Theorem: 1 can be Expressed as Sum of 4 Distinct Unit Fractions in 6 Ways

The number $1$ can be expressed as the sum of $4$ distinct unit fractions in $6$ different ways:
{{begin-eqn}}
{{eqn | l = 1
      | r = \frac 1 2 + \frac 1 3 + \frac 1 7 + \frac 1 {42}
}}
{{eqn | r = \frac 1 2 + \frac 1 3 + \frac 1 8 + \frac 1 {24}
}}
{{eqn | r = \frac 1 2 + \frac 1 3 + \frac 1 9 + \frac 1 {18}
}}
{{eqn | r = \frac 1 2 + \frac 1 3 + \frac 1 {10} + \frac 1 {15}
}}
{{eqn | r = \frac 1 2 + \frac 1 4 + \frac 1 5 + \frac 1 {20}
}}
{{eqn | r = \frac 1 2 + \frac 1 4 + \frac 1 6 + \frac 1 {12}
}}
{{end-eqn}}

Proof:
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 $w$:
{{begin-eqn}}
{{eqn | l = \dfrac 1 2
      | r = \dfrac 1 w + \dfrac 1 x + \dfrac 1 y
}}
{{eqn | l = \dfrac 1 2
      | o = <
      | r = \dfrac 1 3 + \dfrac 1 4 + \dfrac 1 5
}}
{{eqn | l = \dfrac 1 2
      | o = <
      | r = \dfrac 1 4 + \dfrac 1 5 + \dfrac 1 6
}}
{{eqn | l = \dfrac 1 2
      | o = <
      | r = \dfrac 1 5 + \dfrac 1 6 + \dfrac 1 7
}}
{{eqn | l = \dfrac 1 2
      | o = >
      | r = \dfrac 1 6 + \dfrac 1 7 + \dfrac 1 8
}}
{{end-eqn}}
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:
{{begin-eqn}}
{{eqn | l = \dfrac 1 x
      | o = <
      | r = \dfrac {w - 2} {2 w}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = x
      | o = >
      | r = \dfrac {2 w} {w - 2}
      | c = 
}}
{{end-eqn}}
As $x < y$:
{{begin-eqn}}
{{eqn | l = x
      | o = <
      | r = \dfrac 1 {\dfrac {w - 2} {2 w} - \dfrac 1 x}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \dfrac 1 x
      | o = <
      | r = \dfrac {w - 2} {2 w} - \dfrac 1 x
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \dfrac 2 x
      | o = <
      | r = \dfrac {w - 2} {2 w}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = x
      | o = <
      | r = \dfrac {4 w} {w - 2}
      | c = 
}}
{{end-eqn}}
Therefore solutions exist only in the domain:
:$\dfrac {2 w} {w - 2} < x < \dfrac {4 w} {w - 2}$
and:
:$w < x$
Case $w = 3$:
{{begin-eqn}}
{{eqn | l = 6
      | o = <
      | r = x < 12
      | rr= \text {and } 3 < x
      | c = 
}}
{{eqn | ll= \leadsto
      | l = 6
      | o = <
      | r = x < 12
      | c = 
}}
{{end-eqn}}
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 $w = 4$:
{{begin-eqn}}
{{eqn | l = 4
      | o = <
      | r = x < 8
      | rr= \text {and } 4 < x
      | c = 
}}
{{eqn | ll= \leadsto
      | l = 4
      | o = <
      | r = x < 8
      | c = 
}}
{{end-eqn}}
Integer solutions in the above domain can again be found by inspection:
:$\tuple {5, 20}, \tuple {6, 12}$
Case $w = 5$:
{{begin-eqn}}
{{eqn | l = \dfrac {10} 3
      | o = <
      | r = \dfrac {20} 3
      | rr= \text {and } 5 < x
      | c = 
}}
{{eqn | ll= \leadsto
      | l = 5
      | o = <
      | r = \dfrac {20} 3
      | c = 
}}
{{end-eqn}}
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:
{{begin-eqn}}
{{eqn | l = 1
      | r = \frac 1 2 + \frac 1 3 + \frac 1 7 + \frac 1 {42}
}}
{{eqn | r = \frac 1 2 + \frac 1 3 + \frac 1 8 + \frac 1 {24}
}}
{{eqn | r = \frac 1 2 + \frac 1 3 + \frac 1 9 + \frac 1 {18}
}}
{{eqn | r = \frac 1 2 + \frac 1 3 + \frac 1 {10} + \frac 1 {15}
}}
{{eqn | r = \frac 1 2 + \frac 1 4 + \frac 1 5 + \frac 1 {20}
}}
{{eqn | r = \frac 1 2 + \frac 1 4 + \frac 1 6 + \frac 1 {12}
}}
{{end-eqn}}
{{qed}}