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

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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

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:

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

$\blacksquare$

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