# Egyptian Fractions

An **Egyptian fraction** is the sum of finitely many rational numbers, each of which can be expressed in the form \(\dfrac{1}{q},\) where \(q\) is a positive integer.

For example, the Egyptian fraction \(\dfrac{61}{66}\) can be written as \(\dfrac{61}{66} = \dfrac12 + \dfrac13 + \dfrac{1}{11}.\)

#### Contents

## Formal Definition

For any rational number \( \frac{m}{n} \), where \(m, n\) are integers, there exist unit fractions (with numerator 1, and integer denominators) such that the sum of these unit fractions is equal to \(\frac{m}{n} \). As an example, \[\dfrac{2}{3}= \dfrac{1}{2} + \dfrac{1}{6}. \] However, the representation of \(\frac{m}{n} \) in this form is not unique because of the identity \[ 1 = \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{6}. \] Thus we have \[\dfrac{2}{3}= \dfrac{1}{2} + \dfrac{1}{6} = \dfrac{1}{2} + \dfrac{1}{6}\left(\dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{6}\right) = \dfrac{1}{2} + \dfrac{1}{12} + \dfrac{1}{18} + \dfrac{1}{36} \] as well, and we can keep on decomposing any fraction as such.

## History

## Methods for Decomposing an Egyptian Fraction

Some useful algebraic identities for the Egyptian fraction decomposition are: \[\begin{align} \dfrac{a}{ab - 1} &= \dfrac{1}{b} + \dfrac{1}{b(ab - 1)} \\ \dfrac{1}{a} &= \dfrac{1}{a + 1} + \dfrac{1}{a(a + 1)}. \end{align} \] These are easily proved by the laws of addition of fractions.

Find the Egyptian fraction representation of \( \dfrac{8}{11} \).

We have \[\dfrac{8}{11} = \dfrac{6}{11} + \dfrac{2}{11}.\] Now, \(11 = 2 \cdot 6 - 1. \) Thus, using the identity \[ \dfrac{a}{ab - 1} = \dfrac{1}{b} + \dfrac{1}{b(ab - 1)},\] we have \[\begin{align} \dfrac{6}{11} &= \dfrac{1}{2} + \dfrac{1}{22} \\\\ \dfrac{2}{11} &= \dfrac{1}{6} + \dfrac{1}{66} \\\\ \Rightarrow \dfrac{8}{11} &= \dfrac{1}{2} + \dfrac{1}{6} + \dfrac{1}{22} + \dfrac{1}{66}.\ _\square \end{align} \]

## Relevance to Modern Number Theory

Egyptian fractions show up in all sorts of places! Below are some problems that involve Egyptian fraction representations. First, we prove a useful lemma:

Lemma:

If \(r\) and \(s\) are two integers, such that \( \gcd(r,s) = 1 \) and \( a_1, a_2, \ldots, a_n \) are integers such that \[\frac{r}{s} = \sum\limits_{i = 1} ^n \frac{1}{a_i}, \] then \(s\) is a divisor of \( a_1\times a_2 \times \cdots \times a_n \).

By summing up the fractions on the right-hand side, we have \[\frac{r}{s} = \frac{\sum\limits_{cyc} {(a_1\times a_2 \times \cdots \times a_{n-1})}}{a_1\times a_2 \times \cdots \times a_n}. \]

Clearly, \(\sum\limits_{cyc} {(a_1\times a_2 \times \cdots \times a_{n-1})} = t\) is an integer. Thus, we have \[ st = r (a_1\times a_2 \times \cdots \times a_n), \] which means \(r (a_1\times a_2 \times \cdots \times a_n)\) is a multiple of s. But \( s \) cannot divide \( r \) as \( \gcd(r,s) = 1 \).Therefore, \( s \) divides \( a_1\times a_2 \times \cdots \times a_n \). \(_\square\)

[IMO 1979 Problem 1]

If \(p\) and \(q\) are positive integers such that \[ \frac{p}{q}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+ \cdots -\frac{1}{1318}+\frac{1}{1319}, \] prove that \(p\) is divisible by \(1979\).

We have

\[\begin{align} \dfrac{p}{q} &= 1 + \dfrac{1}{2}+ \dfrac{1}{3} + \dfrac{1}{4}+ \cdots + \dfrac{1}{1318}+\dfrac{1}{1319} - 2\left(\dfrac{1}{2} + \dfrac{1}{4} + \cdots + \dfrac{1}{1318}\right) \\ &= 1 + \dfrac{1}{2}+ \dfrac{1}{3} + \dfrac{1}{4}+ \cdots + \dfrac{1}{1318}+\dfrac{1}{1319} - \left(1 + \dfrac{1}{2}+ \dfrac{1}{3} + \dfrac{1}{4}+ \cdots + \dfrac{1}{659}\right) \\ &= \dfrac{1}{660}+ \dfrac{1}{661} + \dfrac{1}{662}+ \cdots + \dfrac{1}{1318}+\dfrac{1}{1319}. \end{align}\]

Now, \[ \dfrac{1}{660} + \dfrac{1}{1319} = \dfrac{660 + 1319}{660 \cdot 1319} = \dfrac{1979}{660 \cdot 1319}.\]

Similarily, \[ \dfrac{1}{661} + \dfrac{1}{1318} = \dfrac{661 + 1318}{661 \cdot 1318} = \dfrac{1979}{661 \cdot 1318}. \]

We can pair up all the fractions as such (as there are an even number of them).

We can now write \[ \frac{p}{q} = 1979 \cdot \frac{r}{s}, \] where \[\frac{r}{s} = \dfrac{1}{660 \cdot 1319} + \dfrac{1}{661 \cdot 1318} + \cdots + \dfrac{1}{989 \cdot 990}. \]

Hence, \[ p = 1979 \cdot \frac{qr}{s}. \]

But \( 1979 \) is prime! Therefore, as \( p \) is an integer and \(s \) does not divide \( 1979 \) \((\)as \( s \) is a divisor of \( 660 \times 661 \times \cdots \times 1319\), using our lemma above\()\), \(p \) is a multiple of \(1979\). \(_\square\)

The next problem is from the USA Mathematical Olympiad.

[USAMO 2010 Problem 5]

Let \( p \) be an odd prime, and let \(q = \frac{3p-5}{2}\). If \[ S_q = \frac{1}{2\cdot 3 \cdot 4} + \frac{1}{5\cdot 6 \cdot 7} + \cdots + \frac{1}{q(q+1)(q+2)} \] and \(\frac{1}{p}-2S_q = \frac{m}{n}\) for integers \(m\) and \(n\), then \(m - n\) is divisible by \(p\).

**Cite as:**Egyptian Fractions.

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