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: 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.
A: First note that we can assume $a,b,c \in \mathbb{Q}$, without loss of generality. A rational solution can then be scaled to integers.
Combining into a single fraction gives the quadratic in $a$
\begin{equation*}
a^2+\frac{b^2-nbc+c^2}{b+c}a+bc=0
\end{equation*}
and, for rational solutions, the discriminant must be a rational square.
Thus, the quartic
\begin{equation*}
D^2=b^4-2(n+2)b^3c+(n^2-6)b^2c^2-2(n+2)bc^3+c^4
\end{equation*}
must have rational solutions.
This quartic is birationally equivalent to the elliptic curve
\begin{equation*}
V^2=U^3+(n^2-12)U^2+16(n+3)U
\end{equation*}
with
\begin{equation*}
\frac{b}{c}=\frac{V+(n+2)U}{2(U-4(n+3))}
\end{equation*}
The elliptic curve is singular when $n=-2,-3,6$. $n=6$, for example, corresponds to $a=b=c=K$ as a solution. 
If $n \ne -2,-3,6$, the curve has $5$ finite torsion points at $(0,0)$, $(4,\pm 4(n+2))$ and $(4n+12,\pm 4(n+2)(n+3))$ none of which give a solution. For $n=7$, there are a further $6$ torsion points which lead to the solutions $(a,b,c)=(1,1,2)$ and $(a,b,c)=(1,2,2)$.
Thus, if $n \ne -3,-2,6,7$, we need the elliptic curve to have rank greater than zero to find a solution. Computations using the Birch and Swinnerton-Dyer conjecture suggest the first few solutions are with $n=8,11,12,15,\ldots$.
For example, the $n=15$ curve has a generator $(-36,468)$ which gives the solution $a=2, b=3, c=15$. As $n$ gets larger, the size of solutions generally increases.
A: 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 $(1,1,1),(1,1,2),(1,2,3)$
