the power of cyclic subgroups (a proposal)
- From: Gottfried Helms <helms@xxxxxxxxxxxxx>
- Date: Wed, 07 Dec 2005 21:52:31 +0100
Hi,
I'm trying to put some ideas about the power of cyclic-
subgroups together into a page at my site. Because I'm
yet rather unfamiliar with proving and arguing sufficiently
I'd like comments for this part.
What I want to do is discuss the primefactorization of
the term
n
a - 1
with one goal to illustrate the restrictions which lay
on that term being a perfect power:
n m
a - 1 = b
============================================================
Intro:
For a typical primefactor p, not a factor of (a-1), one
can write
n
~ ( f + {n,p})
n g
a - 1 = x * p (1)
with g and f each depending on a and the selected p such that
<legend: -------------------------------------------------------
g = order of the multiplicative cyclic subgroup of p to the base a
or "index g, (g>0), of first occurence" of
g
a - 1 = 0 (mod p)
f = the power to which p occurs in that expression, such that
g f
a - 1 = x * p
n
~ =1 if g divides n, else 0 (may also be serially written as the more
g usual "n|g" to save print-space)
{n,p} = the power to which p occurs in n
-------------------------------------------------------------- /legende>
Example for a=2, n=42 , p=7
42
~ ( 1 + {42,7})
42 3
2 - 1 = x * 7 (according to eq 1)
3
g = 3, since 7 divides 2 - 1 first time,
f = 1, since it is in there only to the power of 1,
{42,7} = 1 since 7 divides 42 only to the power of 1
so
42 1*(1 + 1) 2
2 - 1 = x * 7 = x*7
=============================================================================
I restate the first equation:
For a typical primefactor p, not a factor of (a-1), one can write
n
~ ( f + {n,p})
n g
a - 1 = x * p (1)
and for a typical primefactor q, being a factor of (a-1), this
comes out to be
n
~ ( f + {n,q})
n 1 f + {n,q}
a - 1 = y * q = y * q (2)
since for any n we have q in the product at least to the
same power as in (a-1).
If q divides also n, we have that respected in the {n,q}-term of
an q-primefactor to account for higher powers of q in the equation.
----------------------------------------------------------------
This two equations put together is
____ ____
n || f + {n,p} || f + {n,q}
a - 1 = || p || q (3)
(a-1)~p=0 (a-1)~q =1
g(p)>1
n~g(p)=1
and since "if g(p) divides n then does not p", the term {n,p} at
the exponents of the p-primefactors vanish and we even can reduce
this to
____ ____
n || f || f + {n,q}
a - 1 = || p || q (4)
(a-1)~p=0 (a-1)~q =1
g(p)>1
n~g(p)=1
-------------------------------------------------------------------
For most cases the exponent f for p is 1; if f is greater than 1,
then for a=2 these p are called wieferich primes; for a=3 they
could be called Miramanoff-primes; anyway they occur very rare,
and especially rare with high values. For instance for a=2 there
are only two primes with f=2 known (p=1093,p=3511) and none with
f=3.
For a=19 I found p=7 with f=3, and if primefactors p occur
in that term at all, they must *all* occur to the *same* exponent m,
if we want to have a^n - 1 = b^m , a pefect power.
Thus to get a higher power than f (while rarely f>1) for the type-p
primefactors, we would select an n, which contains not only g(p)
but also powers of p.
For a certain p with grouporder g at the given base, n must then
configured like
m-1
n = g(p) * p
to get p only to the power of m into the result - and for
more factors p one had to add such factors - and the additional
needs to get them to the higher power of m. The next problem
is then, that *for each combination* of factors in n at least
one new factor p or q is introduced - a chain seemingly without
end, much obviously excessive for each m>2.
One other possibility to get the thing work and prevent infinite
growth of n and collecting new factors p, were to try to
prevent such primefactors completely.
But this meant to have an a^n -1 which contains only factors of
a^1 - 1, which is impossible. So we stay with the restriction,
that we had to find a solution in (a,n) having only primefactors
p where f equals n, and this seems very unlikely for -say- n>7.
--------------------
Next look at the primefactors of type q (these are dividing (a-1)).
Again they can only occur to the power f (with which they already
occur in (a-1)); and only such q, which divide n to a certain power
k, can also have k additional repetitions, what means: occur in
power of m.
But what does this mean? An example:
To get a high f at the beginning, q must be small against a, so
if we choose a=81 then q=2 occurs with the power f=4 in (a-1), since
81-1 = 2^4 * 5.
