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NNTP-Posting-Date: Sat, 28 Sep 2024 03:48:30 +0000
Subject: Re: group theory question
Newsgroups: sci.math
References: <vd7ior$u0fi$1@dont-email.me> <vd7k2o$u0fi$2@dont-email.me>
From: Mike Terry <news.dead.person.stones@darjeeling.plus.com>
Date: Sat, 28 Sep 2024 04:48:28 +0100
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On 28/09/2024 01:53, Peter Fairbrother wrote:
> That cam out wrong, should be e^(2^n) mod p maybe?
> 
> e to the power (2 ^n) mod p
> 
> I don't know how to do multi-level superscripts in newsgroups, sorry.

You could say e^(2^c).  Given standard precedence rules that is the same as without the brackets, 
i.e. e^2^c.   [That's what you wrote I believe.]

But... my newsreader unhelpfully displays the latter using superscripts, which makes it look like 
(e^2)^c, which is incorrect.  Well, that's my problem I guess, not the poster's.

Note: most posters here don't go for top-posting, preferring responses intermixed with the original 
quoted text.  Just saying, because some people will get cross with top posters!

> 
> 
> Peter F
> 
> 
> On 28/09/2024 01:31, Peter Fairbrother wrote:
>> Is the set e^2^n mod p (where e is a generator and element of the multiplicative group mod p, p is 
>> prime and n=0 to p) equal to the set of quadratic residues of the group?

No - you could just try out a couple of low p examples to see it doesn't work.  E.g. p=7.

generators:     3 and 5
qres:           1,2,4

e:              3       5
               ---     ---
e^(2^0)         3       5
e^(2^1)         2       4
e^(2^2)         4       2
e^(2^3)         2       4
e^(2^4)         4       2
....

both are missing qr: 1

(Note we can calculate e^(2^n) = e^(2*2^(n-1)) = e^(2^(n-1))^2 iteratively by squaring, taking mod p 
after each iteration.  In the above table 3^2 = 2 [mod 7], 2^2 = 4 [mod 7], 4^2 = 2 [mod 7], then 
pattern repeats...)

Obviously looking at e^(2n) gives quadratic residues, e.g. 3^0 = 1, 3^2 = 2, 3^4 = 4, which is 
iteratively multiplying by e^2 = 2 [mod 7], but iteratively /squaring/ doesn't work.

Using a spreadsheet for testing is an easy way to investigate this sort of thing...


Regards,
Mike.