TITLE: Solutions of $x^p+y^q=y^r+z^p=z^q+x^r$
QUESTION [7 upvotes]: I'm struggling with the following problem from Terence Tao's "Solving Mathematical Problems":

Find all positive reals $x,y,z$ and all positive integers $p,q,r$ such that
  $$x^p+y^q=y^r+z^p=z^q+x^r.$$

Obviously, taking $x=y=z=1$ we can have $p,q,r$ arbitrary. Also, I've found the symmetries of the problem
$$x,y,z,p,q,r \mapsto x,y,z,p,q,r \\
x,y,z,p,q,r \mapsto x,z,y,r,q,p   \\
x,y,z,p,q,r \mapsto y,x,z,q,p,r   \\
x,y,z,p,q,r \mapsto y,z,x,r,p,q   \\
x,y,z,p,q,r \mapsto z,x,y,q,r,p   \\
x,y,z,p,q,r \mapsto z,y,x,p,r,q$$
I was hoping that all solutions follow a simple rule, like $x=y=z$, but unfortunately the solution
$$10,10,\sqrt{190},2,2,1 $$
shows it is not the case. Can I get any help on this please? Thanks!
EDIT:
I've noticed that equality of two exponents (i.e. $p=q$) implies equality of two basis (i.e. $x=y$), and then the solution is 
$$z=(2x^p-x^r)^{1/p} $$
where $2x^p-x^r>0$. (Of course, different exponents being equal gives rise to different forms of the solution)
I've also noticed that you could have solutions with $x,y,z,p,q,r$ all distinct:
$$x=\frac{1}{2} \left(\frac{\sqrt{69}}{4}-\frac{3}{4}\right),y=\frac{1}{2},z=\frac{1}{2} \left(\frac{\sqrt{69}}{4}-\frac{1}{2}\right),p=1,q=2,r=3.$$
This makes me believe that the problem is very general and hence difficult.

REPLY [4 votes]: Apparently, the problem was indeed too complicated, as stated in the errata here:

In page 66, problem 44, there is a $=2$ missing at the end of the string of equations, thus $x^p+y^q=y^r+z^p=z^q+x^r=2$.

We still have the solution $x=y=z=1$ for arbitrary $p,q,r$. Say we have a solution $x',y',z'$ where WLOG $x'>1$. That would force $y'<1$ from the first expression, which, in turn forces $z'>1$ from the second expression, which forces $x'<1$ from the third, which is a contradiction. 
Thus the only solution is $x=y=z=1$ and $p,q,r \in \mathbb{N}$.
[P.S. 
I don't think there is be a closed form for the solution without the $=2$ at the end, but this is based on solely numerical evidence.]