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:
$$\frac{a+b}{c}+\frac{b+c}{a}+\frac{a+c}{b} = n?$$

A linear diophantine equation
$$ax+by=c$$
where $c$ is divisible by $d=gcd(a,b)$, has solution
$$a=a_{0}+\frac{b}{d}\cdot k$$
$$b=b_{0}-\frac{a}{d}\cdot k$$
However, your equation is not linear, since
$$\frac{a+b}{c}+\frac{b+c}{a}+\frac{a+c}{b}=n$$
$$\frac{\left(a+b\right)ab+\left(b+c\right)bc+\left(a+c\right)ac}{abc}=n$$
Note here you are interested in combinations of $\left(a,b,c\right)$ such that this quotient is an integer $n$. Also,
$$a^{2}b+ab^{2}+b^{2}c+bc^{2}+a^{2}c+ac^{2}=abcn$$
proves it's not possible to represent $\left(a,b,c\right)$ in the same terms of $k$ as a linear diophantine equation, so they don't share a straight-forward relationship, but this last equation is very interesting, since it is contained in the cubic expansion $\left(a+b+c\right)^{3}$.
Then,
$$\left(a+b+c\right)^{3}=a^{3}+b^{3}+c^{3}+3\left(a^{2}b+ab^{2}+b^{2}c+bc^{2}+a^{2}c+ac^{2}\right)+6abc$$
$$\left(a+b+c\right)^{3}=a^{3}+b^{3}+c^{3}+3\left(abcn\right)+6abc$$
$$\left(a+b+c\right)^{3}=a^{3}+b^{3}+c^{3}+3abc\left(n+2\right)$$
$$\left(a+b+c\right)^{3}-\left(a^{3}+b^{3}+c^{3}\right)=3abc\left(n+2\right)$$
As you can see, this difference is divisible by 3, so using congruence is an excellent idea:
$$\left(a+b+c\right)^{3}\equiv a^{3}+b^{3}+c^{3}\;\left(mod\;\;3\right)$$
So, given that $gcd\left(a,b,c\right)=1$, every combination of $\left(a,b,c\right)$ where the cube of the sum is congruent to the sum of the cubes $\left(mod\;\;3\right)$ is a possible answer.
For instance, $\left(1,1066,3977\right)$ and $\left(1,1598,4182\right)$ are valid answers, but it's difficult to trace an easy linear relation between $1$ and the other values.
