Proving or disproving the only co-prime solutions 
Prove or disprove that the only solutions of
$$\begin{align}
xy = z\pmod {x+y}   \\
xz = y \pmod {x+z}\\
yz = x \pmod {y+z}\\
\end{align}
$$ in positive co-prime integers $x,y,z$ are $(1,1,1)$ and $(5,7,11)$.

My work:
The system of congruence is clearly symmetrical so we can assign the parameter labels so that $a<b<c$. In explicit form, the system of congruences can be written as
$$\begin{align}
xy = z+c(x+y)   \\
xz = y+b(x+z)\\
yz = x+a(y+z)\\
\end{align}
$$
where $a,b,c$ are integers. I tried adding and factoring these equations but nothing useful came out. I am stuck here.
Any help is greatly appreciated.
 A: This is not an actual proof but just an idea
The equations that you wrote imply that
$$\left(\frac{x-1}{a}\right)\left(\frac{y-1}{b}\right)\left(\frac{z-1}{c}\right)-\left(\frac{x-1}{a}\right)-\left(\frac{y-1}{b}\right)-\left(\frac{z-1}{c}\right)=2$$
An equation of this form i.e. $$ABC-A-B-C=2$$
has the positive integer solutions $(2,2,2)$, $(1,2,5)$, and $(1,3,3)$, and the integer solution $(-1,-1,-1)$.  However, not all triples $(a,b,c)$ that satisfy
the original set of congruences lead to integer quantities in the brackets.
Another interesting implication of the three coupled equations is
$$\left(\frac{z-x}{by}\right)=\frac{\left(\frac{z-y}{ax}\right)+\left(\frac{y-x}{cz}\right)}{1+\left(\frac{z-y}{ax}\right)\left(\frac{y-x}{cz}\right)}$$
which shows that the normalized distances [$a$ to $b$] and [$b$ to $c$] add up to the total distance [$a$ to $c$] in accord with the addition rule for velocities in special relativity.  (This can be seen most clearly by setting $x=\frac{1}{z}$, $y=\frac{1}{y}$, and $z=\frac{1}{x}$. )
For any integer solution $A,B,C$ of the original congruences, define
$$\begin{align}
g=\textrm{GCD}(A,B,C)\\
x=\frac{A}{g}\\
y=\frac{B}{g}\\
z=\frac{C}{g}
\end{align}
$$
and put $M = LCM(x+y,x+z,y+z)$.  It's clear that $x,y,z$ are pairwise
co-prime and that infinitely many other solution triples are given by
$$\begin{align}
A' = x(Mk+g)\\
B' = y(Mk+g)\\
C' = z(Mk+g)\\
\end{align}
$$
where $k$ is any integer.  A solution $(xg,yg,zg)$ with $g < M$ is called a minimal solution. Examination of the minimal solutions with $g=1$ shows that certain values
of $(x+y+z)$ occur frequently.
The basic equations imply that if $(x,y,z)$ are co-prime then $x$ divides
$bc-1$, $y$ divides $ac-1$, and $z$ divides $ab-1$, but this doesn't seem to be sufficient to prove that $(1,1,1)$ and $(5,7,11)$ are the only positive
co-prime solutions.  It appears that for most $($all$?)$ $g > 1$ there are
infinitely many minimal solutions, but I can't prove that either.
