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?
Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
|
1
2
|
|||||||||||||||
|
|
3
|
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: |
||||||||||||||
|
You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
|
1
|
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 |
|||||||||||||||
|
