A Note on Unit Fractions and its Representations as Sum of Unit Fractions

Published by Ganit Charcha | Category - Math Articles | 2015-01-14 02:00:47

A rational number or a fraction where the numerator is $1$ and the denominator is any positive integer $n,$ is called a unit fraction and is denoted as $\frac{1}{n}$. Unit fraction $\frac{1}{n}$, is also called the reciprocal of the positive integer $n$.
Let us now consider the following problem whch looks so innocent.

Problem: Find three distinct positive integers whose reciprocals add upto $1$.
Solution: Since, $2 < 3 < 4 < 5 < \ldots$, so $\frac{1}{2} > \frac{1}{3} > \frac{1}{4} > \frac{1}{5} > \ldots$. Note that $$\frac{1}{3} + \frac{1}{4} + \frac{1}{5} = \frac{47}{60} < 1$$ Without considering the unit fraction $\frac{1}{2}$, therefore, if we take any other three unit fractions $\frac{1}{a}$, $\frac{1}{b}$ and $\frac{1}{c}$ such that $a, b, c \neq 2$, then $S = \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \leq \frac{47}{60} < 1.$
So, one of the reciprocals has to be $\frac{1}{2},$ if their sum is to be equal to $1$. Without loss of generality, let us take $a = 2$. Therefore, $S - \frac{1}{a} = \frac{1}{b} + \frac{1}{c} = \frac{1}{2}$. Again, note that $\frac{1}{4} + \frac{1}{5} = \frac{9}{20} < \frac{1}{2}$.
So, following a similar reasonin as above, it is easily seen that, if $\frac{1}{b} + \frac{1}{c} = \frac{1}{2},$ then one of $\frac{1}{b}$, $\frac{1}{c}$ has to be $\frac{1}{3}$. Without loss of generality, we consider $\frac{1}{b} = \frac{1}{3}$. Therefore, $\frac{1}{c} = \frac{1}{2} - \frac{1}{3} = \frac{1}{6}$.
Therefore, the three integers are $2$, $3$ and $6$ such that $\frac{1}{2} + \frac{1}{3} + \frac{1}{6} = 1$.

We have adopted here a methodical approach and a bit of reasoning to solve the problem. We will now see if we can solve the problem in some intelligent way.
Note that, $\frac{1}{2m} = \frac{1}{6m} + \frac{1}{3m}$ for any $m > 0$. In other words, {\bf for any $m > 0$, any unit fraction with even denominator $2m$, can be expressed as the sum of two unit fractions. And the two unit fractions are $\frac{1}{3m}$ and $\frac{1}{6m}$}.
So, $\frac{1}{2} = \frac{1}{3} + \frac{1}{6}$ and this helps us to see very easily that $\frac{1}{2} + \frac{1}{3} + \frac{1}{6} = \frac{1}{2} + \frac{1}{2} = 1$. So, we have got the solution to the above problem very easily using the above trick.

Given the trick above, naturally one question comes to our mind. Whether there is any other way of representing this?
Look $\frac{1}{6m}$ resembles to the form described above, so it can be expressed as $\frac{1}{6m} = \frac{1}{18m} + \frac{1}{9m}$ and hence $$\frac{1}{2m} = \frac{1}{6m} + \frac{1}{3m} = \frac{1}{18m} + \frac{1}{9m} + \frac{1}{3m}$$ and this can be further expresed as sum of four unit fractions by decomposing $\frac{1}{18m}$ as sum of two unit fractions.
As for example, for $m = 1$, $$\frac{1}{2} = \frac{1}{6} + \frac{1}{3} = \frac{1}{18} + \frac{1}{9} + \frac{1}{3} = \frac{1}{54} + \frac{1}{27} + \frac{1}{9} + \frac{1}{3}$$. We can therefore restate the above trick as follows.
For any $m > 0$, any unit fraction with even denominator $2m$, can be expressed as the sum of any number of other unit fractions.
Our next question is, whether this can be true for any unit fraction. Let, $n$ be any integer. Then, $$\frac{1}{n} = \frac{1}{n(n + 1)} + \frac{1}{n + 1}$$ and each of the individual unit fractions can be further decomposed and so we have the following Theorem.
Theorem: For any integer $n > 0$, the unit fraction $\frac{1}{n}$ can be expressed as the sum of any number of other unit fractions.
We will now take a look at the different ways of representing an unit fraction $\frac{1}{n}$ as sum of two other unit fractions. Let us take an example first. For example, $$\frac{1}{12} = \frac{1}{24} + \frac{1}{24} = \frac{1}{18} + \frac{1}{36} = \frac{1}{16} + \frac{1}{48} = \frac{1}{15} + \frac{1}{60} = \frac{1}{14} + \frac{1}{84} = \frac{1}{13} + \frac{1}{156} = \frac{1}{21} + \frac{1}{28} = \frac{1}{20} + \frac{1}{30}$$ So, there are $8$ dfferent ways of representing $\frac{1}{12}$ as sum of two unit fractions we have been able to found out. Question is whether there is any way of systematically generating these representations of an unit fraction.