Now we had to find some power n in 81^n - 1 which also makes 5
having the exponent 4. This could only happen, if n contains
5 to the power of 3, means n=k*125, so we had as a candidate
81^5^3 - 1 containing (2*5) to the power of m=4, having
81^5^3 - 1 = x * (2*5)^4
So with the factors which are unavoidable by being factors
of (a-1), and finding an n which makes their exponents equal,
we have still an x, which consists of primefactors of the
type p, which do *not* divide 81-1, do *not* divide n=5^3,
and thus have the {n,p}-term of their exponents zero, which means,
they have only their exponent f, which exceeds 1 only in the
rare wieferich-like cases.
To get them (if they are not "wieferich"-like) involved to the power
of 4, it is required, that n contains also them all as factors to
the power of 3;
say the primefactor p=11 may be involved in x, it is not wieferich,
and its group-order must divide n (so only such p can occur, whose
group-order is a multiple of 5), so g(11,81) may be assumed g=5.
Then we have in a^n = 81^5^3 the factor 11 to the power of 1; and to
screw this up to 4 we need 11 to the power of 3 in n, so n grows
again to n=5^3 * 11^3 and we have
3 3
5 11 4 4 4
81 - 1 = x * 11 * 2 * 5
This extends arbitrarily.
Another important point is, that all factors occuring
on the rhs are occuring at the left side either in the
(a-1)-part or in the exponent (there with one power less),
so the left side is far more quickly growing than the
right side.
----------------------------
Now the things can be simplified a bit, considering, that
a composite n in a^n - 1 can be split up into a pure prime-
power of another power of a:
a^n - 1 = (a^k)^r - 1 = c^r - 1 (with r being prime)
so the criteria on the p and q-factorization are more
restrictive again.
From (4) we then have:
____ ____
n || f || f + {n,q}
a - 1 = || p * || q (4)
(a-1)~p=0 (a-1)~q =1
g(p)>1 g(q)=1
n~g(p)=1
f + {r,q} f
where q = q since q does not divide r,if q<>r
except for possibly the one primefactor q=r, which occurs, if
the exponent r divides also c-1,
in that case we have q^(f+{r,r}) = q^(f+1) else we have q^(f+0)
____ ____
r || f || f + r~q
c - 1 = || p * || q (5)
(c-1)~p=0 (c-1)~q =1
g(p)=r g(q)=1
Here we see immediately, that we cannot have any primefactor
p or q to a higher power than 1, except either they are wieferich-
primes to the given base, or q=r, then the power of *one primefactor
only* on the rhs can be f+1.
Analoguously this can be discussed in the reverse view; if it is asked,
whether
n m
a - 1 = b
then it can be asked too, whether
m n
b + 1 = a
and b,m changing places with a,n we have
n m
a + 1 = b
and from this again
r m
c + 1 = b
where we get different p-factors; namely we can get only
such primefactors p, whose grouporder in c^p -1 is of the form
g(p)=2*(2k+1)
==============================================================================
The existence, or the limit of the occurence of wieferich-
type-primes cannot be judged with this method, but it can
be shown with relatively simple means, which crucial role
they play in such exponential diophantine problems. Before
Mihailovics' proof 2002 of the catalan-conjecture there were
intermediate results, that the number of solutions could only
be finite for any selection of a, or (accordingly to a small
article of G.Frey) it was 1850 known due to Lebesgue that
p 2
a - b = 1
with p>1 has only the trivial solution with b=0, and 1965
Chao Ko has proved the inverse, that
2 p
b - a = 1
has only the solution (a,b,p) = (2,3,3), and for the
more general case with a variable and higerh value than
2 in the exponent, G.Frey mentions Tijdeman with the proof
of an effective computable numerical upper bound for p.
The first result that Mihailescu in 2000 achieved, was to prove,
that in
p q
a - b =1
there is a relation between the exponents p and q such that
q-1
p = 1 (mod q²)
and
x = 0 (mod q²)
which (symmetrically applied) means, that p and q must build
a pair of double-wieferich primes, if there were a solution.
(finally in April 2002 Mihailescu proved the complete
catalan-conjecture accordingly to G.Frey, but I don't know
that article and doubt, I could even chew that)
---------------------------------------------
This consideration of cyclic subgroups seems to give an
interesting insight in such problems and important conditions
implied with their formulation. The last Fermat-theorem can
be studied quite similar like this before; instead of 1 in
c^p - 1 = b^m one has to insert d^n and the logic is completely
the same as exercised above. Even then value of m is restricted
to a certain value, namely it must equal p, such that
c^p - d^p = b^p , and again the double-wieferich-pair plays an
important role as a restriction to any considered solution.
( Informations about the Catalan-history taken from
G.Frey "Der Satz von Preda Mihalescu..." in IMV,Nr 192
(2003, p 1-11; also online available)
Gottfried Helms
.
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