Conference rusure::math

Since this is too easy, I must be missing something about the question.

     +    ___________
x =  -  \/ 1 + 92y^2

     +    ________________
y =  -  \/ (x^2 - 1) / 92
    

    x=1 y=0

    a quick computer search yielded x=1151 y=120

	Yes, all integral (or just as easily, all non-negative
	integral) solutions are desired.

	Dan

	For what its worth ... sqrt(92) is 9 + something, where the
	continued fraction representation of something is the
	repeating block [1 1 2 4 2 1 1 18 ... ].

	Dan

	If x^2 - 92y^2 = 1, then x/y is a pretty good approximation
	to sqrt(92).  This gives the idea of looking at the best
	rational approximations of sqrt(92), and these come from
	the continued fraction expansion of sqrt(92).  If you learned
	continued fractions as [a0 a1 a2 ...] = 1/(a0 + 1/(a1 + 1/(a2 + ...
			     ________________
	then sqrt(92) = 9 + [1 1 2 4 2 1 1 18 ...] where the overlined
	block repeats.  If you learned [a0 a1 a2 ...] = a0 + 1/(a1 + 1/(a2 + ...
			   ________________
	then sqrt(92) = [9 1 1 2 4 2 1 1 18 ...] where again it is the
	overlined block that repeats.  It is easier to use the latter
	form here, so I will.  Note they are reciprocals of each other,
	so you just exchange the numerator and denominator to go from
	one to the other.

	Computing the successive approximations to the continued
	fraction according to the formulas

		x(n) = a(n) * x(n-1) + x(n-2)
		y(n) = a(n) * y(n-1) + y(n-2)

	yields

	   n      a(n)      x(n)      y(n)      x^2 - 92y^2
	------------------------------------------------------
	  -2		      0		1
	  -1		      1		0
	   0	   9	      9		1	    -11
	   1	   1	     10		1	      8 
	   2	   1	     19		2	     -7
	   3	   2	     48		5	      4
	   4	   4	    211	       22	     -7
	   5	   2	    470	       49	      8
	   6	   1	    681	       71	    -11
	   7	   1	   1151	      120	      1 (Eureka!)
	   8	  18	  21399	     2231	    -11
	   9	   1	  22550	     2351	      8
	  10	   1	  43949	     4582	     -7
	  11	   2	 110448	    11515	      4
	  12	   4	 485741	    50642	     -7

	Notice how the successive rational approximations x/y alternate
	being greater than then less than sqrt(92).  Anyway, this shows
	that (1151, 120) is a solution over the positive integers to
	x^2 - 92y^2 = 1 (although we knew that already, from .4).

	Dan

	Suppose now that (x,y) is a solution to x^2 - 92y^2 = 1.
	This factors into (x + y sqrt(92))(x - y sqrt(92)) = 1.
	Raising to the n-th power, you also have that
	(x + y sqrt(92))^n (x - y sqrt(92))^n = 1.  Now, if
	you use the binomial theorem and separate the even and
	odd powers of sqrt(92), you will see there is an A and B
	(dependent on n) such that (x + y sqrt(92))^n = A + B sqrt(92)
	and (x - y sqrt(92))^n = A - B sqrt(92).  Then you will
	also have (A + B sqrt(92)) (A - B sqrt(92)) = A^2 - 92B^2 = 1.

	So given one nontrivial solution, we can generate an
	infinite sequence of others.  Let A(1) = x, B(1) = y.
	Then (x + y sqrt(92))^(n+1) = (A(n) + B(n) sqrt(92))(x + y sqrt(92))
	which multiplies out to

		A(n+1) = x A(n) + 92 y B(n)
		B(n+1) = y A(n) + x B(n)

	Note: using n=0 for the exponent corresponds to the trivial
	solution (1,0), and using n=1 for the exponent corresponds
	to the solutiion (x,y).  We need this first solution to start
	generating the others.  Here our first solution is (x,y) =
	(1151,120).  Thus our infinite sequence of solutions (A,B)
	is given by

		A(0)=1, B(0)=0

		A(n+1) = 1151 A(n) + 11040 B(n)
		B(n+1) =  120 A(n) +  1151 B(n)

	For n = 0,1,2,3 this gives (1,0), (1151,120) and the new
	solutions (2649601,276240) and (6099380351,635904360).

	This gives infinitely many solutions.  Are there any
	others?  :-)  Would the A(n) for n > 1 have been found
	if the continued fraction table in .-1 were continued?

	Dan

>	This gives infinitely many solutions.  Are there any
>	others?  :-)  Would the A(n) for n > 1 have been found
>	if the continued fraction table in .-1 were continued?

	You'd have seen those A values in the continued fraction
	tables at lines n=7, 15, 23, 31, ...  (note that the
	values of x^2 - 92y^2 are cyclic with period 8).

	The coefficients 1151, 11040, 120, and 1151 can be
	directly derived by expanding the x(n), y(n) recurrences
	in your previous note.

	Does this yield all (positive) solutions?  Yes, but I don't
	remember why.

>>	Does this yield all (positive) solutions?  Yes, but I don't
>>	remember why.

	See .0 ... if you're a mathematician you'll remember why
	within the year. :-)

	My vague recollection of this part of the proof is that
	you bracket a solution (X,Y) between (A(n),B(n)) and
	(A(n+1),B(n+1)), i.e., A(n) <= X < A(n+1) and B(n) <= Y < B(n+1),
	then show it must be (A(n),B(n)).  But the proof was only
	for the special case where the "92" differed from a square
	by one (I forget whether it was a square plus one or a square
	minus one).  Anyway that's the approach I would try first.
	Looking at the structure of the units of Z[sqrt(92)] would
	come next.

	Dan

	p.s. .7 and .8 were from memory, not from solving it :-)

	The original usenet signature has now expanded to:

>> --
>> Chris Long, 272 Hamilton St. Apt. 1, New Brunswick NJ  08901  (201) 846-5569
>>
>> "A person who can within a year solve x^2 - 92y^2 = 1 is a mathematician."
>> Brahmagupta, 650 AD

	Dan

        re .0
        
>> "A person who can within a year solve x^2 - 92y^2 = 1 is a mathematician."
        
        Less than six months to go. :-)
        
        Dan