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Difference between revisions of "2008 Indonesia MO Problems/Problem 3"

(Created page with "==Solution 1== Summing up the equation <math>\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}</math> yields the result <math>\frac{a^2b+ab^2+b^2c+bc^2+c^2a+ca^2}{abc}=z\in \mathbb...")
 
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Let <math>b>1</math>. If <math>b=c</math>, then <math>\gcd(b,c)=b>1</math>, and they won't be co-prime. As a result, <math>b</math> is strictly less than <math>c</math>. Since <math>b|c+a</math>, and <math>a=1</math>, we have <math>b|c+1</math>. Similarly, <math>c|b+1</math>. But <math>b<c</math>. Thus, <math>c=b+1</math>. Using the fact that <math>b|c+1</math>, we have <math>b|b+2\implies b|2</math>. Hence, <math>b=2</math>, and <math>c=3</math>.
 
Let <math>b>1</math>. If <math>b=c</math>, then <math>\gcd(b,c)=b>1</math>, and they won't be co-prime. As a result, <math>b</math> is strictly less than <math>c</math>. Since <math>b|c+a</math>, and <math>a=1</math>, we have <math>b|c+1</math>. Similarly, <math>c|b+1</math>. But <math>b<c</math>. Thus, <math>c=b+1</math>. Using the fact that <math>b|c+1</math>, we have <math>b|b+2\implies b|2</math>. Hence, <math>b=2</math>, and <math>c=3</math>.
  
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Plugging in the answers <math>(1,1,1)</math>, <math>(1,2,3)</math>, and <math>(1,1,2)</math> all yield valid results.
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Plugging in the answers <math>(1,1,1)</math>, <math>(1,2,3)</math>, and <math>(1,1,2)</math> all yield valid results of <math>k=6</math>, <math>k=8</math>, and <math>k=7</math>, respectively.

Latest revision as of 12:10, 17 September 2024

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 $\gcd(a,c)=1$.

$a|b^2c+bc^2\implies a|bc(b+c)$, but because $\gcd(a,b)=1$ and $\gcd(a,c)=1$, $a|b+c$. Similarly, $b|c+a$ and $c|a+b$.

\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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