Conference rusure::math

    Let the point be (a/b,c/d).  Then we must have:

	    a�/b� +     c�/d� = rational�
	(a-b)�/b� +     c�/d� = rational�
	(a-b)�/b� + (c-d)�/d� = rational�
	    a�/b� + (c-d)�/d� = rational�

    Peter:
    
    Would you please explain .1?  What are a, b, c, and d?  What are
    the sums?  Are the boxes on my VT100 superscript two's?
    
    John

.2> Would you please explain .1?

    Of course.
    
    I assumed the points of the square are at (0,0), (0,1), (1,0), and (1,1),
    and the desired point has rational coordinates (a/b, c/d).  The equations
    are simply a starting point for solving the problem.  On later reflection,
    I realize that these assumptions may not be valid (they may help find an
    example, but won't prove that no such square and point exist).

    Yes, those little boxes are superscript 2's.

	re .1 -< the four corners themselves is a trivial solution >-
        
        re .2 -< I don't understand .1's title either >-
        
        It looks like .1's title means each corner of the unit
        square is a trivial solution, but nonadjacent corners are
        not a rational distance away from each other.
        
        The assumption of where the corners are is okay, as you
        can translate and rotate any other unit square there. 
        But it isn't obvious that a solution point must
        necessarily have rational coordinates.
        
        Dan

.0> Is there a point in the plane that is at a rational distance
.0> from all 4 corners of a unit square?

I've made some progress on this.  Now can anyone tell me whether there is
a rational x such that both sqrt(2 x� + 2) and sqrt(8 x� + 2) are rational?
Equivalently: find integers c and d such that 2 c� + 2 d� and 8 c� + 2 d�
are both perfect squares.

    There are no c and d such that
    
    2c� + 2d� = a� and 8c� + 2d� = b�, with a,b,c,d all integers > 0.
    
    
    Proof
    
    Suppose there is such an a,b,c,d.
    
    Then either d is even or d is odd.
    
    If d is even (and equal to 2e), then there is another solution with
    smaller value of (c+d).
    
    	2c� + 8e� = a�,  2c� + 2e� = f�, (with 2f = b. f is an integer)
    
    There must be a solution with least value of (c+d), which must
    therefore have d odd. I.e. If there are any solutions then there is at
    least one with d odd.
    
    However if d is odd then 4c� + d� is also odd, and therefore b is
    irrational and hence not integer. Therefore there are no solutions with
    d odd.
    
    We have a contradiction which shows that the supposition is false.
    
    Andrew

I agree with .6.   I put in some time on this problem in the past, but
never managed to solve it.   I'd be interested to see Peter's progress.

Andrew.

Find rational s1, s2, s3, and s4, all greater than 1, such that:

	(s1� - 1) s2 s3 (s4� - 1) + (s2� - s2� s3� + s3� - 1) s1 s4 = 0
								(eq.1)
and

	(s2� - 2 s2 - 1) s3 (s4� - 1)

		+ (s2� s3� - 2 s2� s3 - s2� - s3� + 2 s3 + 1) s4 = 0
								(eq.2)

Then choose p_i and q_i so that s_i = p_i/q_i (ex: the q_i = 1, and p_i = s_i).

Compute k1 thru k4 from the equations (this is trivial).

	a = k1 (p1� - q1�) = 1 - k2 (p2� - q2�)
	  = k3 (p3� - q3�) = 1 - k4 (p4� - q4�)

	b = 2 k1 p1 q1 = 2 k2 p2 q2 = 1 - 2 k3 p3 q3 = 1 - 2 k4 p4 q4

The point is at (a,b), and the square has corners (0,0), (0,1), (1,0), (1,1).


So all I need is a way to solve those first two equations in rationals.


The derivation of the above is fairly simple.  The 'distance squared' from
(a,b) to each of the corners is in the form of a Pythagorean triangle.  E.g.:

	d3� = a� + (1-b)�

Use the general form for generating right triangles; the k must be allowed
to be rational (rather than simply an integer); viz:

		    a = k3 (p3� - q3�)

		1 - b = 2 k3 p3 q3

		   d3 =  k3 (p3� + q3�)

There are three other sets of equations like the above.  Solve for the k
and eliminate them.  Use s_i = p_i/q_i to eliminate the p's and to simplify.
This leaves the two equations.   Two equations in four unknowns... would be
easier to solve if they weren't all quadratic.

