Sum of 4 Unit Fractions that equals 1

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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

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