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