# Only Number which is Sum of 3 Factors is 6

## Contents

## Theorem

The only positive integer which is the sum of exactly $3$ of its distinct coprime divisors is $6$.

## Corollary

$1$ can be expressed uniquely as the sum of $3$ distinct unit fractions:

- $1 = \dfrac 1 2 + \dfrac 1 3 + \dfrac 1 6$

## Proof

Let $n$ be such a positive integer with corresponding divisors $a, b, c$ such that:

- $a + b + c = n$

We note that the set $\set {k a, k b, k c}$ satisfy the same properties trivially as divisors of $k n$.

Hence the specification that $\set {a, b, c}$ is a coprime set.

Without loss of generality, suppose $a < b < c$.

Since $a, b, c$ are strictly positive, $n \ne c$.

Suppose $\dfrac n c \ge 3$.

Then:

\(\displaystyle n\) | \(=\) | \(\displaystyle a + b + c\) | |||||||||||

\(\displaystyle \) | \(<\) | \(\displaystyle c + c + c\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle n\) |

which is a contradiction.

Hence:

\(\displaystyle \frac n c\) | \(=\) | \(\displaystyle 2\) | |||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle a + b + c\) | \(=\) | \(\displaystyle 2 c\) | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle a + b\) | \(=\) | \(\displaystyle c\) | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle n\) | \(=\) | \(\displaystyle 2 a + 2 b\) |

Since $a, b$ are divisors of $n$:

- $a \divides \paren {2 a + 2 b}$
- $b \divides \paren {2 a + 2 b}$

which reduces to:

- $a \divides 2 b$
- $b \divides 2 a$

Suppose $b$ is odd.

Then by Euclid's Lemma, we would have $b \divides a$.

By Absolute Value of Integer is not less than Divisors, this gives $b \le a$, which is a contradiction.

Thus $b$ is even.

Suppose $a$ is even.

Then $a, b, c$ are all even.

So $\gcd \set {a, b, c} \ne 1$, which is a contradiction.

Therefore it must be the case that $a$ is odd.

Then by Euclid's Lemma, we have:

- $a \divides \dfrac b 2$

and:

- $\dfrac b 2 \divides a$

By Absolute Value of Integer is not less than Divisors, this gives:

- $\dfrac b 2 = a$

Because $\gcd \set {a, b, c} = 1$, we must have $a = 1$.

Hence the set $\set {1, 2, 3}$ is obtained.

$\blacksquare$

## Sources

- 1986: David Wells:
*Curious and Interesting Numbers*... (previous) ... (next): $6$ - 1997: David Wells:
*Curious and Interesting Numbers*(2nd ed.) ... (previous) ... (next): $6$