How many positive integer solutions does the equation x^2+y^2+z^2-xz-yz = 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?



Diophantine equation problem 
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 $u=(y+z-x)/2,v=(x-y+z)/2,w=(x+y-z)/2$ being half-integers. 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 wolfram|alpha:


[solve x x + y y + z z - x z - y z == 1 over the integers](http://www.wolframalpha.com/input/?i=solve+x+x+%2B+y+y+%2B+z+z+-+x+z+-+y+z+%3D%3D+1+over+the+integers) 


