Re: group theory question

On 28/09/2024 04:48, Mike Terry wrote:

On 28/09/2024 01:53, Peter Fairbrother wrote:

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!

I shall beat myself up immediately.
[...]

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

Ok, thanks for the help - I should have asked:
Is the set S = g^(2^n) mod p (where g is any generator / element of the multiplicative group mod p, p is prime and n=1 to p) plus the element 1 equal to the set QR of quadratic residues of the aforementioned group?

(Note we can calculate g^(2^n) = g^(2*2^(n-1)) = g^(2^(n-1))^2 iteratively by squaring, taking mod p after each iteration. 

Yes, so we would never get a non-QR in S. But do we get all the QRs?
A proof?
Peter Fairbrother