Re: group theory question

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.