum of 4 Unit Fractions that equals 1
Theorem
There are $14$ ways to represent $1$ as the sum of exactly $4$ unit fractions.
Proof
Let:
- $1 = \dfrac 1 a + \dfrac 1 b + \dfrac 1 c + \dfrac 1 d$
where:
- $a \le b \le c \le d$
and:
- $a \ge 2$
$a = 2$
Let $a = 2$.
$b = 2$
Let $b = 2$.
Then:
- $\dfrac 1 a + \dfrac 1 b = 1$
leaving no room for $c$ and $d$.
Hence there are no solutions where $a = 2$ and $b = 2$.
$\Box$
$b = 3$
Let $b = 3$.
\(\ds \dfrac 1 a + \dfrac 1 b\) | \(=\) | \(\ds \dfrac 1 2 + \dfrac 1 3\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {3 + 2} 6\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 5 6\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \dfrac 1 c + \dfrac 1 d\) | \(=\) | \(\ds 1 - \dfrac 5 6\) | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 1 6\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \dfrac 1 c\) | \(<\) | \(\ds \dfrac 1 6\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds c\) | \(>\) | \(\ds 6\) |
Thus we try $c = 7, 8, \ldots$ in turn.
$c = 7$:
\(\ds \dfrac 1 a + \dfrac 1 b + \dfrac 1 c\) | \(=\) | \(\ds \dfrac 1 2 + \dfrac 1 3 + \dfrac 1 7\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {21 + 14 + 6} {42}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {41} {42}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \dfrac 1 d\) | \(=\) | \(\ds 1 - \dfrac {41} {42}\) | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 1 {42}\) |
Thus we have:
- $(1): \quad 1 = \dfrac 1 2 + \dfrac 1 3 + \dfrac 1 7 + \dfrac 1 {42}$
$c = 8$:
\(\ds \dfrac 1 a + \dfrac 1 b + \dfrac 1 c\) | \(=\) | \(\ds \dfrac 1 2 + \dfrac 1 3 + \dfrac 1 8\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {12 + 8 + 3} {24}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {23} {24}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \dfrac 1 d\) | \(=\) | \(\ds 1 - \dfrac {23} {24}\) | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 1 {24}\) |
Thus we have:
- $(2): \quad 1 = \dfrac 1 2 + \dfrac 1 3 + \dfrac 1 8 + \dfrac 1 {24}$
$c = 9$:
\(\ds \dfrac 1 a + \dfrac 1 b + \dfrac 1 c\) | \(=\) | \(\ds \dfrac 1 2 + \dfrac 1 3 + \dfrac 1 9\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {9 + 6 + 2} {18}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {17} {18}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \dfrac 1 d\) | \(=\) | \(\ds 1 - \dfrac {17} {18}\) | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 1 {18}\) |
Thus we have:
- $(3): \quad 1 = \dfrac 1 2 + \dfrac 1 3 + \dfrac 1 9 + \dfrac 1 {18}$
$c = 10$:
\(\ds \dfrac 1 a + \dfrac 1 b + \dfrac 1 c\) | \(=\) | \(\ds \dfrac 1 2 + \dfrac 1 3 + \dfrac 1 {10}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {15 + 10 + 3} {30}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {28} {30}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {14} {15}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \dfrac 1 d\) | \(=\) | \(\ds 1 - \dfrac {14} {15}\) | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 1 {15}\) |
Thus we have:
- $(4): \quad 1 = \dfrac 1 2 + \dfrac 1 3 + \dfrac 1 {10} + \dfrac 1 {15}$
$c = 11$:
\(\ds \dfrac 1 a + \dfrac 1 b + \dfrac 1 c\) | \(=\) | \(\ds \dfrac 1 2 + \dfrac 1 3 + \dfrac 1 {11}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {33 + 22 + 6} {66}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {61} {66}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \dfrac 1 d\) | \(=\) | \(\ds 1 - \dfrac {61} {66}\) | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 5 {66}\) |
But $\dfrac 5 {66}$ is not a unit fraction.
Thus $a = 2, b = 3, c = 11$ does not lead to a solution.
$c = 12$:
\(\ds \dfrac 1 a + \dfrac 1 b + \dfrac 1 c\) | \(=\) | \(\ds \dfrac 1 2 + \dfrac 1 3 + \dfrac 1 {12}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {6 + 4 + 1} {12}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {11} {12}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \dfrac 1 d\) | \(=\) | \(\ds 1 - \dfrac {11} {12}\) | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 1 {12}\) |
Thus we have:
- $(5): \quad 1 = \dfrac 1 2 + \dfrac 1 3 + \dfrac 1 {12} + \dfrac 1 {12}$
Let $c > 12$.
