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 $a = 2, b = 3$.

$b = 5$
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: