Is there some odd prime $p$ and odd number $q$ such that $q|p-1$ and $ord_{p}(2)|ord_{q}(2)$? I'm fairly sure there isn't, so I'm mostly looking for a proof of that, although a counterexample would also be greatly appreciated.
Asked
Active
Viewed 140 times
3
-
$q=5, p=31$ works, I think. – Aphelli Jan 09 '19 at 23:49
-
I don't think so - $ord_{5}(2)=4$, $ord_{31}(2)=5$? Unless I'm seriously misunderstanding something. – Zachary Barbanell Jan 09 '19 at 23:52
-
@Mindlack No, $ord_{31}(2)=5$ and $ord_5(2)=4$. – David Jan 09 '19 at 23:52
-
In fact a Mersenne prime cannot work as we then have $p=2^k-1$ and $ord_p(2)=k$ and $q\mid 2^{k-1}-1$ and $ord_q(2)\le k-1$. – David Jan 09 '19 at 23:54
-
Right, I was mistaken, thanks. – Aphelli Jan 09 '19 at 23:56
-
@David how do you get from $q | 2^{k-1} - 1$ to $ord_{q}(2) <= k-1$? – Zachary Barbanell Jan 10 '19 at 00:00
-
Definition of order: it is the smallest $m$ such that $q\mid 2^m-1$. – David Jan 10 '19 at 00:48
-
Oh, of course. I didn't think of using larger divisibility values to set an upper bound. – Zachary Barbanell Jan 10 '19 at 00:50
-
@TheZachMan Isn't the notation $ord_2(p)$, if we want to find the smallest positive $k$ with $2^k\equiv 1\mod p$ ? – Peter Jan 10 '19 at 08:11
1 Answers
2
There are examples : One of it is $$p=251$$ $$q=125$$ With the following PARI/GP - program you can search for more examples :
? forprime(p=3,10^4,fordiv(p-1,q,if(Mod(q,2)<>0,if(Mod(znorder(Mod(2,q)),znorder
(Mod(2,p)))==0,print([p,q])))))
[251, 125]
[1459, 729]
[5419, 301]
[5419, 387]
[5419, 903]
[5419, 2709]
?
By replacing "10^4" by a larger value, you can find even more examples.
Peter
- 86,576
-
That's interesting that the lowest two both have $q = k^{3}$ and each $p$ that works works with $p-1/2$ – Zachary Barbanell Jan 11 '19 at 03:10