Then:
- $1 - \dfrac 1 a + \dfrac 1 b + \dfrac 1 c > \dfrac 1 {12}$
and so:
- $d < c$
and so no further solutions can be found where $a = 2, b = 3, c > 12$.
Hence there are exactly $5$ solutions where $a = 2, b = 3$.
$\Box$
$b = 4$
Let $b = 4$.
\(\ds \dfrac 1 a + \dfrac 1 b\) | \(=\) | \(\ds \dfrac 1 2 + \dfrac 1 4\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {2 + 1} 4\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 3 4\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \dfrac 1 c + \dfrac 1 d\) | \(=\) | \(\ds 1 - \dfrac 3 4\) | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 1 4\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \dfrac 1 c\) | \(<\) | \(\ds \dfrac 1 4\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds c\) | \(>\) | \(\ds 4\) |
Thus we try $c = 5, 6, \ldots$ in turn.
$c = 5$:
\(\ds \dfrac 1 a + \dfrac 1 b + \dfrac 1 c\) | \(=\) | \(\ds \dfrac 1 2 + \dfrac 1 4 + \dfrac 1 5\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {10 + 5 + 4} {20}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {19} {20}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \dfrac 1 d\) | \(=\) | \(\ds 1 - \dfrac {19} {20}\) | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 1 {20}\) |
Thus we have:
- $(6): \quad 1 = \dfrac 1 2 + \dfrac 1 4 + \dfrac 1 5 + \dfrac 1 {20}$
$c = 6$:
\(\ds \dfrac 1 a + \dfrac 1 b + \dfrac 1 c\) | \(=\) | \(\ds \dfrac 1 2 + \dfrac 1 4 + \dfrac 1 6\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {12 + 6 + 4} {24}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {22} {24}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {11} {12}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \dfrac 1 d\) | \(=\) | \(\ds 1 - \dfrac {11} {12}\) | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 1 {12}\) |
Thus we have:
- $(7): \quad 1 = \dfrac 1 2 + \dfrac 1 4 + \dfrac 1 6 + \dfrac 1 {12}$
$c = 7$:
\(\ds \dfrac 1 a + \dfrac 1 b + \dfrac 1 c\) | \(=\) | \(\ds \dfrac 1 2 + \dfrac 1 4 + \dfrac 1 7\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {14 + 7 + 4} {28}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {25} {28}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \dfrac 1 d\) | \(=\) | \(\ds 1 - \dfrac {25} {28}\) | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 3 {28}\) |
But $\dfrac 3 {28}$ is not a unit fraction.
Thus $a = 2, b = 4, c = 7$ does not lead to a solution.
$c = 8$:
\(\ds \dfrac 1 a + \dfrac 1 b + \dfrac 1 c\) | \(=\) | \(\ds \dfrac 1 2 + \dfrac 1 4 + \dfrac 1 8\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {4 + 2 + 1} 8\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 7 8\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \dfrac 1 d\) | \(=\) | \(\ds 1 - \dfrac 7 8\) | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 1 8\) |
Thus we have:
- $(8): \quad 1 = \dfrac 1 2 + \dfrac 1 4 + \dfrac 1 8 + \dfrac 1 8$
Let $c > 8$.
Then:
- $1 - \dfrac 1 a + \dfrac 1 b + \dfrac 1 c > \dfrac 1 8$
and so:
- $d < c$
and so no further solutions can be found where $a = 2, b = 4, c > 8$.
Hence there are exactly $3$ solutions where $a = 2, b = 4$.
$\Box$
$b = 5$
Let $b = 5$.
\(\ds \dfrac 1 a + \dfrac 1 b\) | \(=\) | \(\ds \dfrac 1 2 + \dfrac 1 5\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {5 + 2} {10}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 7 {10}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \dfrac 1 c + \dfrac 1 d\) | \(=\) | \(\ds 1 - \dfrac 7 {10}\) | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 3 {10}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \dfrac 1 c\) | \(<\) | \(\ds \dfrac 3 {10}\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds c\) | \(\ge\) | \(\ds 5\) |
Thus we try $c = 5, 6, \ldots$ in turn.
