Date: Sat, 24 Aug 1996 19:26:02 -0700 (PDT)
To: rusin@math.niu.edu (Dave Rusin), ksbrown@seanet.com
From: ksbrown@ksbrown.seanet.com (Kevin Brown)
Subject: Re: Question about non-euclidian geometry

At 12:32 AM 8/22/96 CDT, Dave Rusin wrote:
> ...you have some problems in which you noted a one-parameter family of 
> solutions to a diophantine equation, and asked if there were any other 
> such families defined with polynomials.  I'd just like to point out 
> that you are essentially asking if the diophantine equation admits 
> solutions in the ring  Z[X]. In general, when you think there are no 
> such families of solutions, you ask more generally if there are any 
> solutions in the _fields_  Q(x) or perhaps Z_p(x) ; of course, if the 
> answer is "no" then there are no parameterized solutions in integers. 

Yes, in the particular case of the equation x^3 + y^3 + z^3 = t^3  the
complete rational solution is given by

             x = q [ 1 - (a - 3b)(a^2 + 3b^3) ]
             y = q [ (a + 3b)(a^2 + 3b^2) - 1 ]
             z = q [ (a^2 + 3b^2)^2 - (a + 3b) ]
             t = q [ (a^2 + 3b^2)^2 - (a - 3b) ]

where q,a,b are any rational numbers.  So if we set q equal to the inverse
of  [(a^2 + 3b^2)^2 - (a-3b)]  we have rational solutions of the equation
on my web page, x^3 + y^3 + z^3 = 1.  However, I think the problem of 
finding the _integer_ solutions is more difficult.  If t is allowed to be
any integer (not just 1) then Ramanujan gave the integer solutions

                      x = 3n^2 + 5nm - 5m^2
                      y = 4n^2 - 4nm + 6m^2
                      z = 5n^2 - 5nm - 3m^2
                      t = 6n^2 - 4nm + 4m^2

but none of these is a solution of x^3 + y^3 + z^3 = 1.  This still leaves 
me wondering if there are any other parametric integer solutions of this
equation.

Regards,
Kevin Brown

==============================================================================
[One solution is as follows  --djr]
   (1 - 72*b^3)^3 + (6*b - 144*b^4)^3 + (144*b^4)^3 = 1