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A number of the form $\frac{1}{n}$, where $n$ is an integer greater than $1$, is called a unit fraction.

Noting that $\frac{1}{2} = \frac{1}{3} + \frac{1}{6}$ and $\frac{1}{3} = \frac{1}{4} + \frac{1}{12}$, find a general result of the form $\frac{1}{n} = \frac{1}{a} + \frac{1}{b}$ and hence prove that any unit fraction can be expressed as the sum of two other distinct unit fractions.

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$$\frac{1}{N+1}+\frac{1}{N(N+1)}=\frac{N}{N(N+1)}+\frac{1}{N(N+1)}=\frac{N+1}{N(N+1)}=\frac{1}{N}.$$


Here's a question for further investigation: is the above decomposition of a unit fraction into a pair of distinct unit fractions unique? After all, there is more than one way to split a unit fraction into a triplet of unit fractions.

$$\frac12=\frac14+\frac16+\frac{1}{12},$$

but also

$$\frac12=\frac13+\frac18+\frac{1}{24}.$$

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  • $\begingroup$ Beautifully done in one line +1 $\endgroup$ – Asimov Sep 6 '14 at 15:26
  • $\begingroup$ Can you make the $N$ lowercase? $\endgroup$ – Cole Johnson Sep 7 '14 at 17:02
  • $\begingroup$ @ColeJohnson The person who asked the question used uppercase, so I did too. But why on earth does it matter? $\endgroup$ – David H Sep 7 '14 at 17:11
  • $\begingroup$ @DavidH variables are commonly lowercase, and the original question has been edited. $\endgroup$ – Cole Johnson Sep 7 '14 at 17:41
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In reply to DavidH's question "is the above decomposition of a unit fraction into a pair of distinct unit fractions unique?":

If $n$ is not prime, then we can write $n = n_1n_2$ with $n_1 \neq 1 \neq n_2$, and then we have the decomposition

$$\frac{1}{n} = \frac{1}{n_1}\frac{1}{n_2} = \frac{1}{n_1}\Big(\frac{1}{n_2+1} + \frac{1}{n_2(n_2+1)}\Big) = \frac{1}{n_1(n_2+1)} + \frac{1}{n_1n_2(n_2+1)} \\ = \frac{1}{n+n_1} + \frac{1}{n(n_2+1)}$$

which is different to the decomposition

$$\frac{1}{n} = \frac{1}{n+1} + \frac{1}{n(n+1)}$$

Now suppose $n$ is prime. Suppose

$$\frac{1}{n} = \frac{1}{n+a} + \frac{1}{c}$$

Assume also, wlog, that $n + a < c$, which tells us that $a < n$. Then

$$\frac{1}{c} = \frac{1}{n} - \frac{1}{n+a} = \frac{a}{n(n+a)}$$

But $n$ is prime, so $a$ and $n$ are coprime, as are $a$ and $a + n$, so we must have $a = 1$ and $c = n(n+1)$, i.e. the decomposition

$$\frac{1}{n} = \frac{1}{n+1} + \frac{1}{n(n+1)}$$

is unique when $n$ is prime.

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$\frac{1}{n}$ - $\frac{1}{n + k}$ = $\frac{k}{n(n+k)}$

$\frac{k}{n(n+k)}$ = $\frac{1}{\frac{n(n+k)}{k}}$

So, as long as $\frac{n(n+k)}{k}$ is a whole number, we can say that $\frac{1}{\frac{n(n+k)}{k}}$ is a unit fraction and since

$\frac{1}{n}$ = $\frac{k}{n(n+k)}$ + $\frac{1}{n + k}$

$\frac{1}{n}$ can be expressed as a sum of two unit fractions as long as k has some value such that $\frac{n(n+k)}{k}$ is a whole number.

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