$c = 5$:
\(\ds \dfrac 1 a + \dfrac 1 b + \dfrac 1 c\) | \(=\) | \(\ds \dfrac 1 2 + \dfrac 1 5 + \dfrac 1 5\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {7 + 2 + 2} {10}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 9 {10}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \dfrac 1 d\) | \(=\) | \(\ds 1 - \dfrac 9 {10}\) | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 1 {10}\) |
Thus we have:
- $(9): \quad 1 = \dfrac 1 2 + \dfrac 1 5 + \dfrac 1 5 + \dfrac 1 {10}$
$c = 6$:
\(\ds \dfrac 1 a + \dfrac 1 b + \dfrac 1 c\) | \(=\) | \(\ds \dfrac 1 2 + \dfrac 1 5 + \dfrac 1 6\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {15 + 6 + 5} {30}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {26} {30}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {13} {15}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \dfrac 1 d\) | \(=\) | \(\ds 1 - \dfrac {13} {15}\) | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 2 {15}\) |
But $\dfrac 2 {15}$ is not a unit fraction.
Thus $a = 2, b = 5, c = 6$ does not lead to a solution.
$c = 7$:
\(\ds \dfrac 1 a + \dfrac 1 b + \dfrac 1 c\) | \(=\) | \(\ds \dfrac 1 2 + \dfrac 1 5 + \dfrac 1 7\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {35 + 14 + 10} {70}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {59} {70}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \dfrac 1 d\) | \(=\) | \(\ds 1 - \dfrac {59} {70}\) | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {11} {70}\) |
But $\dfrac {11} {70}$ is not a unit fraction.
Thus $a = 2, b = 5, c = 7$ does not lead to a solution.
Let $c > 7$.
\(\ds 1 - \dfrac 1 a + \dfrac 1 b + \dfrac 1 c\) | \(>\) | \(\ds \dfrac {11} {70}\) | ||||||||||||
\(\ds \) | \(>\) | \(\ds \dfrac {10} {70}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 1 7\) |
and so:
- $d < c$
and so no further solutions can be found where $a = 2, b = 5, c > 7$.
Hence there is exactly $1$ solution where $a = 2, b = 5$.
$\Box$
$b = 6$
Let $b = 6$.
\(\ds \dfrac 1 a + \dfrac 1 b\) | \(=\) | \(\ds \dfrac 1 2 + \dfrac 1 6\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {3 + 1} 6\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 4 6\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 2 3\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \dfrac 1 c + \dfrac 1 d\) | \(=\) | \(\ds 1 - \dfrac 3 3\) | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 1 3\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \dfrac 1 c\) | \(<\) | \(\ds \dfrac 1 3\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds c\) | \(>\) | \(\ds 3\) |
But we already have that $c \ge b$.
Thus:
- $c \ge 6$
Thus we try $c = 6, 7, \ldots$ in turn.
$c = 6$:
\(\ds \dfrac 1 a + \dfrac 1 b + \dfrac 1 c\) | \(=\) | \(\ds \dfrac 1 2 + \dfrac 1 6 + \dfrac 1 6\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {3 + 1 + 1} 6\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 5 6\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \dfrac 1 d\) | \(=\) | \(\ds 1 - \dfrac 5 6\) | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 1 6\) |
Thus we have:
- $(10): \quad 1 = \dfrac 1 2 + \dfrac 1 6 + \dfrac 1 6 + \dfrac 1 6$
Let $c > 6$.
Then:
- $1 - \dfrac 1 a + \dfrac 1 b + \dfrac 1 c > \dfrac 1 6$
and so:
- $d < c$
and so no further solutions can be found where $a = 2, b = 6, c > 6$.
Hence there is exactly $1$ solution where $a = 2, b = 6$.
$\Box$
$b > 6$
Let $b > 6$.
Then:
\(\ds \dfrac 1 a + \dfrac 1 b + \dfrac 1 c + \dfrac 1 d\) | \(<\) | \(\ds \dfrac 1 2 + \dfrac 1 6 + \dfrac 1 6 + \dfrac 1 6\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 1\) |
Hence there are no solutions such that $a = 2, b > 6$
$\Box$
$a = 3$
Let $a = 3$.
$b = 3$
Let $b = 3$.
\(\ds \dfrac 1 a + \dfrac 1 b\) | \(=\) | \(\ds \dfrac 1 3 + \dfrac 1 3\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 2 3\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \dfrac 1 c + \dfrac 1 d\) | \(=\) | \(\ds 1 - \dfrac 2 3\) | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 1 3\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \dfrac 1 c\) | \(<\) | \(\ds \dfrac 1 3\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds c\) | \(>\) | \(\ds 3\) |
Thus we try $c = 4, 5, \ldots$ in turn.
