1 can be Expressed as Sum of 4 Distinct Unit Fractions in 6 Ways/Proof 1

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 $w$:

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:

As $x < y$:

Therefore solutions exist only in the domain:


 * $\dfrac {2 w} {w - 2} < x < \dfrac {4 w} {w - 2}$

and:
 * $w < x$

Case $w = 3$:

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 $w = 4$:

Integer solutions in the above domain can again be found by inspection:
 * $\tuple {5, 20}, \tuple {6, 12}$

Case $w = 5$:

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: