# 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 $4$ distinct unit fractions in $6$ different ways:

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

## Proof 1

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 $w$:

 $\ds \dfrac 1 2$ $=$ $\ds \dfrac 1 w + \dfrac 1 x + \dfrac 1 y$ $\ds \dfrac 1 2$ $<$ $\ds \dfrac 1 3 + \dfrac 1 4 + \dfrac 1 5$ $\ds \dfrac 1 2$ $<$ $\ds \dfrac 1 4 + \dfrac 1 5 + \dfrac 1 6$ $\ds \dfrac 1 2$ $<$ $\ds \dfrac 1 5 + \dfrac 1 6 + \dfrac 1 7$ $\ds \dfrac 1 2$ $>$ $\ds \dfrac 1 6 + \dfrac 1 7 + \dfrac 1 8$

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:

 $\ds \dfrac 1 x$ $<$ $\ds \dfrac {w - 2} {2 w}$ $\ds \leadsto \ \$ $\ds x$ $>$ $\ds \dfrac {2 w} {w - 2}$

As $x < y$:

 $\ds x$ $<$ $\ds \dfrac 1 {\dfrac {w - 2} {2 w} - \dfrac 1 x}$ $\ds \leadsto \ \$ $\ds \dfrac 1 x$ $<$ $\ds \dfrac {w - 2} {2 w} - \dfrac 1 x$ $\ds \leadsto \ \$ $\ds \dfrac 2 x$ $<$ $\ds \dfrac {w - 2} {2 w}$ $\ds \leadsto \ \$ $\ds x$ $<$ $\ds \dfrac {4 w} {w - 2}$

Therefore solutions exist only in the domain:

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

and:

$w < x$

Case $w = 3$:

 $\ds 6$ $<$ $\ds x < 12$ $\ds \text {and } 3 < x$ $\ds \leadsto \ \$ $\ds 6$ $<$ $\ds x < 12$

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 $w = 4$:

 $\ds 4$ $<$ $\ds x < 8$ $\ds \text {and } 4 < x$ $\ds \leadsto \ \$ $\ds 4$ $<$ $\ds x < 8$

Integer solutions in the above domain can again be found by inspection:

$\tuple {5, 20}, \tuple {6, 12}$

Case $w = 5$:

 $\ds \dfrac {10} 3$ $<$ $\ds \dfrac {20} 3$ $\ds \text {and } 5 < x$ $\ds \leadsto \ \$ $\ds 5$ $<$ $\ds \dfrac {20} 3$

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:

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

$\blacksquare$

## Proof 2

There are $14$ ways to represent $1$ as the sum of exactly $4$ unit fractions.

This includes repeated unit fractions.

The full list is:

 $\text {(1)}: \quad$ $\ds$  $\ds \dfrac 1 2 + \dfrac 1 3 + \dfrac 1 7 + \dfrac 1 {42}$ $\text {(2)}: \quad$ $\ds$  $\ds \dfrac 1 2 + \dfrac 1 3 + \dfrac 1 8 + \dfrac 1 {24}$ $\text {(3)}: \quad$ $\ds$  $\ds \dfrac 1 2 + \dfrac 1 3 + \dfrac 1 9 + \dfrac 1 {18}$ $\text {(4)}: \quad$ $\ds$  $\ds \dfrac 1 2 + \dfrac 1 3 + \dfrac 1 {10} + \dfrac 1 {15}$ $\text {(5)}: \quad$ $\ds$  $\ds \dfrac 1 2 + \dfrac 1 3 + \dfrac 1 {12} + \dfrac 1 {12}$ $\text {(6)}: \quad$ $\ds$  $\ds \dfrac 1 2 + \dfrac 1 4 + \dfrac 1 5 + \dfrac 1 {20}$ $\text {(7)}: \quad$ $\ds$  $\ds \dfrac 1 2 + \dfrac 1 4 + \dfrac 1 6 + \dfrac 1 {12}$ $\text {(8)}: \quad$ $\ds$  $\ds \dfrac 1 2 + \dfrac 1 4 + \dfrac 1 8 + \dfrac 1 8$ $\text {(9)}: \quad$ $\ds$  $\ds \dfrac 1 2 + \dfrac 1 5 + \dfrac 1 5 + \dfrac 1 {10}$ $\text {(10)}: \quad$ $\ds$  $\ds \dfrac 1 2 + \dfrac 1 6 + \dfrac 1 6 + \dfrac 1 6$ $\text {(11)}: \quad$ $\ds$  $\ds \dfrac 1 3 + \dfrac 1 3 + \dfrac 1 4 + \dfrac 1 {12}$ $\text {(12)}: \quad$ $\ds$  $\ds \dfrac 1 3 + \dfrac 1 3 + \dfrac 1 6 + \dfrac 1 {6}$ $\text {(13)}: \quad$ $\ds$  $\ds \dfrac 1 3 + \dfrac 1 4 + \dfrac 1 4 + \dfrac 1 6$ $\text {(14)}: \quad$ $\ds$  $\ds \dfrac 1 4 + \dfrac 1 4 + \dfrac 1 4 + \dfrac 1 4$

The result follows from inspection: solutions $(1)$, $(2)$, $(3)$, $(4)$, $(6)$ and $(7)$ are the only solutions of the above such that the denominators of the summands are distinct.

$\blacksquare$