I've tried solving for one variable with the quadratic formula, and substituting
it into the other.  This yields one equation in 3 variables, but also requires
that the discriminant be a rational square.  Some of these discriminants have
strong patterns; they might be coerced to be the difference of two squares,
so that the general form for generating right triangles may be applied, but
no luck yet.  Besides, I already have an equation in 3 variables -- and another
in four to boot!.

    Since the equation is of the form p(x,y,z,w)=0 you can further restrict
    your variables to be integers, not just rationals.  That may help to
    develop restrictions (e.g., if s3 is even, so must s4 by (2)).
    
    A large enough set of restrictions may pave the way for direct search
    on a handy 200 MHz machine...
    
      John

re .7	Thanks.

re .8	This assumes that the point is at rational coordinates.  By other
	means, it's easy to show that it is.

re .9	How does an "equation of the form p(x,y,z,w)=0" restrict the
	variables to integers?

	I chose to write the four distances as a/e, b/e, c/e, d/e, where
a,b,c,d,e are +ve integers and the highest common divisor of them all is 1.
This reduces the number of variables, whilst allowing me to work with integers
rather than rationals.   I also chose to locate the vertices of the square at 
(�1,�1), for greater symmetry.

	After a little bit of munging, I am left with the following
characterization of the problem.   (e4) is a fairly obvious consequence of
(e1-3), and is used later, so I've stuck it in.	

(e1)	[a,b,c,d,e] = 1
(e2)	a� + d� = b� + c� (= k, say)
(e3)	k� + 32e^4 = 8ke� + 2a�d� + 2b�c�	<=== the key quartic
(e4)	a,b,c,d are all odd

-------------------------------------------------------------------------------

	I haven't yet solved the full problem, but here is one sub-problem
that I've cracked.

	One likely location of solution points, it seemed to me, is on the 
axes, since the number of criteria to be satisfied is reduced.   So let's
assert that a=b, c=d and watch what happens...

(f1)	a,d,e are pairwise coprime
(f2)	(a�-d�)� - 8(a�+d�)e� + 32e^4 = 0
(f3)	a,d both odd
(f4)	e is even	(from (f2))

	Now, let's regard (f2) as a quadratic for a� & d� in turn.

a� = d� + 4e� � 4e_/(d� - e�)
d� = a� + 4e� � 4e_/(a� - e�)

	Since the discriminants must be integral, let's write...

	a�-e� = P�
	d�-e� = Q�

	Substituting back into the quadratics for a� & d�...

	P� = Q� + 4e� � 4eQ = (Q�2e)�
	Q� = (P�2e)�

	So choosing signs of P & Q suitably, have P = R+e, Q = R-e, where R is
some arbitrary integer.

	a� = e� + (R+e)�
	d� = e� + (R-e)�

	Now, we want to use generating formula for primitive Pythagorean 
triples.   e is even, which keeps things simple.

	e = 2FG = 2HJ
	R+e = F�-G�
	R-e = H�-J�

where...

	F,G are coprime and one is even
	H,J are coprime and one is even

	Now equate the expressions for e and for R...

(f5)	FG = HJ
(f6)	F�-G�-2FG = H�-J�+2HJ

	Using the lemma given in Note 1154.1 to deal with expressions of the
form of (f5).   (I've never seen this lemma written in a book, but it's jolly 
useful in quadratic number theory!)

	F = pq
	G = rs
	H = pr
	J = qs	where p,q,r,s are pairwise coprime, and exactly one is even.

	Now substitute into (f6)...

	p�q� - r�s� - 2pqrs = p�r� - q�s� + 2pqrs

	(q�-r�)(p�+s�) = 4pqrs

	p or s is the even one, else 4 doesn't divide LHS.

	[4ps,p�+s�] = 1 so 4ps | (q�-r�)
	Similarly [qr,q�-r�] = 1 so qr | (p�+s�)

	Thus, the RHS | LHS, but in fact RHS = LHS, so...

(f7)	qr = p�+s�
(f8)	4ps = (q+r)(q-r)

	Now we apply a version of that Lemma again, to (f8), noticing that...

	[q+r,q-r] = 2

	so, we write...

	q+r = 2tu
	q-r = 2vw
	p = tv
	s = uw	where t,v,u,w are pairwise coprime, and one is even.
	
	q = tu+vw
	r = tu-vw

	substitute back in (f7)

=>	t�u� - v�w� = t�v� + u�w�

	t�(u�-v�) = w�(u�+v�)

	Xt� = u�+v�
	Xw� = u�-v�

	X = [u�+v�,u�-v�] = 1 or 2.