Let, $n = a.b$ holds for the integer $n$ and $a > 0$, $b > 0$ are the two factors of $n$. Note that $$\frac{1}{n} = \frac{a + 1}{(ab)(a+1)} = \frac{1}{b(a+1)} + \frac{1}{n(a+1)}$$ and also$$\frac{1}{n} = \frac{b + 1}{(ab)(b+1)} = \frac{1}{a(b+1)} + \frac{1}{n(b+1)}$$ It is therefore easily understandable that each factor of $n$ contributes to one such decomposition of $\frac{1}{n}$ as sum of two unit fractions. If $c(n)$ is the number of divisors of $n$, then number of representations of $\frac{1}{n}$ as sum of two unit fractions is at least $c(n)$. Remember that for any integer $n$, $c(n)$ can be found out using prime factorization theorem.
Let us take $n = 12,$ the divisors of which are $1$, $2$, $3$, $4$, $6$ and $12$. We can see now the following.

1.  Representation $\frac{1}{12} = \frac{1}{24} + \frac{1}{24}$ is being contributed by divisor $1$.
2.  Representation $\frac{1}{12} = \frac{1}{18} + \frac{1}{36}$ is being contributed by divisor $2$.
3.  Representation $\frac{1}{12} = \frac{1}{16} + \frac{1}{48}$ is being contributed by divisor $3$.
4.  Representation $\frac{1}{12} = \frac{1}{15} + \frac{1}{60}$ is being contributed by divisor $4$.
5.  Representation $\frac{1}{12} = \frac{1}{14} + \frac{1}{84}$ is being contributed by divisor $6$.
6.  Representation $\frac{1}{12} = \frac{1}{13} + \frac{1}{156}$ is being contributed by divisor $12$.

Note that there is still two other ways of representing and those are $$\frac{1}{12} = \frac{7}{12.7} = \frac{3 + 4}{12.7} = \frac{1}{21} + \frac{1}{28}$$ and $$\frac{1}{12} = \frac{5}{12.5} = \frac{2 + 3}{12.5} = \frac{1}{20} + \frac{1}{30}$$ And these two ways of representing $\frac{1}{12}$ provides us clue to formulate another rule of representing a unit fraction as sum of two other unit fractions.
Let, $S$ be the set of divisors of the integer $n$, including $1$ and the number $n$ itself. Let, $a \in S$ be any divisor. If there exists two other divisors $c \in S$ and $d \in S$, $c \neq d \neq a$ and also ($c \neq 1$, $d \neq 1$) such that $a + 1 = c + d$, then $$\frac{1}{n} = \frac{a+1}{n(a+1)} = \frac{c+d}{n(a+1)} = \frac{1}{\frac{n}{c}(a+1)} + \frac{1}{\frac{n}{d}(a+1)}$$ Denoting the number of representations of $\frac{1}{n}$ as sum of two unit fractions as $r(n)$, we can clearly see that $$r(n) \geq c(n)$$ It is easy to verify that for any prime $p$, $r(p) = c(p) = 2$. We leave it to the reader to verify this.

