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If $n$ is the smallest integer for which $p | 2^n-1$, ($p>2$ is a prime), then whether there is an integer $k (k < p)$, $s.t.$ $p^2 | 2^{kn}-1$. It's obvious to notice that $p^2 | 2^{pn}-1$

I try to use some properties of multiplicative order, but still have no idea. Can someone please give me some inspirations?

Thanks!

Bill Dubuque
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Bill Wang
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1 Answers1

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Sometimes there is such a $k$, sometimes there isn't.

$3|(2^6-1)$, and also $9|(2^6-1)$ so you don't even need to increase the exponent.

$7|(2^3-1)$, but in $\mathbb{Z}/49\mathbb{Z}$ the multiplicative order of $8$ is $7$.

Chris Sanders
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  • in Z/49Z the multiplicative order of 8 is 7 means $7^2|2^{21}-1$, still satisfy the conjecture right? – Bill Wang Feb 19 '22 at 07:21
  • no, because in this case $k=p$ – Chris Sanders Feb 19 '22 at 07:23
  • sorry made a mistake, I edited the question, $p>2$ is a prime, $n$ needs to be the smallest integer for which $p|2^n-1$. In such case, does $k$ still exist? – Bill Wang Feb 19 '22 at 07:59
  • doesn't make a difference: $3$ is the smallest positive integer $n$ such that $7|(2^n-1)$ – Chris Sanders Feb 19 '22 at 08:03
  • {2,4,8,16,32,15,30,11,22,44,39,29,9,18,36,23,46,43,37,25,1} is the list of $2^n$ in Z/49Z, so such $k$ does not exist for $p=7$. Thank you for your example. While now for 3, 2 is the smallest integer such that $3|(2^2-1)$, and there is no $k$ less than 6, $s.t. 9|2^k-1$ – Bill Wang Feb 19 '22 at 08:23
  • A better example is $p=1093$. $1093>2$ is prime, $n=1092$ is the smallest positive $n$ such that $p$ divides $2^n-1$, and $p^2$ divides $2^n-1$. See https://en.wikipedia.org/wiki/Wieferich_prime – Gerry Myerson Feb 19 '22 at 08:33