	Either way, you derive

	t� = u�+v�
	w� = u�-v�

or something equivalent to it.   At last we have a pair of "linear quadratics".
Are there any solutions?   Well, t,u & w are in arithmetic progression,
and we know (from note 1154 again) that 3 squares in ap will have difference:

	N�EH(E�-H�) = v�

where N,E and H are arbitrary +ve integers such that E > H are odd and 
pairwise coprime.   For this to equal v�, *each* of E, H, �(E+H), and �(E-H)
must be a square (since they are pairwise coprime), call them respectively 
t', w', u' & v'.   (You can see what's coming next.)

	t'� = u'�+v'�
	w'� = u'�-v'�

	0 < ... < v' =< v/4, so by infinite descent, there is no solution to 
this subproblem.

Phew!

	So I repeat, I haven't solved the general problem, but I've shown that
a plausible subproblem has no solution.

Cheers,
Andrew.

PS: Any confirmation gratefully received.

Summary of problem: 

(a,b,c,d,e,k are all +ve integers)

(e1)	[a,b,c,d,e] = 1
(e2)	a� + d� = b� + c� (= k, say)
(e3)	k� + 32e^4 = 8ke� + 2a�d� + 2b�c�	<=== the key quartic
(e4)	a,b,c,d are all odd

-------------------------------------------------------------------------------

	(e2) tranforms to

	(a+c)(a-c) = (b+d)(b-d)

	We can write:

	a+c = 2jfg
	a-c = 2jhi
	b+d = 2jfh
	b-d = 2jgi	(and (f,i) = (h,g) = 1)

	hence...

	a = j(fg+hi)
	c = j(fg-hi)
	b = j(fh+gi)
	d = j(fh-gi)

Substituting,

	k = j�(f�+i�)(g�+h�)
	a�d� + b�c� = 2j^4f�g�h�i�[ (f/i - i/f)� + (g/h - h/g)� ]

	Examining (e3) mod j, it's clear that j | e.   By (e1) then, j = 1.

	k� + 32e^4 = 8ke� + 2a�d� + 2b�c�

	regarding this as a quadratic in e�, we get:

	e� = [k � _/(-k� + 4a�d� + 4b�c�)]/8

	Call the discriminant m�.

--->	Amazingly, m splits!!!

	m� = - (f^4-6f�i�+i^4)(g^4-6g�h�+h^4)
	   = - (f� - 2fi - i�)(f� + 2fi - i�)(g� - 2gh - h�)(g� + 2gh - h�)

	This quadratic seems to be cropping up a lot in this problem.   I
wonder why.   Gosh, I wish I understood mathematics...

	I think the right approach now is experimental.   Let's take various
pairs (x,y) and look at Q(x,y) = f^4 - 6f�i� + i^4.

re .11,

>	One likely location of solution points, it seemed to me, is on the 
>axes, since the number of criteria to be satisfied is reduced.   So let's
>assert that a=b, c=d and watch what happens...

If a and b refer to opposite corners, then a=b puts the point
on a diagonal.  If a and b refer to adjacent corners, then a=b
puts the point on an axis.  Do your a,b,c,d go in order around
the square?  Just checking.  Mine didn't. :-)

The point also can't lie along a diagonal, as then the distances
to the two corners on that diagonal either add to or differ by
sqrt(2) (in a unit square).

Has anyone ruled out the point being along the line extending a
side of the square?

Dan

About a week and a half ago, I made some substantial progress on this,
but I was waiting, hoping that it would lead to something even better.
But the result is impressive such as it is....

I can find a family of points such that they are at rational distances
from three of the corners, and the fourth distance is arbitrarily close
to a rational number.

It gets better...

I've found a point that's at a rational distance from each of the four sides!

   Re:           <<< Note 345.14 by TRACE::GILBERT "Ownership Obligates" >>>

degenerate solution:

   side = 0, 
   corners = {(0,0), (0,0), (0,0), (0,0)}
   required point = (0,0)

I've been working on it, and have gotten it down to *one* equation.

   2                4     4                2     2                 3   2
(g1  - 4 g1 + 8) (z1  + g1 ) + 16 g1 z1 (g1  - z1 ) + (8 - 2 g1) g1  z1

The problem is to find z1 and g1 so that this is a perfect square, where
g1 must be of the form s1 - 1/s1.


If it can be massaged into the sum of two squares, that'd help greatly.

        The set of points such that the distance to each of the
        four corners of the unit square is algebraic, is a dense
        subset of the plane.
        
        Dan