Conference rusure::math

    Let T be the 2 by 2 matrix

	| 1  -3 |
	| 3   1 |

    Then successive powers of T will map the points

	{ (1,3), (-1,3), (1,-3), (-1,-3), (3,1), (3,-1), (-3,1), (-3,-1) }

    to integer points in the plane, and hence to rational points on the
    circle since T is orthogonal.

    I'm not sure, but any one of the points above may suffice...

    - Jim

    (1,3),(-1,3),(1,-3),(-1,-3)
    
    do you want more ? :-)
    
    we solve x^2+y^2=10
    
     x^2=10-y^2
    
       x=sqrt(10-y^2)=sqrt(10-a^2/b^2)=1/b * sqrt(10 b^2-a^2)
                                       
       write a program , try different a for some b untill we get a
       rational x.
    
    i.e the question is for what a,b  is   1/b * SQRT(10 b^2-a^2) is
    rational ?
    
    one is a=3,b=1 (and sign combinations)
    
    any others..? i leave it as an excersise to the reader

    The points:
    
    (3b�+2b-3)/(b�+1) , (b�-6b-1)/(b�+1)
    
    where b is any rational. This is a first try, got by considering
    
    	X�+Y�=10a�    (x=X/a,y=y/a)
    
    so (X+3a)(X-3a)=(a+Y)(a-Y)
    
    therefore (X+3a)=b(a+Y) and b(X-3a)=(a-Y) and so on.
    
    I'm not very happy with that as a proof though.
    
    Dick

    There's a mistake in .1, that should be even powers of T...

    But there are really more points than that.  If a^2 + b^2 = c^2 is
    any pythagorean triple, then the matrix

	| a/c -b/c |
	| b/c  a/c |

    is a rotation that maps rational points to rational points, and so
    successive powers would generate "all" the points from the list of
    8 points mentioned in .1

    This is indirect geometric reasoning, and is probably not as neat as
    a more algebraic way of doing it, such as .3.

    - Jim

        re .4
        
>    There's a mistake in .1, that should be even powers of T...
        
        The matrix in .1 isn't made up of unit vectors, so you
        need to divide the elements by sqrt(10).  Then since that
        makes the entries irrational, you need to limit yourself
        to even powers of T/sqrt(10), i.e., to powers of
        
        	| -0.8 -0.6 |
        	|  0.6 -0.8 |
        
        (which is just an example of the general form you used in
        .4).
        
        Dan

If you replace the circle by a Supercircle (x^4 + y^4 = 1; see note 583)
that passes through (0,1) etc., you get a curve that, except for where it
crosses the x-y axes, passes through *no* points with both coordinates
rational. If they were both rational, they would be contradictions of
Fermat's Last Theorem for n=4. 

re .3: That's it.
    
    The result looks nicer when you want x and y in: x�+y� = a�+b�
    (you basically solved it this way, with 10 = 3�+1�).

.3> I'm not very happy with that as a proof though.
    
    Well, the 'a' isn't needed.  But the clever introduction of 'b' is crucial.


    Can you find all x, y such that (1+x�)(1+y�) = a square?

>    Can you find all x, y such that (1+x�)(1+y�) = a square?

Apart from x=y, I don't see a pattern yet; but there are for starters

	(1+1^2)*(1+41^2) = 58^2
	(1+2^2)*(1+38^2) = 85^2

        re .7, .8
        
	The solutions with 0 <= y < x < 200, including the two
        that Lynn found, are:
        
        	x	y	n
        
		117	3	370
		41	7	290
		41	1	58
		38	2	85
		7	1	10
        
        Dan

Can you find all *rational* x, y such that (1+x�)(1+y�) = a *rational* squared?

(Sorry, I just assumed everyone would know I was working with rationals).

