To: rusin@math.niu.edu (Dave Rusin) Subject: Re: ellipsoids hitting Z^n Date: Thu, 02 Nov 95 10:23:10 +0100 From: [Permission pending] > It's not clear what an appropriate generalization would be. > None of the ellipses x^2/a^2 + (y-1/2)^2/(0.249999)=1 has any > integral points, despite the fact that their areas are unbounded. The generalization obviously cannot use the volume of the body but rather the semiaxis lengths. (Also the trivial theorem about the sphere I formulated in terms of the radius, not the volume.) What you are saying is that if there is a theorem of the type P(a1,a2) ==> any ellipse with semiaxes a1,a2 contains a lattice point (where P(a1,a2) is some binary predicate) then P(a1,a2) must also imply a1>=1/2, a2>=1/2. Obviously, theorems of this type must exist. My question was what is the best (i.e., least restrictive) known choice for P(a1,...,an). Petr ============================================================================== Date: Thu, 2 Nov 95 13:50:39 CST From: rusin (Dave Rusin) To: [Permission pending] Subject: Re: ellipsoids hitting Z^n >The generalization obviously cannot use the volume of the body >but rather the semiaxis lengths. (Also the trivial theorem about the sphere >I formulated in terms of the radius, not the volume.) >What you are saying is that if there is a theorem of the type > > P(a1,a2) ==> any ellipse with semiaxes a1,a2 contains a lattice point > >(where P(a1,a2) is some binary predicate) >then P(a1,a2) must also imply a1>=1/2, a2>=1/2. > >Obviously, theorems of this type must exist. My question was what >is the best (i.e., least restrictive) known choice for P(a1,...,an). Oops, I see now what you meant; sorry to be careless. I don't know what the theorem would be but I guess you could work out the strongest possible result for small n. For example, for n=2 what you are asking is for a description of this set: S={ (a1, a2) in (0,oo)^2 s.t. all ellipses with these semiaxes meet Z^2 } The example I mentioned last time shows that all such (a1,a2) lie in (1/2, oo)^2. Conversely, for R=sqrt(1/2), any (a1, a2) in (R, oo)^2 lies in S. If (a1, a2) lies in S, then so does (a1', a2') if a1' >= a1 and a2' >=a2. Clearly S is symmetric: (a1,a2) is in S iff (a2, a1) is in S. So now I see what the question is: determine the precise shape of the curve in the (a1, a2)-plane, which separates S from its complement. Here's how I guess one could approach the answer. It suffices to determine the shape of S in the half-quadrant where a1 >= a2. So fix an a1; let us find the set of all the a2 <= a1 for which (a1, a2) is _not_ in S. These are the a2 for which there is some ellipse with these semi-axes not meeting Z^2. There is no loss of generality in assuming the center of the ellipse lies in the square [0,1]^2, so not meeting Z^2 is the same as not meeting any of the lattice points whose distance from the origin is at most a1+sqrt(2) -- in particular, it is a finite set. For each of these finitely many lattice points z, let U(z) be the set { (a2, theta, h, k) in (0,a1]x[0, 2pi)x[0,1)x[0,1) such that z lies in the ellipse obtained by rotating the ellipse (x-h)^2/a1^2 + (y-k)^2/a2^2=1 through an angle theta } The complement of the union of these U(z) is the set of ellipses with semimajor axis a1 and not meeting Z^2. The projection of this complement to the first coordinate is the set of "bad" a2's. I would be curious to see what the final shape of the set S would be, but not curious enough to carry out the proposed method of the previous paragraph! Possibly, however, there is a more-easily-described region inside S which would be sufficient for your purposes. If you can guess such a region S, a proof along the lines sketched above may be possible. dave