# Diophantine equations, Tzanakis articles

Gottfried Helms Annette.Warlich at t-online.de
Thu Sep 29 11:59:56 CEST 2005

```In my previous post some argumentation seems to be
messed a bit. I just set up an improvement; since
I've seen no comments so far, I assume, it is not
of too much interest, so I only post the link to
my website:

http://141.51.96.22/divers/3_hoch_m_minus_1.htm

I focused the argumentation to the following extract:

Q: for which m is the following expression a purely even power of n:
(3^m-1)/2 = n = (p^a*q^b*r^c *...z^h)^2
or more general
(k^m-1)/(k-1) = n = (p^a*q^b*r^c *...z^h)^2

1) the prime-factors p,q,r,... of n are linked to n's primefactors by their
group orders (base 3 or base k)

2) the relevant group orders are defined by the complete list of divisors of m

3) because of the requirement, that n is a square, not only the group orders
but also the primefactors p,q,r... of n themselves must -for the general case-
belong to the divisors of m - except if p,q,r are wieferich-analog primes

4) the group order and its associated prime(s) are coprime
4.1) the requirement, that each primefactor of n occurs as square, forces
the associated prime itself into the list of group-orders. The only
exceptions are primes, which are wieferich-analog primes.

5) from 1) to 4.1) follows, that this requirement most likely leads to an
infinitude of factors -in form of a tree-, which can only be limited by
the occurence of factors of n which are wieferich-primes

6) The wieferich-primes are very rare compared to general primes, and more
rare compared to the combinatorical inflation of factors, which occurs
once non-wieferich primes are involved.
Even if m links only two wieferich primes to the product n by its factorial
composition, then it requires a third wieferich-prime having exactly the
group-order of the product - besides of that each of the two group-orders
must be prime and unique and a group order of a wieferich prime.

7) Unique group orders are also rare; I defined "uniqueness" of group-orders
as "uniquely occuring group orders with a specific base" ; for instance,
with base 2 the only unique group orders define the mersenne-primes.
Each single none-unique group-order as factor in m leads to several
primefactors in n associated with that group order - again a case of
inflation of factors of n
So the tree-limiting wieferich-analog primes must also be mersenne-analog
primes in the same base (in the case of the current question this is base 3)

8) From 6) and 7) then follows, that very likely
only for m (being the group order of some p)
m being prime
and m being a unique group order (p being a mersenne-analog prime base 3)
and m being the group order of p (p being also a wieferich-analog prime)
n could be a square.
(For base 2 there is no known wieferich prime which is also mersenne-prime)

Hope, any of the probable shortcomings in the above are not too trivial ... ;-)

Gottfried Helms

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