Problem 3.4. Find all triples of positive integers $a \leq b \leq c$ that are primitive (i.e., they have no common prime factor) such that each of them divides the sum of the other two.
## Solution Problem 3.4

Suppose $a=b$. Since $a$ and $b$ have no common factors, it must be $a=b=1$. Since $c$ divides $a+b=2$, this gives the triples $(1,1,1)$ and $(1,1,2)$.

Now suppose $a<b$. Since $c$ divides $a+b<c+c=2c$, it must be $a+b=c$. But then, since $b$ divides $a+c=2a+b$, it follows that $b$ divides $2a$, and since $b$ has no common factors with $a$, $b$ divides 2. $b$ cannot be 1 since it is greater than $a$, so the only possible triple in this case is $(1,2,3)$.