Diophantine equations, Tzanakis articles
Gottfried Helms
Annette.Warlich at t-online.de
Wed Sep 28 15:15:02 CEST 2005
What do you think about the following display of the problem:
(Ascii-Font, allow long lines)
Factors of (3^m-1)/2 _____________________________________________________
where the ~ - symbol is used like in a fraction;
the value of the operation is 1 if denominator divides nominator, 0 else.
If two denominators are coprime, and the nominators are equal, then the
multiplication of two such fractions allows simply multiplication of the
denominators, otherwise the new denominator is their lcm
Let the primefactorization of your expression (3^m-1)/2 be displayed in the following
form:
m m m m m m m m m m m m m m
m ~ (1 + ~ + ~~ + ...) ~ (1 + ~ + ~~ +...) ~ (1 + ~ +...) ~ (2 + ~ + ) ~ (1 + ~ + ...) ~ (1 + ~ + ...)
3 - 1 2 2 2² 4 5 5² 6 7 5 11 3 13 16 17
------- = 2 5 7 11 13 17 ...
2
This shows very suggestively the interferences of some primefactors regarding their
group-order (cyclic multiplicative group base 3)
For instance, for m containing powers of 4 we have to consider the prime-factors 2,5,17
for the overal expression (and likely some more), that means: all those, whose cofactors
of the exponent-parentheses divide that selected power of 4 with their "denominator".
The parentheses are simply constructed; they have the obvious structure analog to a
geometric series; the "denominator" of their cofactors are an integral part of the
phi-function, for a prime-factor p it is the length of the cyclic subgroup base b (here:3)
(may be denoted as d(p,3) or L(p,3)), and with c(p,3) as the number of such cyclic
subgroups it is
phi(p) = d(p,3) * c(p,3)
Now apply this to the original question: can the expression equal a perfect even power?
It can be answered by inspection of each term separately and see, under which condition
all occuring exponents are either even or zero.
Assume we want to see, if the expression can ever contain a perfect even power of 2.
We try to find an m, which lets the exponent of the first prime-term sum up to be even.
Consider the exponent of the prime-factor 2:
m m m
~ (1 + ~ + ~~ + ...) // exponent of primefactor 2 for the expression
2 2 2²
This can be even if we have, that m is divisible by 2 (to make the cofactor 1) and then
we also need an even number of terms in the parenthese since each term can only
contribute 0 or 1; that means: for all M containing powers of 2 with exponent 1,3,5,7
this expression produces an even exponent for the primefactor 2 and thus an even
powered prime-factor 2 for the overall expression in question.
But as I said, one has also to consider the other primefactors, which have powers of 2
in the denominator of the cofactors in their exponent-cofactors; the next one is p=5
m m m
~ (1 + ~ + ~~ +...) // exponent of primefactor 5 of the expression
4 5 5²
If we have m containing 2^3,2^5,... then we will have 5 as a prime-factor
in the overall expression.
To make the exponent of the primefactor 5 even, m must also contain
5^1 or 5^3 or 5^5 and so on, just an odd power of 5
Next we have to check the primefactor 17.
m m
~ (1 + ~ + ...) // exponent of primefactor 17 of the expression
16 17
It applies the same as with the prime-factor 5: if m contains 2^4 then it
therefor must also contain an odd power of 5 and an odd power of 17, which
in turn mutiplies such factors to the basic expression.
One can see, that the question can be expressed in relatively simple terms
of the group-orders of primes base 3.
But the interferences between the group-order properties for the affected
primefactors of the overall-expresion are still more complicated with growing m.
For instance, if we test an m containing 2^3, and m needs also an odd
power of 5 because of that, then also the prime-factor 11 gets relevant,
since its exponent-cofactor contains a 5 in the denominator.
That means: once 5 is in m (which is needed, when 2^3 or higher power of 2
is in m), then we need also an even power of 11 in m (or -luckily- a zero
power is allowed since the first summand 2 is already even).
So this all must be checked, when we start with the sole assumption, that the
basic expression contains a power of 2.
But what, if we start assuming m contains a power of 5?
