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 $b = 2$.
==== $b = 3$ ====
Let $b = 3$.
Thus we try $c = 7, 8, \ldots$ in turn.
$c = 7$:
Thus we have:
:$(1): \quad 1 = \dfrac 1 2 + \dfrac 1 3 + \dfrac 1 7 + \dfrac 1 {42}$
$c = 8$:
Thus we have:
:$(2): \quad 1 = \dfrac 1 2 + \dfrac 1 3 + \dfrac 1 8 + \dfrac 1 {24}$
$c = 9$:
Thus we have:
:$(3): \quad 1 = \dfrac 1 2 + \dfrac 1 3 + \dfrac 1 9 + \dfrac 1 {18}$
$c = 10$:
Thus we have:
:$(4): \quad 1 = \dfrac 1 2 + \dfrac 1 3 + \dfrac 1 {10} + \dfrac 1 {15}$
$c = 11$:
But $\dfrac 5 {66}$ is not a unit fraction.
Thus $a = 2, b = 3, c = 11$ does not lead to a solution.
$c = 12$:
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 $a = 2, b = 3$.
==== $b = 4$ ====
Let $b = 4$.
Thus we try $c = 5, 6, \ldots$ in turn.
$c = 5$:
Thus we have:
:$(6): \quad 1 = \dfrac 1 2 + \dfrac 1 4 + \dfrac 1 5 + \dfrac 1 {20}$
$c = 6$:
Thus we have:
:$(7): \quad 1 = \dfrac 1 2 + \dfrac 1 4 + \dfrac 1 6 + \dfrac 1 {12}$
$c = 7$:
But $\dfrac 3 {28}$ is not a unit fraction.
Thus $a = 2, b = 4, c = 7$ does not lead to a solution.
$c = 8$:
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 $a = 2, b = 4$.
==== $b = 5$ ====
Let $b = 5$.
Thus we try $c = 5, 6, \ldots$ in turn.
$c = 5$:
Thus we have:
:$(9): \quad 1 = \dfrac 1 2 + \dfrac 1 5 + \dfrac 1 5 + \dfrac 1 {10}$
$c = 6$:
But $\dfrac 2 {15}$ is not a unit fraction.
Thus $a = 2, b = 5, c = 6$ does not lead to a solution.
$c = 7$:
But $\dfrac {11} {70}$ is not a unit fraction.
Thus $a = 2, b = 5, c = 7$ does not lead to a solution.
Let $c > 7$.
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 $a = 2, b = 5$.
==== $b = 6$ ====
Let $b = 6$.
But we already have that $c \ge b$.
Thus:
:$c \ge 6$
Thus we try $c = 6, 7, \ldots$ in turn.
$c = 6$:
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 $a = 2, b = 6$.
==== $b > 6$ ====
Let $b > 6$.
Then:
Hence there are no solutions such that $a = 2, b > 6$
=== $a = 3$ ===
Let $a = 3$.
==== $b = 3$ ====
Let $b = 3$.
Thus we try $c = 4, 5, \ldots$ in turn.
$c = 4$:
Thus we have:
:$(11): \quad 1 = \dfrac 1 3 + \dfrac 1 3 + \dfrac 1 4 + \dfrac 1 {12}$
$c = 5$:
But $\dfrac 2 {15}$ is not a unit fraction.
Thus $a = 3, b = 3, c = 5$ does not lead to a solution.
$c = 6$:
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 $a = 3, b = 3$.
==== $b = 4$ ====
Let $b = 4$.
But we already have that $c \ge b$.
Thus:
:$c \ge 4$
Thus we try $c = 4, 5, \ldots$ in turn.
$c = 4$:
Thus we have:
:$(13): \quad 1 = \dfrac 1 3 + \dfrac 1 4 + \dfrac 1 4 + \dfrac 1 6$
$c = 5$:
But $\dfrac {13} {60}$ is not a unit fraction.
Thus $a = 3, b = 4, c = 5$ does not lead to a solution.
Let $c > 5$.
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 $a = 3, b = 4$.
==== $b > 4$ ====
Let $b > 4$.
Then:
Hence there are no solutions such that $a = 3, b > 5$.
=== $a = 4$ ===
Let $a = 4$.
Let $d > 4$.
Then:
Thus the only solution where $a = 4$ is:
:$(14): \quad 1 = \dfrac 1 4 + \dfrac 1 4 + \dfrac 1 4 + \dfrac 1 4$
=== $a > 4$ ===
Suppose $a > 4$.
Then:
Hence there are no solutions such that $a > 4$.
=== Summary ===
Hence our $14$ solutions:
== Historical Note ==
== Sources ==
* 
* : $14$
Category:1
Category:Unit Fractions
Category:Recreational Mathematics