Conference rusure::math

    True.  Even packing points as close together as possible (i.e., at
    least one unit apart), an infinite number of points implies that
    for any point, there are other points arbitrarily far away from it.

    Consider two points having a minimal (integral) distance between them.
    We can choose a point far enough away from the points so that the only
    way the distances between the two points and the 'far' point are integers
    is if the three points are colinear.  Thus, all the points must be colinear.

    > Consider two points having a minimal (integral) distance between them.
    > We can choose a point far enough away from the points so that the only
    > way the distances between the two points and the 'far' point are integers
    > is if the three points are colinear.

    Aren't there triangles with sides {1, 2^n, 2^n} for all n?
    
    John

It's not too difficult to show that for any finite number of points, it's 
possible for them to be non-colinear and all have integer distances in
pairs. To do this you only need find a configuration for which just one 
point is off the line. 

Assume there are n points. Put one point on the y-axis at coordinates (0,
p) where p is the product of the shorter legs of any n-1 primitive
pythagoream triangles (of which there are infinitely many). Position the
remaining n-1 points on the x-axis at appropriate multiples of the longer
legs of those triangles. They all will be at integer coordinates with each
other, and all will form pythagorean triangles with the first point and
(0,0). 

You could extend this proof to infinitely many points except that p, and 
thus the x-coordinates of all the other points, would be infinite. Not sure
if that's kosher. 

Lynn Yarbrough

    Pick any two points.  Then think about where the rest of the points
    are.
    
    John

>    Pick any two points.  Then think about where the rest of the points
>    are.

That doesn't help. You can't always generalise from a [large] finite set to 
an infinite set. See [.-2].

    Pick any two points, A and B.
    
    Let AB be the distance from A to B.
    
    For any other point X, let XA and XB be the distances to A and B.
    
    Where are all the points X such that
    
    		XA = XB + AB ?
    		XA = XB + AB - 1 ?
    		etc. ?

    John (trying to help, Lynn)

    Given two points A and B, find all points at integral distances
    from these points.  To do this, draw concentric circles with
    radii 0, 1, 2, 3, ... around each of these points; then the
    intersections of the circles are the desired points.

    However, I don't see how that helps answer the original question.

    I recall some earlier note that asked for a set of 7 points in a
    plane such that all pair-wise distances were integral.  I presume
    that such a configuration of 6 points was known.

    Pick any two points, A and B.  Let AB be the distance from A to B.
                       
    For any other point X, let XA and XB be the distances to A and B.
    Then XA and XB are integers, and their difference is <= AB.  For
    each possible difference d (d = 0, 1, 2, ..., AB), there is a hyperbola
    containing all points whose difference-of-distances is d.  Since
    there are given an infinite number of points, and a finite number
    of hyperbolas, one of the hyperbolas must have an infinite number
    of points, all integral distances from one another.  Call it "H".
    
    Now throw away all of the given points that are not on H, and repeat:
    
    Pick any two points, C and D on H.  Let CD be the distance from C to D.
    
    For any other point X, let XC and XD be the distances to C and D.
    Then XC and XD are integers, and their difference is <= CD.  For
    each possible difference d (d = 0, 1, 2, ..., CD), there is a hyperbola
    containing all points whose difference-of-distances is d.  Since
    there are given an infinite number of points, and a finite number
    of hyperbolas, one of the hyperbolas must have an infinite number
    of points, all integral distances from one another.  Call it "HH".
    
    But now the intersection of hyperbolas H and HH contains an infinite
    number of points.
    
    Claim:  H = HH = straight line.
    
    John
    
    P.S.  re 665.*:  sorry the hints were not helpful.

    The work in .9 just doesn't make sense to me.  Regardless of the
    internal logic, we find ourselves concluding that a hyperbola is
    a straight line.  That doesn't work.
    
    Let's see, here:

>     For any other point X, let XA and XB be the distances to A and B.
>     Then XA and XB are integers, and their difference is <= AB.  For
>     each possible difference d (d = 0, 1, 2, ..., AB), there is a hyperbola
>     containing all points whose difference-of-distances is d. 
    
    This needs to be nailed down more.  Certainly for any N there is
    a hyperbola containing all points whose distance from some FOCUS
    is N.  But where is the FOCUS in this logic?  X?  In that case
    there is no law that says your entire set must intersect X, so
    the "infinite number of points" does not follow.  If you let your
    FOCUS wander over the entire set, then you have an infinite number
    of hyperbolas, so THAT side of the argument is flawed.
    
    I don't think this approach will get you there.
    
      John

    But part of the argument sounds useful.  Can we show that a hyperbola
    can only contain an infinite number of points with pair-wise integral
    distances if that hyperbola is degenrate -- a straight line?

    Only if the distance is infinite.  In which case its "integrity"
    is in doubt.

    Re .10:  sorry the reasoning was so unclear.  I was trying to describe
    (AB + 1) different hyperbolas.  Each hyperbola has two foci:  A
    and B.  The hyperbola for (d = 0) is degenerate, it's the straight
    line equidistant from A and B.  The hyperbola for (d = AB) is also
    degenerate; it's most of the straight line that passes through A
    and B.
    
    John