The first matching primefactor of the overall-result is 11 - since its
group-order (which is in the denominator of the cofactor of its exponent)
is 5 (with base 3).
m m
~ (2 + ~ + ) // exponent of primefactor 11 of the expression
5 11
If m contains 5, then m needs also an even power of 11, but that can also
mean a zero-power!
Since if m contains 5 and is not divisible by 11, then the exponent of the
primefactor 11 of the basic expression term is already 2 for the first summand
in the parenthese only.
Such cases - occuring with base=2 instead of 3- are called wieferich primes.
The only known two (with respect to base 2) are 1093 and -I think- 3511;
for every base wieferich-analogs seem to exist, like this 11 for base 3
and as far as I know, it is not established, how many and whether there could
be even infinitely many for some base (but I have very limited knowledge here.)
A second question at this point is also - if there is another base-3 wieferich-:
is its group-order (base 3) also a multiple of the group-order of another prime?
That becomes then important, because that would start a new series of interferences
with other prime-factors.
A third question, btw, which you alreadymentionedin your post: are there higher
numbers in the parentheses instead of 2, like 3,4,... as here at the primefactor
11 ? I never saw that question even discussed in the material which were available
to me. So I would like to know, whether this is possibly excluded by some simple
principle and thus not mentioned in that articles.
One more remark related to the group-order 5 of p=11 (base 3): the question is
not infinitely complicated for some type of m, since the group-order of any
prime cannot fall below a certain limit, which is determined by the mersenne-analogs
of a base; for instance for base 2 the group-orders of primes >2^p-1 cannot be equal
or below p (but... that bound 2^p-1 is not really small when you have to consider
from the viewpoint of a selected p....)
Hmm - that's for a first shot.
---------------------
It may be of interest to consider the "conjugate" expression
m
3 + 1
Besides of its own it is interesting because the product of this expression and your
given one results again in that of the first form again:
(3^m + 1 )*(3^m - 1)/2 = (3^(2m) - 1)/2
and the analoguous form of the primefactorizations goes here:
m m m m m m m m m m m m m m m
(~ - ~ ) + 1 (~ - ~ )(1 + ~ + ...) (~ - ~ )(1 + ~ +...) ~ (2 + ~ + ) ~ (1 + ~ + ...) (~ - ~ )(1 + ~ + ...)
m 1 2 2 4 5 3 6 7 oo 11 oo 13 8 16 17
3 + 1 = 2 5 7 11 13 17
Here we have, that the primefactors 11,13 and others do *not* occur in the expression,
thus for systematics I denoted the denominator of their exponential-cofactor with oo
(meaning infinity).
Since multiplication of the initial terms is only addition of the exponents of their
primefactors one can either simply see, that this notion agrees with a substitution
of m by 2m after multiplication resp addition of prime-exponents.
May be, for the "infinity"-terms it would be better to write the subtraction
m m m m m m m m m m m m m m m m m
(~ - ~ ) + 1 (~ - ~ )(1 + ~ + ...) (~ - ~ )(1 + ~ +...) (~ - ~ )(2 + ~ + ) (~ - ~)(1 + ~ + ...) (~ - ~ )(1 + ~ + ...)
m 1 2 2 4 5 3 6 7 5 5 11 3 3 13 8 16 17
3 + 1 = 2 5 7 11 13 17
and then to fit "notationally" the first expression term by term:
m m m m m m m m m m m m
m ~ (1 + ~ +..) ~ (1 + ~ + ...) ~ (1 + ~ +...) ~ (2 + ~ + ) ~ (1 + ~ + ...) ~ (1 + ~ + ...)
3 - 1 2 2 4 5 6 7 5 11 3 13 16 17
------- = 2 5 7 11 13 17
2
to have an vertical addition and/or subtraction-scheme with easily accessible columns.
It is interesting, which prime-factors do *not* occur in the 3^m+1 - primefactorization:
this are -at a first glance- those, whose group order (occuring in the denominator of
its exponent-cofactor) is not divisible by 2.
And also: the factorization of 3^m+1 seems to be a simple scheme, expressible once you
have the prime-factorization of the first one, your basic expression.
----------------------
Well this all may not answer much of your question, but I hope the insight it gives
is interesting for someone here.
Gottfried Helms
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