Sum of 3 Unit Fractions that equals 1

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Theorem

There are $3$ ways to represent $1$ as the sum of exactly $3$ unit fractions.


Proof

Let:

$1 = \dfrac 1 a + \dfrac 1 b + \dfrac 1 c$

where:

$0 < a \le b \le c$

and:


Aiming for a contradiction, suppose $a = 1$.

Then:

$1 = \dfrac 1 1 + \dfrac 1 b + \dfrac 1 c$

and so:

$\dfrac 1 b + \dfrac 1 c = 0$

which contradicts the stipulation that $b, c > 0$.

So there is no solution possible when $a = 1$.


Therefore $a \ge 2$.


$a = 2$

Let $a = 2$.


$b = 2$

Let $b = 2$.

Then:

$\dfrac 1 a + \dfrac 1 b = 1$

leaving no room for $c$.

Hence there are no solutions where $a = 2$ and $b = 2$.

$\Box$


$b = 3$

Let $b = 3$.

\(\ds \dfrac 1 a + \dfrac 1 b\) \(=\) \(\ds \dfrac 1 2 + \dfrac 1 3\)
\(\ds \) \(=\) \(\ds \dfrac {3 + 2} 6\)
\(\ds \) \(=\) \(\ds \dfrac 5 6\)
\(\ds \leadsto \ \ \) \(\ds \dfrac 1 c\) \(=\) \(\ds 1 - \dfrac 5 6\)
\(\ds \) \(=\) \(\ds \dfrac 1 6\)
\(\ds \leadsto \ \ \) \(\ds \dfrac 1 c\) \(=\) \(\ds \dfrac 1 6\)
\(\ds \leadsto \ \ \) \(\ds c\) \(=\) \(\ds 6\)


Thus we have:

$(1): \quad 1 = \dfrac 1 2 + \dfrac 1 3 + \dfrac 1 6$

$\Box$


$b = 4$

Let $b = 4$.

\(\ds \dfrac 1 a + \dfrac 1 b\) \(=\) \(\ds \dfrac 1 2 + \dfrac 1 4\)
\(\ds \) \(=\) \(\ds \dfrac {2 + 1} 4\)
\(\ds \) \(=\) \(\ds \dfrac 3 4\)
\(\ds \leadsto \ \ \) \(\ds \dfrac 1 c\) \(=\) \(\ds 1 - \dfrac 3 4\)
\(\ds \) \(=\) \(\ds \dfrac 1 4\)
\(\ds \leadsto \ \ \) \(\ds c\) \(=\) \(\ds 4\)


Thus we have:

$(2): \quad 1 = \dfrac 1 2 + \dfrac 1 4 + \dfrac 1 4$


$b > 4$

Let $b > 4$.

Then:

\(\ds \dfrac 1 a + \dfrac 1 b + \dfrac 1 c\) \(<\) \(\ds \dfrac 1 2 + \dfrac 1 4 + \dfrac 1 4\)
\(\ds \) \(=\) \(\ds 1\)

Hence there are no solutions such that $a = 2, b > 4$.

$\Box$


$a = 3$

Let $a = 3$.


$b = 3$

Let $b = 3$.

\(\ds \dfrac 1 a + \dfrac 1 b\) \(=\) \(\ds \dfrac 1 3 + \dfrac 1 3\)
\(\ds \) \(=\) \(\ds \dfrac 2 3\)
\(\ds \leadsto \ \ \) \(\ds \dfrac 1 c\) \(=\) \(\ds 1 - \dfrac 2 3\)
\(\ds \) \(=\) \(\ds \dfrac 1 3\)
\(\ds \leadsto \ \ \) \(\ds c\) \(=\) \(\ds 3\)


Thus we have:

$(3): \quad 1 = \dfrac 1 3 + \dfrac 1 3 + \dfrac 1 3$

$\Box$


$a > 3$

Let $a > 3$.

Then:

\(\ds \dfrac 1 a + \dfrac 1 b + \dfrac 1 c\) \(\le\) \(\ds \dfrac 1 a + \dfrac 1 a + \dfrac 1 a\)
\(\ds \) \(<\) \(\ds 1\)

Hence there are no solutions for $a>3$.

$\Box$


Summary

Hence our $3$ solutions:

\(\text {(1)}: \quad\) \(\ds \) \(\) \(\ds \dfrac 1 2 + \dfrac 1 3 + \dfrac 1 6\)
\(\text {(2)}: \quad\) \(\ds \) \(\) \(\ds \dfrac 1 2 + \dfrac 1 4 + \dfrac 1 4\)
\(\text {(3)}: \quad\) \(\ds \) \(\) \(\ds \dfrac 1 3 + \dfrac 1 3 + \dfrac 1 3\)

$\blacksquare$


Sources

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