Is it possible to find all integer solutions to this system of equations, inequalities and inequations? Given the following three equations, assuming all unknowns are integers:
$$
x y =\left(x + y - z \right) a  
\\
 x z =\left(x + z - y \right) b  
\\
 y z =\left(y + z - x \right) c  
$$
And the following constraints on the values for $x$, $y$, and $z$:
$$
0 < z \lt x + y \\
0 < y \lt x + z \\
0 < x \lt y + z \\
x \ne y \\
x \ne z \\
y \ne z 
$$
There are no constraints on the values of $a$, $b$, and $c$ beyond that they are integers.
Is there any tractable way to find all of the sets of integer solutions that satisfy the constraints, or lacking that, a way to prove that no integer solutions for all unknowns that satisfy the constraints exist?
 A: There certainly are integer solutions. Let $p$, $q$, $r$ be any integers satisfying the triangle inequalities and let $L$ be the least common multiple of $p+q-r$, $q+r-p$, $r+p-q$. Then $x=Lp$, $y=Lq$, $z=Lr$ will do, with $a = \dfrac{pqL}{p+q-r}$, $b = \dfrac{prL}{p+r-q}$, $c = \dfrac{qrL}{q+r-p}$, which are all integers by construction of $L$. For example, if $p=4$, $q=5$, and $r=7$, then $L=24$ so $x=96$, $y=120$, $z=168$, $a=240$, $b=112$, $c=105$
A: Defining $x':=c$, $y':=b$, $z':=a$ for better notational symmetry ... Note that the system and constraints on $x$, $y$, $z$ require $x'$, $y'$, $z'$ (already assumed to be integers) to be strictly-positive integers.
We can solve the system explicitly. Discarding cases where one or more of $x$, $y$, $z$ is zero, a single case remains:
$$\begin{align}
(x,y,z) &= \left(- \frac{y'z'}{x'} + y' + z' \;,\; x' - \frac{x' z'}{y'} + z' \;,\;  x' + y' - \frac{x' y'}{z'} \right) \tag1\\[6pt]
&= \left(y'z'\left(- \frac1{x'} + \frac1{y'} + \frac1{z'} \right), z'x'\left(\frac1{x'} - \frac1{y'} + \frac1{z'} \right),\;  
x'y'\left( \frac1{x'} + \frac1{y'} - \frac{1}{z'} \right)\right) \tag2
\end{align}$$
Now,

*

*From $(1)$, we see that, for the results be integers, each of $\dfrac{y'z'}{x'}$, $\dfrac{z'x'}{y'}$, and $\dfrac{x'y'}{z'}$ must be integers.


*From $(2)$ we see that, for the results to be positive, the reciprocals $1/x'$, $1/y'$, $1/z'$ must also satisfy their own three-fold triangle inequality.


*For the results to be distinct, we must have, for instance,
$$0 \neq x-y = -z' (x' - y') \left(\frac1{x'}+\frac1{y'}-\frac1{z'} \right) \quad\to\quad x'\neq y'$$
so that $x'$, $y'$, $z'$ themselves must be distinct.
We can state the following:

When (and only when) $x'$, $y'$, $z'$ satisfy the conditions above, $x$, $y$, $z$ will be distinct positive integers that satisfy the three-fold triangle inequality.

(@mcd's answers include the examples $(x',y',z') = (105,112,240)$ and $(x',y',z')=(15,12,20)$ satisfying all of the requirements.)
A: Also, let $p=x+y-z$, $q=x+z-y$, $r = y+z-x$ for positive integer $p$, $q$, $r$ (which is possible because of the inequality constraints). Then $2x=p+q$, $2y=p+r$ and $2z=r+q$, so the equations are $(p+q)(p+r)=4ap$, $(p+q)(r+q)=4bq$, $(p+r)(r+q)=4cr$, which means that $p$ is a factor of $qr$, $q$ is a factor of $pr$ and $r$ is a factor of $pq$. This gives another set of solutions: pick $p = 4st$, $q=4tu$, $r=4us$, so that $a=(s+u)(t+u)$, $b=(s+u)(s+t)$, $c=(t+u)(s+t)$ and $x=2t(u+s)$, $y=2s(u+t)$ and $z=2u(s+t)$. For example, $s=1$, $t=2$, $u=3$ gives $x=16$, $y=10$ and $z=18$, with $a =20$, $b= 12$, $c=15$.
