Conference rusure::math

        Okay, here goes.  We start with
        
        	(sqrt(a) + b)/c
        
        where a,b,c are integers, a and c positive, and let n be
        the greatest integer <= (sqrt(a) + b)/c
        
        Further, let us assume that
        
        	c | (a - b^2)
        
        So (sqrt(a) + b)/c = n + ((sqrt(a) + b)/c - n)
        	= n + (sqrt(a) + (b-cn))/c
        	= n + 1 / c/(sqrt(a) + (b-cn))
        	= n + 1 / (c (sqrt(a) - (b-cn)))/(a - (b-cn)^2)
        	= n + 1 / (sqrt(a) + (cn-b))/( (a - (b-cn)^2)/c )
        	= n + 1 / (sqrt(a) + B)/C
        
        where
        
        	B = cn-b	C = (a - (b-cn)^2)/c
        
        The goal is that C be an integer, i.e., c | a-(b-cn)^2
        
        Well, a-(b-cn)^2 = a - b^2 + 2bcn - c^2n^2
        	= a - b^2 + c(2bn - cn^2)
        
        Since I fortunately assumed that c | a-b^2, it therefore
        follows that c | a-(b-cn)^2, i.e. that C is an integer.
        
        Now what makes this go on, is that you can now show that
        
        	C | a - B^2
        
        This is because a - B^2 = a - (cn-b)^2 = cC.
        
        So if you start with a=any positive integer, b=0 and c=1,
        then c | a-b^2 is satisfied and you get
        
        	sqrt(a) = n + 1 / (sqrt(a) + B)/C
        
        where again a,B,C are integers, a and C positive, and
        C | a-B^2 ... so you can continue this process until some
        triple a,b',c' along the way is repeated.
        
        Finally, if c fails to divide a-b^2, then c will fail to
        divide a-(b-cn)^2.
        
        Dan

What does that sentence mean ?  That only surds arrived at starting with
a pure surd of the form sqrt(a) can be continued from, and that other starting
points will run amuk ?

/Eric

        In .1 I don't believe I ever used the assumption c > 0 (I
        did use c != 0), but that's okay as I didn't show that it
        was preserved in the next step. :-)
        
        So let us assume c > 0.  Then C = (a - (b-cn)^2)/c has a
        positive denominator.  Is the numerator positive?  Well,
        
        	n <= (sqrt(a) + b)/c < n+1
        
        so
        
        	cn <= sqrt(a) + b < cn + 1
                cn-b <= sqrt(a) < (cn-b) + 1
        
        Not knowing the sign of cn-b I don't see how to go any
        further.
        
        Well, we need to have sqrt(a) irrational in order to show
        C != 0, so let's add that to the requirements.  But with
        that addition I've only shown c nonzero implies C
        nonzero, whereas we need c positive implies C positive.
        
        Dan

        re .2,
        
        Yes, something like that.  I meant that if you just pull
        a, b, and c out of a hat, that the process might not
        work.
        
        Dan

Suppose a=2, b=2, and c=3.  Then

	n = |(|sqrt(2)| + 2) / 3| = 1

	a - (b-cn)� = 2 - (2-3)� = 1

We're asked to prove that 3 | 1.  Hmmm.

Dan's point was important, Peter, namely:

	If we start with (aqrt(a)+0)/1, we have b=0, and c=1

and he proved that given this situation, the theorem holds.  Hence you
can't start with ANY arbitrary (sqrt(a)+b)/c.

But it leads to interesting questions which I'll discuss soon, having
to do with the fact that although theorem doesn't hold in those other cases,
those other cases STILL have repeating continued fractions, so it reopens
the question of how to compute the fraction, and about whether it's
still a palindrome (probably not).

/Eric

        Just work backwards from a repeating non-palindrome to
        get an example that doesn't result in a plaindrome. :-)
        
        For example, let x = [a,1,2,3,1,2,3,1,2,3,...]
        	= a + 1/(1 + 1/(2 + 1/(3 + (x-a))))
        	= a + 1/(1 + (x + (3-a))/(2x + (7-2a)) )
        	= a + (2x + (7-2a))/(3x + (10-3a))
        
        It is easy to let a = 0 when you have
        
        	x = (2x + 7)/(3x + 10)
        	3x^2 + 10x = 2x + 7
        	3x^2 + 8x - 7 = 0
        
        	x = (-8 +/- sqrt(148)) / 6
        
        The positive root is (sqrt(148) - 8)/6 = (sqrt(37) + (-4))/3
        Here c | a-b^2 (i.e., 3 | 37 - 16 or 3 | 21) so the
        process of finding the continued fraction for this x
        works but it must result in the nonpalindrome that we
        started with.  (Trust me...besides I worked it out and got
        [0,1,2,3,1,2,3,...].)
        
        Dan

        If you use the more general form (a sqrt(b) + c)/d then
        there shouldn't be any problems computing the continued
        fraction.
        
        Dan

I studied that number theory book a bit more.  They pointed out a simple
trick.  Suppose you have

	(sqrt(a)+b)/c

and the divisibility rule doesn't hold (as Peter showed us).  The book says
merely multiply all three terms by c, giving:

	(sqrt(acc)+bc)/(cc)

Now the divisibility rule holds, and you can proceed the same as before.

/Eric