# Set of 3 Integers each Divisor of Sum of Other Two

## Theorem

There exists exactly one set of distinct coprime positive integers such that each is a divisor of the sum of the other two:

$\set {1, 2, 3}$

## Proof

We note that if $\set {a, b, c}$ is such a set, then $\set {k a, k b, k c}$ satisfy the same properties trivially.

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

We have that:

$5 \times 1 = 2 + 3$ so $1 \divides 2 + 3$
$2 \times 2 = 1 + 3$ so $2 \divides 1 + 3$
$1 \times 3 = 1 + 2$ so $3 \divides 1 + 2$

It remains to be shown that this is the only such set.

We are to find all the sets $\set {a, b, c}$ such that:

$a \divides b + c$
$b \divides a + c$
$c \divides a + b$

where $a \ne b$, $a \ne c$ and $b \ne c$.

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

Since $2 c > a + b > 0$ and $c \divides a + b$, we must have $a + b = c$.

Then it follows that:

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

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.

$\blacksquare$