Path: ...!Xl.tags.giganews.com!local-3.nntp.ord.giganews.com!nntp.brightview.co.uk!news.brightview.co.uk.POSTED!not-for-mail NNTP-Posting-Date: Sat, 28 Sep 2024 03:48:30 +0000 Subject: Re: group theory question Newsgroups: sci.math References: From: Mike Terry Date: Sat, 28 Sep 2024 04:48:28 +0100 User-Agent: Mozilla/5.0 (Windows NT 10.0; Win64; x64; rv:91.0) Gecko/20100101 Firefox/91.0 SeaMonkey/2.53.18.2 MIME-Version: 1.0 In-Reply-To: Content-Type: text/plain; charset=windows-1252; format=flowed Content-Transfer-Encoding: 7bit Message-ID: Lines: 54 X-Usenet-Provider: http://www.giganews.com X-Trace: sv3-r2To3XQEfmohK1sHCqPJhLQjRj8V1NK5ZXZAVlhCo39awI0ihn3hByGon0g2phzfUCbzLXGlJADujXg!iZ006aZitrZkdFpm+ZpvKgr4JqYeOTwfkuvBjNJKZdpoMXl2EA9cnQVxecl4UZ6Fip9jTMUtfPVZ!qrgax/XOtJbq8b1nCg5tx/+VSTA= X-Abuse-and-DMCA-Info: Please be sure to forward a copy of ALL headers X-Abuse-and-DMCA-Info: Otherwise we will be unable to process your complaint properly X-Postfilter: 1.3.40 Bytes: 3042 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.