- --- In primenumbers@yahoogroups.com,

"bhelmes_1" <bhelmes@...> wrote:

> I do not know yet where the mistake in the proof comes from.

http://109.90.3.58/devalco/suf_helmes.htm#4

> (f + g sqrt(A))^2 = (f^2 + A g^2) + (2 f g) sqrt(A) = 1 mod P.

I satisfied your conditions, with composite P, as follows:

> 2 f g = 0 mod P.

f = 0, A*g^2 = 1 mod P. You assumed (incorrectly)

that f was non-zero. Hence everything below these

lines was irrelevant and your "proof" was pure fiction.

David --- In primenumbers@yahoogroups.com, "paulunderwooduk" <paulunderwood@...> wrote:

:

> http://tech.groups.yahoo.com/group/primenumbers/files/Articles/

> I ran various "minimal \lambda+2" tests on Gilbert Mozzo's 20,000 digit PRP, 5890*10^19996+2^66422-3 (x=1), using a 2.4GHz core:

> {

> 0m32.374s pfgw64 (3-prp)

> 1m9.876s pfgw64 -t

> 1m53.535s pfgw64 -tp

> 3m0.483s pfgw64 -tc

> 5m12.972s pfgw64 scriptify

> 4m4.811s gmp (-O3/no pgo)

> 4m9.148 pari-gp

> 1m15s theoretical Woltman implementation

> }

>

I compiled a better version of my code with gmp 5.0.5, on a different box running at 3.6GHz and got some better timings

{

17.505s pfgw (3-prp)

1m1.986s pfgw -tp

1m13.789s gmp (-O3/no pgo)

}

Paul