Oh, *that* problem :-) Well, we can rewrite that as

	   2   2    2   2     2
	(x1 +x2 )(y1 +y2 ) = k

where everything is integer. Now it's clear that any pair of pythagorean
triples will produce a solution, e.g. {3,4,5} and {5,12,13} produces 

	  2  2   2   2     2  2
	(3 +4 )(5 +12 ) = 5 13

or, in the original form,

	      2        2	  2
	(1+4/3 )(1+12/5 ) = (13/3)

or 
	      2	       2		   2
	(1+3/4 )(1+5/12 ) = ((5*13)/(4*12))

Is this set of solutions exhaustive? At the moment I don't see how the 
previous integer solutions are produced, although it's probably obvious...

As Peter points out, the Pythagrean approach assumes that both x and y
terms are squares, which excludes the cases I gave earlier. A more general 
set of solutions takes the form

        (x1^2+x2^2) = r*s^2, (y1^2+y2^2) = r*t^2

where s and t may be divisible by powers of r. Perhaps the Greek we should
be after is Diophantus... 

        Now where did that problem about primitive Pythagorean
        triples with the same z go?
        
        If p and q are primes of the form 4n+1 then they can be
        represented as p = (a^2 + b^2), q = (c^2 + d^2).  That
        will give pq two representations as a sum of two squares,
        which can be multiplied to give the product (pq)^2.  Just
        write p as (a + bi)(a - bi) and q as (c = di)(c - di).
        Use (a + bi)(c + di) and (a + bi)(c - di) to get two
        forms (x + yi) which each satisy x^2 + y^2 = pq.
        
        For example, 5 = 1 + 4 and 13 = 4 + 9 combine to give
        65 = 16 + 49 = 1 + 64.  This yields
        
        	(16 + 49)(1 + 64) = 65^2
        
        or, for example,
        
        	(1 + (7/4)^2)(1 + 8^2) = (65/4)^2
        
        Dan

   re .3 and .7

   A simple way to see that solution is to consider a coordinate system
   whose origin is at the point (3,1), whose X axis also passes thru
   (0,0), and whose Y axis is tangent to the circle at (3,1).

   Then the "rational paramaterization" in .3 is just the slope of
   a line thru the origin in that coordinate frame that meets the
   circle at some other point.

   Thus a line whose slope is zero (the "X" axis) meets the
   point (-3,-1), a line with slope -1 meets (-1,3), infinite slope
   (the "Y" axis) approaches the point (3,1) and so on.

   - Jim

        re .13,
        
        That works with p and q both a sum of two squares; they
        do not *need* to be prime.
        
        Dan

.7> The result looks nicer when you want x and y in: x�+y� = a�+b�

	x = (2bf - a(f�-1))/(f�+1)
	y = (2af - b(f�-1))/(f�+1)

.16>.7> The result looks nicer when you want x and y in: x�+y� = a�+b�
.16>
.16>	x = (2bf - a(f�-1))/(f�+1)
.16>	y = (2af - b(f�-1))/(f�+1)

Rather:

	x = (2bf - a(f�-1))/(f�+1)
	y = (2af + b(f�-1))/(f�+1)
		 ^
		 |
		 |

so that the cross-terms cancel.

Andrew.

.7 paraphrased>    Can you find all rational x,y,z such that (1+x�)(1+y�) = z�?

.8,.9,.10 clarify the problem statement.

.11 addresses the special case where (1+x�) and (1+y�) are *each* the square of
a rational.

.13 addresses the special case where (1+x�) = (1+y�)

I want here to follow up .12 which points the way towards the complete solution.
-------------------------------------------------------------------------------

        I'd like to stick to rationals, in the spirit of the earlier replies
to this note, but I don't see how for *this* problem, since there are some
remarks that I want to make about residue classes.   [Actually, I guess I
could extend the definition residue classes for rationals...]   Anyway, let me 
convert the problem to an equivalent statement of the problem for integers:

        Can you find all integers a,b,c,d,e such that (a,b) = (c,d) = 1, and
(a�+b�)(c�+d�) = e�?