Problem: Find three distinct positive integers with the least possible sum such that the sum of the reciprocals of any two integers among them is an integral multiple of the reciprocal of the third integer.  [RMO 2010]
Solution: Note that if $n = a.b$, then due to this factorization of $n$, $\frac{1}{n}$ can be represented in two ways as sum of two other unit fractions. Let us first consider the decomposition, $$\frac{1}{n} = \frac{1}{b(a+1)} + \frac{1}{n(a+1)}$$ Also note that, $\frac{1}{n} + \frac{1}{b(a+1)} = \frac{2a+1}{n(a+1)}$. So, we have got three integers $n$, $b(a+1)$ and $n(a+1)$, for which we will now take a look at the sum of the reciprocals of the integers $n$ and $n(a+1)$.$$\frac{1}{n} + \frac{1}{n(a+1)} = \frac{a+2}{a\{b(a+1)\}}$$ So, if $a+2$ is divisible by $a$, then we have found out three integers $n$, $b(a+1)$ and $n(a+1)$ which satisfies the property mentioned in the question. But, $a+2$ is divisible by $a$ only if $a = 2$.
Considering $a = 2$, we can readily infer that $n(a+1)$ is divisible by $6$ and in fact $n(a+1) \geq 6$ which implies that $\frac{1}{n(a+1)} \leq \frac{1}{6}$.
Since, $b$ is a positive integer greater or equal to $1$, so $b(a+1) \geq 3$ and hence $\frac{1}{b(a+1)} \leq \frac{1}{3}$. Therefore, $\frac{1}{n} = \frac{1}{b(a+1)} + \frac{1}{n(a+1)} \leq (\frac{1}{6} + \frac{1}{3}) = \frac{1}{2}$ and hence $n \geq 2$. So, the three positive integers $n$, $b(a+1)$ and $n(a+1)$ satisfying the property in question also satisfies $n \geq 2$, $b(a+1) \geq 3$ and $n(a+1) \geq 6$. The sum $n + b(a+1) + n(a+1) \geq 2 + 3 + 6 = 12$ and the equality holds for the integers $2$, $3$ and $6$ respectively. Thus the required numbers with least sum are $2$, $3$ and $6$.

Problem: Find a set of $6$ distinct positive integers, so that the sum of their reciprocals is equal to $1$.
Solution: We will talk about a general strategy for this sort of problems. Note that, if $a < b$, then $\frac{1}{a} > \frac{1}{b}$. This means, $\frac{1}{2} > \frac{1}{3} > \frac{1}{4} > \frac{1}{5} > \ldots$ and so on.
So, we can start with any unit fraction and can add unit fractions on the right while keeping the sum less than $1$. Note that for any starting unit fraction $\frac{1}{a},$ if $\frac{1}{b} (< \frac{1}{a})$ is the first unit fraction on the right of it such that $\frac{1}{a} + \frac{1}{b} < 1$, then for all $i = 1, 2, \ldots$, we have $$\frac{1}{a} + \frac{1}{b + i} < \frac{1}{a} + \frac{1}{b} < 1$$And we need to continue this process until we get the desired number of integers.

Let us now come back to the problem in hand and let us choose $\frac{1}{2}$ to start with. Then, $$\frac{1}{2} + \frac{1}{3} = \frac{5}{6} < 1$$$$\frac{5}{6} + \frac{1}{4} = \frac{13}{12} > 1$$$$\frac{5}{6} + \frac{1}{5} = \frac{31}{30} > 1$$$$\frac{5}{2} + \frac{1}{6} = \frac{6}{6} = 1$$$$\frac{5}{6} + \frac{1}{7} = \frac{41}{42} < 1$$$$\frac{41}{42} + \frac{1}{43} = \frac{1805}{1806} < 1$$ We can now see that if we add the unit fraction $\frac{1}{1806}$ then it will make it $1$. Hence we have got $5$ positive integers $2$, $3$, $7$, $43$ and $1806$ such that $$\frac{1}{2} + \frac{1}{3} + \frac{1}{7} + \frac{1}{43} + \frac{1}{1806} = 1$$We need $6$ distinct positive integers but we have $5$ now, so we can take any of the already found unit fractions and can decompose it as sum of two other distinct unit fractions to get the desired number of unit fractions. Note that, $\frac{1}{3} = \frac{1}{4} + \frac{1}{12}$ and $\frac{1}{7} = \frac{1}{8} + \frac{1}{56}$, so we have $$\frac{1}{2} + \frac{1}{4} + \frac{1}{7} + \frac{1}{12} + \frac{1}{43} + \frac{1}{1806} = 1$$$$\frac{1}{2} + \frac{1}{3} + \frac{1}{8} + \frac{1}{43} + \frac{1}{56} + \frac{1}{1806} = 1$$ We could have even chosen to decompose $\frac{1}{1806} = \frac{1}{1807} + \frac{1}{3263442}$ and hence $$\frac{1}{2} + \frac{1}{3} + \frac{1}{7} + \frac{1}{43} + \frac{1}{1807} + \frac{1}{3263442} = 1$$ Therefore, we have found three sets of distinct positive integers sum of whose reciprocals is equal to $1$. This strategy could be chosen to figure out many more solutions to this problem.