$c = 4$:
\(\ds \dfrac 1 a + \dfrac 1 b + \dfrac 1 c\) | \(=\) | \(\ds \dfrac 1 3 + \dfrac 1 3 + \dfrac 1 4\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {4 + 4 + 3} {12}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {11} {12}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \dfrac 1 d\) | \(=\) | \(\ds 1 - \dfrac {11} {12}\) | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 1 {12}\) |
Thus we have:
- $(11): \quad 1 = \dfrac 1 3 + \dfrac 1 3 + \dfrac 1 4 + \dfrac 1 {12}$
$c = 5$:
\(\ds \dfrac 1 a + \dfrac 1 b + \dfrac 1 c\) | \(=\) | \(\ds \dfrac 1 3 + \dfrac 1 3 + \dfrac 1 5\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {5 + 5 + 3} {15}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {13} {15}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \dfrac 1 d\) | \(=\) | \(\ds 1 - \dfrac {13} {15}\) | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 2 {15}\) |
But $\dfrac 2 {15}$ is not a unit fraction.
Thus $a = 3, b = 3, c = 5$ does not lead to a solution.
$c = 6$:
\(\ds \dfrac 1 a + \dfrac 1 b + \dfrac 1 c\) | \(=\) | \(\ds \dfrac 1 3 + \dfrac 1 3 + \dfrac 1 6\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {2 + 2 + 1} 6\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 5 6\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \dfrac 1 d\) | \(=\) | \(\ds 1 - \dfrac 5 6\) | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 1 6\) |
Thus we have:
- $(12): \quad 1 = \dfrac 1 3 + \dfrac 1 3 + \dfrac 1 6 + \dfrac 1 6$
Let $c > 6$.
Then:
- $1 - \dfrac 1 a + \dfrac 1 b + \dfrac 1 c > \dfrac 1 6$
and so:
- $d < c$
and so no further solutions can be found where $a = 3, b = 3, c > 6$.
Hence there are exactly $2$ solutions where $a = 3, b = 3$.
$\Box$
$b = 4$
Let $b = 4$.
\(\ds \dfrac 1 a + \dfrac 1 b\) | \(=\) | \(\ds \dfrac 1 3 + \dfrac 1 4\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {4 + 3} {12}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 7 {12}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \dfrac 1 c + \dfrac 1 d\) | \(=\) | \(\ds 1 - \dfrac 7 {12}\) | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 5 {12}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \dfrac 1 c\) | \(<\) | \(\ds \dfrac 5 {12}\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds c\) | \(>\) | \(\ds \dfrac {12} 5\) |
But we already have that $c \ge b$.
Thus:
- $c \ge 4$
Thus we try $c = 4, 5, \ldots$ in turn.
$c = 4$:
\(\ds \dfrac 1 a + \dfrac 1 b + \dfrac 1 c\) | \(=\) | \(\ds \dfrac 1 3 + \dfrac 1 4 + \dfrac 1 4\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {4 + 3 + 3} {12}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {10} {12}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 5 6\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \dfrac 1 d\) | \(=\) | \(\ds 1 - \dfrac 5 6\) | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 1 6\) |
Thus we have:
- $(13): \quad 1 = \dfrac 1 3 + \dfrac 1 4 + \dfrac 1 4 + \dfrac 1 6$
$c = 5$:
\(\ds \dfrac 1 a + \dfrac 1 b + \dfrac 1 c\) | \(=\) | \(\ds \dfrac 1 3 + \dfrac 1 4 + \dfrac 1 5\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {20 + 15 + 12} {60}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {47} {60}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \dfrac 1 d\) | \(=\) | \(\ds 1 - \dfrac {47} {60}\) | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {13} {60}\) |
But $\dfrac {13} {60}$ is not a unit fraction.
Thus $a = 3, b = 4, c = 5$ does not lead to a solution.
Let $c > 5$.
\(\ds 1 - \dfrac 1 a + \dfrac 1 b + \dfrac 1 c\) | \(>\) | \(\ds \dfrac {13} {60}\) | ||||||||||||
\(\ds \) | \(>\) | \(\ds \dfrac {12} {60}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 1 5\) |
and so:
- $d < c$
and so no further solutions can be found where $a = 3, b = 4, 5 > 7$.
Hence there is exactly $1$ solution where $a = 3, b = 4$.
$\Box$
$b > 4$
Let $b > 4$.
Then:
\(\ds \dfrac 1 a + \dfrac 1 b + \dfrac 1 c + \dfrac 1 d\) | \(\le\) | \(\ds \dfrac 1 3 + \dfrac 1 5 + \dfrac 1 5 + \dfrac 1 5\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {5 + 3 + 3 + 3} {15}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {14} {15}\) | ||||||||||||
\(\ds \) | \(<\) | \(\ds 1\) |
Hence there are no solutions such that $a = 3, b > 5$.
