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Old 2014-05-02, 21:19   #1
Qubit
 
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Default Square-freeness of Mersenne numbers

(That is, prime-exponent Mersenne numbers.)

I wonder if we shall hear about developments regarding the mentioned open problem (not necessarily in the near future).
In any case, and in a hope to start a discussion about it, here are a few immediate (elementary) results if we assume a square divisor exists:

Let p and q be odd primes s.t. {{p}^{2}}|{{2}^{q}}-1.
(Not all properties below are unique to our assumption.)
1. It is known thatpis a Wieferich prime (moreover, {{2}^{\frac{p-1}{2}}}\equiv 1\left( \bmod {{p}^{2}} \right)).

2. From this theorem: p\equiv 1\left( \bmod q \right),p\equiv \pm 1\left( \bmod 8 \right) (last one implies{{p}^{2}}\equiv 1\left( \bmod 8 \right)).

3. q|\frac{p-1}{2}, thus {{2}^{q}}-1|{{2}^{a}}-1 for a \in \left\{ p-1,\frac{p-1}{2},{{p}^{2}}-1,\frac{{{p}^{2}}-1}{2} \right\}\ (Recall that if x|y then {{2}^{x}}-1|{{2}^{y}}-1).
Because 8|{{p}^{2}}-1, it's also true fora\in \left\{ \frac{{{p}^{2}}-1}{4},\frac{{{p}^{2}}-1}{8} \right\}.

Here are a couple of notes regarding the problem:
http://cybrary.uwinnipeg.ca/people/d...e_numbers.html
(which I haven't thoroughly looked at as of now)

It goes without saying that the facts above won't get us anywhere
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Old 2014-05-02, 22:19   #2
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Quote:
Originally Posted by Qubit View Post
(That is, prime-exponent Mersenne numbers.)

I wonder if we shall hear about developments regarding the mentioned open problem (not necessarily in the near future).
In any case, and in a hope to start a discussion about it, here are a few immediate (elementary) results if we assume a square divisor exists:

Let p and q be odd primes s.t. {{p}^{2}}|{{2}^{q}}-1.
(Not all properties below are unique to our assumption.)
1. It is known thatpis a Wieferich prime (moreover, {{2}^{\frac{p-1}{2}}}\equiv 1\left( \bmod {{p}^{2}} \right)).

2. From this theorem: p\equiv 1\left( \bmod q \right),p\equiv \pm 1\left( \bmod 8 \right) (last one implies{{p}^{2}}\equiv 1\left( \bmod 8 \right)).

3. q|\frac{p-1}{2}, thus {{2}^{q}}-1|{{2}^{a}}-1 for a \in \left\{ p-1,\frac{p-1}{2},{{p}^{2}}-1,\frac{{{p}^{2}}-1}{2} \right\}\ (Recall that if x|y then {{2}^{x}}-1|{{2}^{y}}-1).
Because 8|{{p}^{2}}-1, it's also true fora\in \left\{ \frac{{{p}^{2}}-1}{4},\frac{{{p}^{2}}-1}{8} \right\}.

Here are a couple of notes regarding the problem:
http://cybrary.uwinnipeg.ca/people/d...e_numbers.html
(which I haven't thoroughly looked at as of now)

It goes without saying that the facts above won't get us anywhere
why not just 8|p^2-1 therefore p^2 \eq 7\text{ mod 8} since p^2 for any p can only be 1,4 or 0 mod 8 p can not exist in the whole numbers ?
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Old 2014-05-02, 23:51   #3
Batalov
 
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Quote:
Originally Posted by science_man_88 View Post
why not just 8|p^2-1 therefore p^2 \eq 7 (??) \text{ mod 8} ?
Because your statement contains a trivial arithmetic error.

Remove this error and the statement becomes vacuous:
8|p^2-1 "therefore" p^2 \eq 1\text{ mod 8}
therefore nothing.

p^2 \eq 1\text{ mod 8} was already redundantly stated in (2) and is equivalent to "p is odd". (any odd p, not just p \eq \pm 1\text{ mod 8} )
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