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

Theorem
The number $1$ can be expressed as the sum of $4$ reciprocals of unique positive integers in $6$ different ways:

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:
 * $1 \stackrel{?}{=} \dfrac {1}{3} + \dfrac {1}{4} + \dfrac{1}{5} + \dfrac{1}{6}$

Although we find:


 * $1 > \dfrac {1}{3} + \dfrac {1}{4} + \dfrac{1}{5} + \dfrac{1}{6}$

Therefore, there can be no solutions where $v \geq 3$, as that solution was the largest possible.

Hence, $v = 2$ if there are any solutions.

Repeating the above anaylsis on $w$:
 * $ \dfrac {1}{2} = \dfrac {1}{w} + \dfrac{1}{x} + \dfrac{1}{y}$


 * $ \dfrac {1}{2} < \dfrac {1}{3} + \dfrac{1}{4} + \dfrac{1}{5}$
 * $ \dfrac {1}{2} < \dfrac {1}{4} + \dfrac{1}{5} + \dfrac{1}{6}$
 * $ \dfrac {1}{2} < \dfrac {1}{5} + \dfrac{1}{6} + \dfrac{1}{7}$
 * $ \dfrac {1}{2} > \dfrac {1}{6} + \dfrac{1}{7} + \dfrac{1}{8}$

Potential Solutions located where $w = 3, 4 , 5$.

Now that $v$ and $w$ are known, the variable $y$ can be written in terms of $x$:


 * $y = \displaystyle \frac {1}{\frac{w - 2}{2w} - \frac{1}{x}}$

Solutions are only positive when:


 * $\displaystyle \frac{1}{x} < \displaystyle\frac{w - 2}{2w}$


 * $x > \displaystyle \frac{2w}{w - 2}$

As $x < y$:


 * $x < \displaystyle \frac {1}{\frac{w - 2}{2w} - \frac{1}{x}}$


 * $\displaystyle \frac{1}{x} < \displaystyle \frac{w - 2}{2w} - \frac{1}{x}$


 * $\displaystyle \frac{2}{x} < \displaystyle \frac{w - 2}{2w}$


 * $x < \displaystyle\frac{4w}{w - 2}$

Therefore solutions only exist in the domain:


 * $ \displaystyle \frac{2w}{w - 2} < x < \displaystyle \frac{4w}{w - 2}$ and $w < x$

Case $w = 3$:


 * $6 < x < 12$ and $3 < x$


 * $6 < x < 12$

Integer solutions in the above domain:
 * $ \tuple{7,42}, \tuple{8,24}, \tuple{9,18}, \tuple{10,15}$

Case $w = 4$:


 * $ 4 < x < 8$ and $4 < x$


 * $ 4 < x < 8$

Integer solutions in the above domain:
 * $ \tuple{5,20}, \tuple{6,12}$

Case $w = 5$


 * $ \displaystyle \frac{10}{3} < x < \displaystyle \frac{20}{3}$ and $5 < x$


 * $ 5 < x < \displaystyle \frac{20}{3}$

There are no integer solutions in this domain.

All solutions:
 * $ \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:


 * $ \dfrac {1}{2} + \dfrac {1}{3} + \dfrac{1}{7} + \dfrac{1}{42}$
 * $ \dfrac {1}{2} + \dfrac {1}{3} + \dfrac{1}{8} + \dfrac{1}{24}$
 * $ \dfrac {1}{2} + \dfrac {1}{3} + \dfrac{1}{9} + \dfrac{1}{18}$
 * $ \dfrac {1}{2} + \dfrac {1}{3} + \dfrac{1}{10} + \dfrac{1}{15}$
 * $ \dfrac {1}{2} + \dfrac {1}{4} + \dfrac{1}{5} + \dfrac{1}{20}$
 * $ \dfrac {1}{2} + \dfrac {1}{4} + \dfrac{1}{6} + \dfrac{1}{12}$

Also see

 * 2520 equals Sum of 4 Divisors in 6 Ways