        Let f(a,b) be defined as follows:
                (i) f is square-free
		(ii) a�+b� = f.g� for some integer g.

        Then a,b,c,d (& e = _/[(a�+b�)(c�+d�)]) give us a solution iff:
f(a,b) = f(c,d).

        So, let's take f a constant for the moment, and look for solutions
a,b,g to:
        a�+b�=fg�.

        Considering this equation mod 4, we can see that g is odd [else both
a and b are even, yet we know that (a,b)=1.]   So g� == 1 mod 8.   Now, at
least one of a & b is odd.   So a�+b� == 1, 2 or 5 mod 8.   f must also be of
this form.

        We know from a well-known theorem that if f is of this form, it can be
represented as the sum of two squares, for instance:
        f = k�+l�.

        Hence:
        a�+b�=(k�+l�)g�.
=>      (a+kg)(a-kg) = (lg+b)(lg-b)

        (a+kg,a-kg,lg+b,lg-b) | (2a,2b) = 2.

        We can transpose k & l if necessary to ensure that a==k mod 2, and
b==l mod 2.   Thus, the highest common factor of the four terms above *is* 2.
Using my most ubiquitous lemma:

        a+kg = 2mn
        a-kg = 2pq
        lg+b = 2mp
        lg-b = 2nq

        where m,n,p,q are non-zero integers, such that (m,q)=(p,n)=1.
Thus:
        a = mn+pq
        b = mp-nq
subject to:
        k(mp+nq) = l(mn-pq)

        Since f is square-free, (k,l)=1, and so there exists s, non-zero
such that:
        ks = (mp+nq)
        ls = (mn-pq)
Hence:
        m = (pk+nl)s/(n�+p�)
        q = (nk-pl)s/(n�+p�)
                                                                          
There is a unique value of s, depending on p,n,k,l such that m & q can be
integers, while (m,q)=1.   We don't need to compute it;  it's enough to
know that it exists and is unique.   Thus:

        a =  [2npk+(n�-p�)l]s/(n�+p�)
        b =  [2npl+(p�-n�)k]s/(n�+p�)

Looks familiar, doesn't it?

But all we're interested in are the ratios a/b, which is:
                                                                                
      +----------------------------------------+
      |  x = [2npk+(n�-p�)l] / [2npl+(p�-n�)k] |
      +----------------------------------------+

Remember how we get this: start with integers k&l such that k�+l� is square-
free.   Then any integers p&n such that (p,n)=1 will give us a "valid" rational
x.   Any two such pairs, p&n and p'&n' say, taken for the same k&l, will give
us x & x' such that

        (1+x�)(1+x'�) is then the square of a rational.

I look forward to reading Jim Roth's geometric interpretation of this.

Cheers,
Andrew.

Wow.

.7 paraphrased>    Can you find all rational x,y,z such that (1+x�)(1+y�) = z�?

I did it this way.  First, note that (1+x�)(1+y�) = (xy+1)� + (x-y)� = a square.
Wlg, this implies (xy + 1) = �(x - y)(v - 1/v) for some v (see below), and hence
z = �(x - y)(v + 1/v).  Solving for y, and expanding z:

	y = ((v�-1)x - 2v)/(2vx + (v�-1))
	z = �(x-y)(v + 1/v) = (v�+1)(x�+1)/(2vx + (v�-1))

C'est tout.

I've found the following very useful.  In general,
	(X)� + (Y)� = a square	=> (X) = �(Y)(v-1/v), for some rational v

The above is a special case of:
	(X)� + (Y) = a square	=> (X) = �(v-(Y)/v), ""

	I'm very impressed with the brevity of all these "rational" proofs
that you've been generating recently.   It seems that this is the right
language to be talking for many of the puzzles that we have here:  the integer
approach that I've been using for many of these number theoretical puzzles
seems to introduce many unnecessary variables and case-by-case analyses.
Hmmm.

Cheers,
Andrew.