How many solutions exist for a non-linear system How many solutions exist to the following system:
$$
\begin{eqnarray}
xy+xz &=& 54+x^2 \\
yx+yz &=& 64+y^2  \\
xz+yz &=&  70+z^2
\end{eqnarray}
$$
I have guessed that the solutions, if exist, have to be integer, then we get:
$$
\begin{eqnarray}
x(y+z-x) &=& 2\cdot 3\cdot 3\cdot 3 \\
y(x+z-y) &=& 2\cdot2\cdot2\cdot2\cdot2\cdot2  \\
z(x+y-z) &=&  2\cdot5\cdot7
\end{eqnarray}
$$
Looking at the third equation, there are 8 possibilities for the value of $z$ and the situation is much worse for the other two equations. 
My questions are:


*

*is my guess about the solutions is correct, and how one can prove it?

*how to proceed without exhaustive testing for each of the possibilities?

*is there a better way to approach this problem?


Thanks!

Firstly as you had factorized:
  $x(y+z-x) = 54$.
  $y(z+x-y) = 64$.
  $z(x+y-z) = 70$.
Also the differences between equations are:
  $(x-y)z = (x-y)(x+y) - 10$.
  $(y-z)x = (y-z)(y+z) - 6$.
  $(x-z)y = (x-z)(x+z) - 16$.
Which are equivalent to:
  $(x-y)(x+y-z) = 10$.
  $(y-z)(y+z-x) = 6$.
  $(x-z)(z+x-y) = 16$.
And combining with the factorizations gives:
  $\frac{x-y}{z} = \frac{10}{70}$.
  $\frac{y-z}{x} = \frac{6}{54}$.
  $\frac{x-z}{y} = \frac{16}{64}$.
You can already easily guess an answer at this point: $7,8,9$.
But continuing we can always scale $(x,y,z)$ so that $z = 7$ and so:
  $x-y = 1$ and $9(y-7) = x$.
Whence you get the scaled solution and unscale to get the 'general' solution.
It turns out that only the two obvious integer triples satisfy the original.
