# 2008 Indonesia MO Problems/Problem 3

## Solution 1

Summing up the equation yields the result

. 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 , , , are pairwise relatively prime, this implies that , , and .

, but because and , . Similarly, and .

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

WLOG, .

Suppose .

Since they are pairwise relatively prime, and all ,,, is not equal to nor , and is not equal to . A strict order of can be made. Since and , we have . We also know that and , which implies that . Thus, the only way , while , is if .

Plugging in , we get , and . Because , we know that . Thus, in order for and , the only possible way is for . However, we have previously established that , and combined with the fact that , and are not co-prime, which is a contradiction.

Hence, .

Case 1:

Let . since , we have . The only options will be and . This gives us the answers and

Case 2:

Let . If , then , and they won't be co-prime. As a result, is strictly less than . Since , and , we have . Similarly, . But . Thus, . Using the fact that , we have . Hence, , and .

Plugging in the answers , , and all yield valid results of , , and , respectively.