<s>[INST] Find all the integers which are of form $\dfrac{b+c}{a}+\dfrac{c+a}{b}+\dfrac{a+b}{c}, a,b,c\in \mathbb{N}$, any two of $a,b,c$ are relatively prime. 
I have a question which askes to find all the integers which can be
expressed as
$\displaystyle \tag*{} \dfrac{b+c}{a}+\dfrac{c+a}{b}+\dfrac{a+b}{c}$
where $a,b,c\in \mathbb{N} $  and any two of $a,b,c$ are relatively
prime.

My approach:
Since they told any two of $a,b,c$ are relatively prime, so:
$\displaystyle \tag*{} \begin{align} \text{gcd}(a,b) &= 1 \\ \text{gcd}(b,c) &= 1 \\ \text{gcd}(c,a) &= 1\end{align}$
We have:
$\displaystyle \tag*{} \dfrac{b+c}{a}+\dfrac{c+a}{b}+\dfrac{a+b}{c} = \dfrac{b}{a}+\dfrac{c}{a}+\dfrac{c}{b}+\dfrac{a}{b}+\dfrac{a}{c}+\dfrac{b}{c}$
And we have to prove:
$\displaystyle \tag*{} \frac{a}{b} + \frac{b}{a}+ \frac{c}{a} + \frac{a}{c}+ \frac{b}{c} + \frac{c}{b} \in \mathbb{N} $
We have the following that if $\text{gcd}(m,n)=1$ and $m,n,k \in \mathbb{Z}$ then
$\displaystyle \tag*{} \frac{m}{n} + \frac{n}{m} = k$
has only $1$ positive solution, which is $m=n=1$
So, the only solution I found is $a=b=c=1$.
Is there anything I am missing? Any help would be appreciated, thank you. :)
 [/INST] $\frac{ab(a+b)+bc(b+c)+ca(c+a)}{abc}\in \mathbb{N}$, therefore $abc|ab(a+b)+bc(b+c)+ca(c+a)$. Because $ab(a+b)$ and $ca(c+a)$ is a multiple of $a$, $a|bc(b+c)$ is true. $bc$ is coprime with $a$, thus $a|b+c$. Using the same ways, it can be proven that $$a|b+c \\ b|a+c \\ c|a+b$$ WLOG $a\geq b\geq c$. Therefore $b+c$ is $a$ or $2a$. If $a=b+c$, then $b|b+2c$ and $b|2c$. If $b=c$, then $(a,b,c)$ is $(2,1,1)$ and the value is 7. If $b=2c$, $(a,b,c)$ is $(3,2,1)$ and the value is 8. If $2a=b+c$, then $(a,b,c)$ is $(1,1,1)$ and the value is 6. Therefore, all integers possible are 6, 7 and 8.
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