How many positive integer solutions does the equation x^2+y^2+z^2xzyz = 1 have? My guess is (1,0,1), (0,1,1) and (1,1,1). What is proof of that fact that there are none other?

To elaborate on Robin's suggestion, set $x=v+w,y=u+w,z=u+v$, and the equation becomes $$u^2+v^2+2w^2=1,$$ with $u=(y+zx)/2,v=(xy+z)/2,w=(x+yz)/2$ being halfintegers. Now a brute force run through the possibilities is feasible, since $u\leq 1$ and such. For the quick answer, you can use Mathematica: Reduce[x x + y y + z z  x z  y z == 1, Integers] or even wolframalpha: 


Shame they are going to close this, I actually know something about this one. Your form is equivalent to $$ X^2 + Y^2 + Z^2 + Y Z + Z X + X Y $$ meaning there is an invertible integral change of variables. This is a regular form and integrally represents the same numbers as $$ U^2 + V^2 + 2 W^2 $$ which is to say all numbers not of shape $$ 4^k ( 16 m + 14 ) .$$ As it is positive definite, rather than just semidefinite, there are simple bounds on possible values for the variables in your original $$ x^2 + y^2 + z^2  y z  z x \leq M$$ which can be derived either by eigenvalues for the related symmetric Gram matrix,or, as I do, by Lagrange multipliers in maximizing $x^2$ or $y^2$ or $z^2$ subject to the constraint given. So all solutions to $$ x^2 + y^2 + z^2  y z  z x = M$$ can be found fairly quickly even for large $M.$ Well, see my article with Kaplansky and Schiemann, http://zakuski.math.utsa.edu/~kap/Forms/Kap_Jagy_Schiemann_1997.pdf 

