2008 Indonesia MO Problems/Problem 3

Solution 1

Summing up the equation $\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}$ yields the result

$\frac{a^2b+ab^2+b^2c+bc^2+c^2a+ca^2}{abc}=z\in \mathbb{Z}$ . Thus,

\begin{align*} a|(a^2b+ab^2+b^2c+bc^2+c^2a+ca^2)&\implies a|b^2c+bc^2\\ b|(a^2b+ab^2+b^2c+bc^2+c^2a+ca^2)&\implies b|c^2a+ca^2\\ c|(a^2b+ab^2+b^2c+bc^2+c^2a+ca^2)&\implies c|a^2b+ab^2 \end{align*}

Since $a$ , $b$ , $c$ , are pairwise relatively prime, this implies that $\gcd(a,b)=1$ , $\gcd(b,c)=1$ , and $\gcd(a,c)=1$ .

\begin{align*} a&|b+c\\ b&|c+a\\ c&|a+b \end{align*}

WLOG, $a\le b\le c$ .

Suppose $a>1$ .

Since they are pairwise relatively prime, and all $a$ , $b$ , $c > 1$ , $a$ is not equal to $b$ nor $c$ , and $b$ is not equal to $c$ . A strict order of $1<a<b<c$ can be made. Since $c|a+b$ and $b>a>0$ , we have $a+b\ge c$ . We also know that $a<c$ and $b<c$ , which implies that $a+b<2c$ . Thus, the only way $c|a+b$ , while $c\le a+b < 2c$ , is if $a+b = c$ .

Plugging in $c = a+b$ , we get $a|(a+2b)\implies a|2b$ , and $b|2a$ . Because $a<b$ , we know that $2a<2b$ . Thus, in order for $b|2a$ and $b\le 2a <2b$ , the only possible way is for $b=2a$ . However, we have previously established that $a>1$ , and combined with the fact that $b=2a \implies \gcd(a, b) = a$ , $a$ and $b$ are not co-prime, which is a contradiction.

Hence, $a=1$ .

Case 1: $b=1$

Let $b=1$ . since $c|a+b$ , we have $c|2$ . The only options will be $c=1$ and $c=2$ . This gives us the answers $(1,1,1)$ and $(1,1,2)$

Case 2: $b>1$

Let $b>1$ . If $b=c$ , then $\gcd(b,c)=b>1$ , and they won't be co-prime. As a result, $b$ is strictly less than $c$ . Since $b|c+a$ , and $a=1$ , we have $b|c+1$ . Similarly, $c|b+1$ . But $b<c$ . Thus, $c=b+1$ . Using the fact that $b|c+1$ , we have $b|b+2\implies b|2$ . Hence, $b=2$ , and $c=3$ .

Plugging in the answers $(1,1,1)$ , $(1,2,3)$ , and $(1,1,2)$ all yield valid results of $k=6$ , $k=8$ , and $k=7$ , respectively.

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2008 Indonesia MO Problems/Problem 3

Solution 1