$\Box$
$a = 4$
Let $a = 4$.
Let $d > 4$.
Then:
\(\ds \dfrac 1 a + \dfrac 1 b + \dfrac 1 c + \dfrac 1 d\) | \(<\) | \(\ds \dfrac 1 4 + \dfrac 1 4 + \dfrac 1 4 + \dfrac 1 4\) | ||||||||||||
\(\ds \) | \(<\) | \(\ds 1\) |
Thus the only solution where $a = 4$ is:
- $(14): \quad 1 = \dfrac 1 4 + \dfrac 1 4 + \dfrac 1 4 + \dfrac 1 4$
$\Box$
$a > 4$
Suppose $a > 4$.
Then:
\(\ds \dfrac 1 a + \dfrac 1 b + \dfrac 1 c + \dfrac 1 d\) | \(<\) | \(\ds \dfrac 1 4 + \dfrac 1 4 + \dfrac 1 4 + \dfrac 1 4\) | ||||||||||||
\(\ds \) | \(<\) | \(\ds 1\) |
Hence there are no solutions such that $a > 4$.
$\Box$
Summary
Hence our $14$ solutions:
\(\text {(1)}: \quad\) | \(\ds \) | \(\) | \(\ds \dfrac 1 2 + \dfrac 1 3 + \dfrac 1 7 + \dfrac 1 {42}\) | |||||||||||
\(\text {(2)}: \quad\) | \(\ds \) | \(\) | \(\ds \dfrac 1 2 + \dfrac 1 3 + \dfrac 1 8 + \dfrac 1 {24}\) | |||||||||||
\(\text {(3)}: \quad\) | \(\ds \) | \(\) | \(\ds \dfrac 1 2 + \dfrac 1 3 + \dfrac 1 9 + \dfrac 1 {18}\) | |||||||||||
\(\text {(4)}: \quad\) | \(\ds \) | \(\) | \(\ds \dfrac 1 2 + \dfrac 1 3 + \dfrac 1 {10} + \dfrac 1 {15}\) | |||||||||||
\(\text {(5)}: \quad\) | \(\ds \) | \(\) | \(\ds \dfrac 1 2 + \dfrac 1 3 + \dfrac 1 {12} + \dfrac 1 {12}\) | |||||||||||
\(\text {(6)}: \quad\) | \(\ds \) | \(\) | \(\ds \dfrac 1 2 + \dfrac 1 4 + \dfrac 1 5 + \dfrac 1 {20}\) | |||||||||||
\(\text {(7)}: \quad\) | \(\ds \) | \(\) | \(\ds \dfrac 1 2 + \dfrac 1 4 + \dfrac 1 6 + \dfrac 1 {12}\) | |||||||||||
\(\text {(8)}: \quad\) | \(\ds \) | \(\) | \(\ds \dfrac 1 2 + \dfrac 1 4 + \dfrac 1 8 + \dfrac 1 8\) | |||||||||||
\(\text {(9)}: \quad\) | \(\ds \) | \(\) | \(\ds \dfrac 1 2 + \dfrac 1 5 + \dfrac 1 5 + \dfrac 1 {10}\) | |||||||||||
\(\text {(10)}: \quad\) | \(\ds \) | \(\) | \(\ds \dfrac 1 2 + \dfrac 1 6 + \dfrac 1 6 + \dfrac 1 6\) | |||||||||||
\(\text {(11)}: \quad\) | \(\ds \) | \(\) | \(\ds \dfrac 1 3 + \dfrac 1 3 + \dfrac 1 4 + \dfrac 1 {12}\) | |||||||||||
\(\text {(12)}: \quad\) | \(\ds \) | \(\) | \(\ds \dfrac 1 3 + \dfrac 1 3 + \dfrac 1 6 + \dfrac 1 {6}\) | |||||||||||
\(\text {(13)}: \quad\) | \(\ds \) | \(\) | \(\ds \dfrac 1 3 + \dfrac 1 4 + \dfrac 1 4 + \dfrac 1 6\) | |||||||||||
\(\text {(14)}: \quad\) | \(\ds \) | \(\) | \(\ds \dfrac 1 4 + \dfrac 1 4 + \dfrac 1 4 + \dfrac 1 4\) |
$\blacksquare$
Historical Note
According to 1997: David Wells: Curious and Interesting Numbers (2nd ed.), this result is attributed to David Breyer Singmaster.
Sources
- 1994: Richard K. Guy: Unsolved Problems in Number Theory (2nd ed.)
- 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $14$