Integer Solutions to a system of diophantine equations Are there any integer solutions $(x,y,z,a_2,a_3,b_1,b_3,c_1,c_2)$ for: $x + y^{b_1}z^{c_1} = y + x^{a_2}z^{c_2}=z+x^{a_3}y^{b_3}$.
Where $x \le y \le z$ The one thing I could tell about this equation is that if you consider $x + y^{b_1}z^{c_1} = y + x^{a_2}z^{c_2}$ since $y - x \le z$ either $c_1$ or $c_2$ must be $0$.
EDIT: $(x,y,z,a_2,a_3,b_1,b_3,c_1,c_2)$ are all positive integers. Also I realized that if $c_1,c_2 = 0$ for the system of equations to be satisfied it is possible when for $(x,y,a,b)$ if $x^a-x = y^b - y$ who's solutions are here:
http://oeis.org/A057896.
 A: Looking first for a simple solution in positive integers with $x\leq y\leq z$, suppose we try $x=1$ and $a_2=a_3=b_1=b_3=c_1=c_2=1$.  The equation then reduces to:
$$1+yz=y+z=z+y$$
where we also require $1\leq y\leq z$. Rearranging:
$$yz-y=z-1$$
$$y(z-1)=z-1$$
Hence $y=1$ and $z$ can be any integer $\geq 1$. Taking $z=2$ as the simplest case that avoids the trivial $x=y=z$ we have the solution:
$$1+1^12^1=1+1^12^1=2+1^11^1=3$$
A necessary condition for $x + y^{b_1}z^{c_1} = y + x^{a_2}z^{c_2}$ with $c_1,c_2\geq 1$ is that $z|(y-x)$. In the above solution this is trivially satisfied since $y-x=0$. 
Suppose now we add the additional requirement that $x<y<z$. Then $z\nmid (y-x)$ since $0<y-x<z$. Thus the necessary condition can no longer be satisfied.  So we can conclude that the equation:
$$x + y^{b_1}z^{c_1} = y + x^{a_2}z^{c_2}=z+x^{a_3}y^{b_3}$$
has no solutions in positive integers with $x<y<z$.  Given the equation's symmetry in $x,y,z$, we can also state that it has no solution in positive integers with $x,y,z$ all different.
