Diophantine equation $\frac{a+b}{c}+\frac{b+c}{a}+\frac{a+c}{b} = n$ Let $a,b,c$ and $n$ be natural numbers and $\gcd(a,b,c)=\gcd(\gcd(a,b),c)=1$.
Does it possible to find all tuples $(a,b,c,n)$ such that:
$$\frac{a+b}{c}+\frac{b+c}{a}+\frac{a+c}{b} = n?$$

Add +3 both sides 
We get $(a+b+c)(\frac{ab+bc+ca}{abc})=n+3$ 
Since $gcd(ab+bc+ca,abc)=1$ beacuse $(a,b,c)=1$ we should have $(a+b+c)\vdots (abc)$ and then we can get solutions $(1,1,1),(1,1,2),(1,2,3)$
