Correct solutions came from Nancy Schwarzkopf, Jens Voss, Burkart Venzke,
Douglas A. Vander Griend.
The answer is yes. You can take as the center of
your circle the point P = (Ö2,Ö3),
though many (in fact, most) other points work, too.
It's enough to show that there are no two lattice points
which lie on any circle centered at P, for then we can start with a very
small circle around P and slowly increase its radius, picking up exactly
one new point each time.
To that end, suppose that (a,b) and (s,t) both lie on a
circle of the same radius centered at P. Then (a -Ö2)2 + (b -Ö3)2 = (s -Ö2)2 + (t -Ö3)2. Squaring and
collecting like terms we get
a2 +
b2 - s2
- t2 = (2a - 2s)Ö2
+ (2b - 2t)Ö3.
Put a = a2 +
b2 - s2 - t2, b = 2a - 2s, g = 2b
- 2t. It's easy to see that
b, g are
non-zero, so a, b,
g are all non-zero integers. After
rearranging this equation you get,
Ö3 = p + qÖ2, (*)
where p and q are non-zero rational fractions.
Squaring (*) you get 3 = q2 + r2 + 2rqÖ2, from which you conclude quickly that
Ö2 is a rational fraction, which is not
the case!
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Page last updated 15 May 2001.
ã 2001 Alberto L Delgado