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?

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    $\begingroup$ That's the equation of an ellipsoid: a bounded set. Diagonalizing it will give explicit bounds for $x$, $y$ and $z$. By diagonalization I mean a change of coordinates transforming it to the form $ax^2+by^2+cz^2=d$. $\endgroup$ – Robin Chapman May 30 '10 at 17:27
  • $\begingroup$ You may want to change "positive" to "non-negative" in your question, and you may want to consider solutions with more zeroes. I'm afraid I'm voting to close, since your question is a little outside the scope of this site. Please see the FAQ for a list of problem-solving websites. $\endgroup$ – S. Carnahan May 30 '10 at 18:20
  • $\begingroup$ Just to clarify before closing: write your equation (x-z/2)^2 + (y-z/2)^2 + z^2/2 =1 , whence z is either 0 or 1, &c. $\endgroup$ – Pietro Majer May 30 '10 at 18:30
  • $\begingroup$ Cindy, why don't you guess the fourth nonnegative solution $(0,0,1)$? (In my comment to Will's post below I explain how to give all rational points on your ellipsoid. It is this fourth solution through which the lines with rational pair of slopes pass.) $\endgroup$ – Wadim Zudilin Jun 2 '10 at 12:18

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

  • $\begingroup$ Kevin, why does Mathematica prefer $xx$ instead of $x^2$? $\endgroup$ – Wadim Zudilin May 30 '10 at 23:38
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    $\begingroup$ @Wadim: that's just me optimizing my typing. It's faster to type "x x" than "x^2". $\endgroup$ – Kevin O'Bryant May 31 '10 at 1:27
  • $\begingroup$ That's great, Kevin! I'll try to follow your trick (which I guess comes from ancient time). $\endgroup$ – Wadim Zudilin May 31 '10 at 3:02

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,


  • $\begingroup$ Will, I am getting up too early. The quadratics correspond to the Cartan matrix which reminds me about Rogers-Ramanujan identities. Apart from this, the OP sounds like homework. :) $\endgroup$ – Wadim Zudilin May 30 '10 at 21:44
  • $\begingroup$ Hi Wadim, it does give the appearance of homework as stated for the value $1$ on the right hand side. This would be a stretch for a school-child first exposed to "completing the square." I chose to ignore the $1$ and give the merest hint of what happens next. Meanwhile, I think you should go back to bed. Victor Wadimovich depends on you and you need your strength. $\endgroup$ – Will Jagy May 30 '10 at 22:11
  • $\begingroup$ Will, in spite of rain outside I have to teach, so no way to get a longer sleep. On my bus way to uni I wrote a general solution of the equation in rational (by intersecting the ellipsoid with the line $x=t(z-1)$, $y=s(z-1)$ for a rational pair of slopes $t,s$) to see that it does not give a simpler way to produce integer solutions. $\endgroup$ – Wadim Zudilin May 30 '10 at 23:43
  • $\begingroup$ Hello, Cindy. I have put a ton of related material at: zakuski.math.utsa.edu/~kap/forms.html On Wadim's idea, there should not be a really easy way to produce integral representations with arbitrary $M$ on the right-hand-side, see my answer (and the others) to mathoverflow.net/questions/3596 $\endgroup$ – Will Jagy May 31 '10 at 3:04

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