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 2014-05-02, 21:19 #1 Qubit     Jan 2014 2×19 Posts 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 that$p$is 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 for$a\in \left\{ \frac{{{p}^{2}}-1}{4},\frac{{{p}^{2}}-1}{8} \right\}$. 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
2014-05-02, 22:19   #2
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

100000110000002 Posts

Quote:
 Originally Posted by Qubit (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 that$p$is 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 for$a\in \left\{ \frac{{{p}^{2}}-1}{4},\frac{{{p}^{2}}-1}{8} \right\}$. 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 ?

2014-05-02, 23:51   #3
Batalov

"Serge"
Mar 2008
Phi(4,2^7658614+1)/2

100100101010112 Posts

Quote:
 Originally Posted by science_man_88 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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