Conference rusure::math

    Sorry, there was an error in .0
    
    A doesn't have a sixth root, but it does have the cube root
    
    	Y = (49/35 84/35)
    	    (14/35 49/35)
    
    X is merely close to being a sixth root.
    
    The original question still stands.
    
    Dick

    I find
    
    	(a b)^6   ( 5 12 )   ( A B )
    	(c a)   = ( 2  5 ) = ( C A )
    
    where
    	a = 0.391435775293197
        c = 0.438410379701368
        b = c * B/C
                                (A B)             (a b)
    In general, the 6th root of (C A) is given by (c a), where
    a and c are solutions to:
    
    	(C*a^2 + B*c^2) * (B^2*c^4 + 14*B*C*a^2*c^2 + C^2*a^4) = A*C^3
    	2*a*c * (C*a^2 + 3*B*c^2) * (3*C*a^2 + B*c^2) = C^3
    and
    	b = c * B/C
    
    This is worth noting:
    
    	( sqrt(2)  sqrt(12) )^2  ( 5 12 )
        ( sqrt(0.5) sqrt(2) )  = ( 2  5 )
    
    Finally, the cube root in .1 is incorrect.

>    This is worth noting:
>    
>    	( sqrt(2)  sqrt(12) )^2  ( 5 12 )
>       ( sqrt(0.5) sqrt(2) )  = ( 2  5 )
    
    Really? For what value of sqrt(6)?

    Oops.  There was a typo.  It should read:
    
    	( sqrt(2)  sqrt(18) )^2  ( 5 12 )
    	( sqrt(0.5) sqrt(2) )  = ( 2  5 )
    
    Also,
    	( sqrt(3)  sqrt(12) )^2  ( 5 12 )
    	( sqrt(1/3) sqrt(3) )  = ( 2  5 )
    

    Using Gilbert's notation and limiting ourselves to just squaring,
    it is easy to see:
    
    (a b)�    (A B)
    (c a)  =  (C A)
    
    implies  that A = a�+bc. But evaluating the determinant on the left we
    also see that 1 = a�-bc.
             or A+1 = 2a�.
    
    In the given example A=5 but 6=2a� is not solvable over the rationals,
    so there's no second-power solution.
    
    Having a look at the comparable cube/cubic gets pretty hairy and I
    wouldn't try it without a symbolic algebra package. Anybody else
    better equipped?
    
      John

    >                   <<< Note 1863.2 by RTL::GILBERT >>>
    >    Finally, the cube root in .1 is incorrect.

    Yes Peter you are right, I was guilty of two careless mistakes in a
    row. But let me expand on what led up to my false conclusions
    
    I was looking at the equation x^2 - 6y^2 = 1150 which arose in 1837.* .
    Using the reasoning in 1837.4 we only need to find the "generator"
    solutions with y < sqrt(1150*2*2) and the transformation
    
    	A = (5 12)
            (2  5)
    
    will provide the rest.
    
    The generator solns are (34,1) (38,7) (50,15) (70,25) (106,41) (158,63)
    (s1 to s6 say)
    and the first generated solution is A*(34,1) = (182,73).
    (s7 say)
    
    I was looking for relationships between the generator solutions (to try
    and cut down the number of generators) and the
    following appeared:
    
    s2 = X*s1   s4 = X*s2   s6 = X*s4
    s3 = X'*s2   s5 = X'*s3   !s7 = X'*s5!
    
    Where  X = (25/23 24/23)   and   X' = (25.4/23 26.4/23)
               ( 4/23 25/23)              ( 4.4/23 25.4/23)
    
    implying that Y = X'*X is a cube root of A.
    
    My first mistake was to confuse X' with X (must have been a really bad
    day) and the second was to believe the statement in !...!
    
    Other variations on the original equation are more obliging. For
    example
    
    x^2 - 5y^2 = 4 has the transform matrix ( 9 20)
                                            ( 4  9)
    
    which has the cube root (3/2 5/2)
                            (1/2 3/2)
    
    which lets all solutions be derived from the single generator (2,0).
    
    Anyway, sorry about the red herrings. I've obviously got some more
    analysis to do.