Re: Wieferich primes and a wikipedia statement

Am 20.12.2005 00:23 schrieb Chip Eastham:
>
> After looking up the URL of the Wikipedia article from which
> you seem to be quoting:
>
> http://en.wikipedia.org/wiki/Wieferich_prime
>
> it seems clear that the article is garbled/in need of editing.
>
> Instead of "2^q - 1 divides p^2", surely the reverse is meant.
>
Chip -

that was exactly my first impression, and I corrected that
statement in the german version of wikipedia recently.
But I must confess, that I even misread that statement slightly
in a "third" way. Just yesterday I got it clear that -besides
the possible editing error- the given claim is much more
distant from my usual thinking... so possibly I might have
overlooked something.

> After all for something to divide p^2 with p prime leaves very
> little room to maneuver! Especially if p is assumed to be
> "a prime factor" of M_q = 2^q - 1.

That may have legitimated that "...immediately obvious..." in
the article... ;-)

>
> Rather we should ask about the situation when p^2 divides
> Mersenne number M_q = 2^q - 1. I referenced a proof of
> one direction, that this implies p is Wieferich, in my prior
> post.
The squarefreeness of Mersenne-numbers? Or only the fact,
that for a number p dividing 2^q - 1, q must be an integral
divisor of phi(p). or k*q = p-1, which is subject of the
Euler-theorem. If, instead of p only, even p² is assumed to
be a divisor of 2^q-1, then 2^phi(p²) - 1 must be divisible
by p² too, and phi(p²) = (p-1)*p and q must be a divisor of (p-1)*p .
And assumed that q is an integral divisor of p-1, then also
it is a divisor of phi(p²) = (p-1)*p. ...

Ah - I see: since q cannot be p it must be a divisor of
p-1 and then follows q divides p-1 instead and thus p² divides
2^(p-1)-1 (since q is assumed to be a divisor of p-1), thus p
must be wieferich.

Did I get this right?


>
> I did not find a proof of the converse, that if a Wieferich
> prime p divides M_q (for q prime), then p^2 divides M_q.
> However that lies behind the claim that a Mersenne
> number (for prime exponent q) cannot be a Wieferich
> prime, a claim I do see repeated at other Web sites.
>
> regards, chip
>

Gottfried
.

Relevant Pages

  • Re: proof of Wilsons theorem
    ... multiplication modulo p. Let a in Uand suppose ... so p divides. ... that (p-1) is an inverse of itself. ...
    (sci.math)
  • Re: Extended-Precision Division
    ... is pretty simple to implement via a pair of 64-bit / 64-bit divides, ... xor edx,edx mov ebx,div ebx; At this point in time EDX contains the remainder, and EAX ... Shift the divisor up so that it has no leading zero bits. ... Do a single back-multiply and subtract to decide if you need to repeat ...
    (comp.lang.asm.x86)
  • Re: Wieferich primes and a wikipedia statement
    ... >> using w for a wieferich prime, ... >> Gottfried Helms ... >>> It can be shown that a prime factor p of a Mersenne number ... > prime p divides M_q, ...
    (sci.math)
  • Re: a problem in elementary number theory
    ... Find the maximal integer x such that x divides p^4-1 for all prime ... So, since p is odd, we see that all three factors p^2+1, p+1, p-1 are ... More generally use the Carmichael lambda function. ...
    (sci.math)
  • Re: Wieferich primes and a wikipedia statement
    ... >> using w for a wieferich prime, ... >>> It can be shown that a prime factor p of a Mersenne number ... > prime p divides M_q, ... prime" or in a few cases the Wikipedia article ...
    (sci.math)