Only Number which is Sum of 3 Factors is 6

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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