A Diophantine equation, solution and conjecture

["richard miller"]

I posted this about a year ago but didn't publish the analytic solution
with it
since I thought the conjecture would soon be disproved, preferably by
me. A year
later I've not got any further with it so have fun.

Regards,

Richard Miller


Equation
--------

The following Diophatine equation is defined for positive integers a,
b, c, k and
exponent n, where 1 < a < b < c, n > 2, and k >= 0


1)      c^n = k.a^n.b^n + a^n + b^n


Solution
--------

If q_a and q_b denote 'unity roots' mod a^n and mod b^n respectively,
defined by


2)      q_a^n = 1 mod a^n

3)      q_b^b = 1 mod b^n


then equation 1 has integer solutions for c (and consequently k),
parametrised
by integers s and t, as given by


4)      c = s.a^n + q_a.b

5)      c = t.a^n + q_b.a

where integers s and t are solutions to the following linear
Diophantine
equation obtained by equating 4 and 5

6)      s.a^n - b.t^n = q_b.a - q_a.b


Example
-------

   a = 2
   b = 3
   n = 3, i.e. the cubic case,
   q_a = 1
   q_b = 19

we find that a particular solution for s and t is

   s = 1
   t = -1

so that equation 4 then gives c = 11 since

  11 = 1.2^3 + 1.3

and equation 5 also gives c = 11 since

  11 = -1.3^3 + 19.2

hence equation 1 becomes

  11^3 = 6.2^3.3^3 + 2^3 + 3^3

whereby the k = 6 value was obtained by calculating

  6 = (11^3 - 2^3 - 3^2)/(2^3.3^3)


Conjecture
----------

All solutions to equation 1 for integer c are such that

   c > a^n + b

subsequently all solutions to equation 1 for integer k are such that

   k > 0

Commentary
----------

The example above is the only case known where the equality holds since

  11 = 2^3 + 3

all other cases seem to show c >> a^n + b and, consequently, k >> 0

I would have thought this a weak conjecture doomed to failure since
even
if it were relaxed so that only c > a + b then c^n > a^n + b^n and
hence k > 0
which we know has to be the case (Wiles 1994).

A computer analysis has found no counter-example (yet) but neither can
I prove it outright. In particular, the proof of the case where
s < 0, t < 0, q_a > 1, q_b > 1 eludes me.

I have done quite a lot of work on it if anyone wants to discuss but,
as
we know, conjectures are two a penny.