Set of 3 Integers each Divisor of Sum of Other Two

Theorem
There exists exactly one set of positive integers such that each is a divisor of the sum of the other two:
 * $\set {1, 2, 3}$

Proof
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 find sets $\set {a,b,c}$ such that:


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

We only find $\set {a,b,c}$ with the property $\gcd \set {a,b,c} = 1$,

since if $\set {a,b,c}$ is such a set, $\set {k a, k b, k c}$ satisfy the same properties trivially.

The case $\set {1,1,1}$ is trivial as well.

Suppose $a = b$, with $c \ne a$. Then we have:


 * $a \divides a + c$
 * $c \divides a + a$

Which reduces to:


 * $a \divides c$
 * $c \divides 2 a$

As we require $a = \gcd \set {a,c} = 1$, we have $c \divides 2 a = 2$.


 * $c \ne a \implies c = 2$.

Therefore $\set {1,1,2}$ is the second such set.

Now we consider the case where $a,b,c$ are distinct.

, suppose $a < b < c$.

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

Then we have:


 * $a \divides b + a + b$
 * $b \divides a + a + b$

Which reduces to:


 * $a \divides 2 b$
 * $b \divides 2 a$

Suppose $b$ is odd.

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

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 we must have $a$ is odd.

Then by Euclid's Lemma, we would have and $a \divides \dfrac b 2$ and $\dfrac b 2 \divides a$.

By Absolute Value of Integer is not less than Divisors, this gives $\dfrac b 2 = a$.

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

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