How to solve diophantine problems?

[Cameron Laird]

In article <abr0dq$fsm$1 at thccy25.nthu.edu.tw>,
Joshua, Y.J. Lai <g893404 at oz.nthu.edu.tw> wrote:
>
>"Cameron Laird" <claird at starbase.neosoft.com> wrote in message
>news:EF9D274DA8540EF1.D12AA7084B83AAA2.1A9250647A55D036 at lp.airnews.net...
>> In article <TQXD8.31658$Po6.14894 at rwcrnsc52.ops.asp.att.net>,
>> Emile van Sebille <emile at fenx.com> wrote:
>> >Joshua, Y.J. Lai
>> >> I can roughly solve the diophantine problem by using a nest loop
>> >
>> >I'm not familiar with the "diophantine problem" and didn't, in a quick
>> >look, spot anything obvious to me stating it.
>> .
>> .
>> .
>> Rough translation:  a solution in integers (to
>> a system of polynomial equations and constraints).
>
>Thank you for your precise explanation. The problem now I suffer is how can
>I write a new checking loop
>instead of using two FOR LOOPs as nest loop. I am really interested in that.
>I will be very grateful if anyone of you can give me some hints.
>
>

Please explain the problem again.  Are you looking for
an implementation that solves this specific diophantine
problem (perhaps with the coefficients as variables,
with respect to the implementation) while manifesting
only one overt loop, or a general way to express
a program transformation which reduces loop counts for
diophantine problems, or ...?  At this point, you have
complete knowledge--that is, a terminating procedure
--about the solutions of the problems; what more can
you want?
you first expressed
-- 

Cameron Laird <Cameron at Lairds.com>
Business:  http://www.Phaseit.net
Personal:  http://starbase.neosoft.com/~claird/home.html