- --- Paul Leyland <pleyland@...> wrote:
> > I was just mucking around with some weird groups trying to

Subgroup of

> > see if I could

> > improve on William's P+1, and I noticed that in one of my

> > groups I had the

> > choice of either checking gcd(N,x_i-1) _or_ gcd(N, y_i) for

> > factors, where

> > < x_i, y_i > is a point in my group (and <1,0> is my I)

>

> Fascinating. Which group are you using?

Mod x^2+1 (Gaussian integers)

Mod N (obviously).

> I've been playing around with commutative hypercomplex numbers for the last couple of weeks.

I was going to look at higher dementions after this. :-I

> These are like quaternions, but commutative and i^2 = j^2 = -k^2 = 1. They do *not* form a

> division algebra.

I was just trying to strip away the 'try both x and y' fluff from my gp

script to include here, and I think may have broken it! Having said that, I

should probably keep that stuff in so that my bias can be demonstrated.

> I also played a bit with the 2-d algebra defined by i*i = 1. Again, not a division algebra.

I've heard the term in a context of Clifford Algebras, but know very little about them

(I can construct them - double the previous size, and add a twist). I know

IDs, EDs, and UFDs though, is it an alias for one of those?

> Because these things don't form a field, it would seem to be possible to exploit zero divisors

It's an interesting concept. I've not convinced myself that the zero-divisors

> in a factoring algorithm. Not got very far yet.

help find factors though. However, in this field it's worth throwing any idea

around. Hmmm, no, scratch that, I think I have convinced myself. I think I

like it. However, I oughtn't get distracted from my little gaussian babies

until I've got something fluff-free.

Phil

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http://mailplus.yahoo.com > > I also played a bit with the 2-d algebra defined by i*i =

The group table for this algebra is straightforward:

> 1. Again, not a division algebra.

>

> I've heard the term in a context of Clifford Algebras, but

> know very little about them

> (I can construct them - double the previous size, and add a

> twist). I know

> IDs, EDs, and UFDs though, is it an alias for one of those?

* | 1 i -1 -i

---+------------

1 | 1 i -1 -i

i | i 1 -i -1

-1 | -1 -i 1 i

-i | -i 1 i 1

Clearly an abelian group.

You can easily form a commutative ring with unit from it.

A typical element X looks like x+iy but the difference from

the field C is that X^(-1) is (x-iy)/(x^2-y^2), and X is a divisor

of zero when x = \pm y.

I rather suspect that the P-1 analog in this ring is just the P+1 algorithm, but haven't yet proved that.

Paul- --- Paul Leyland <pleyland@...> wrote:
> The group table for this algebra is straightforward:

What's your b_0?

>

> * | 1 i -1 -i

> ---+------------

> 1 | 1 i -1 -i

> i | i 1 -i -1

> -1 | -1 -i 1 i

> -i | -i 1 i 1

>

> Clearly an abelian group.

>

> You can easily form a commutative ring with unit from it.

> A typical element X looks like x+iy but the difference from

> the field C is that X^(-1) is (x-iy)/(x^2-y^2), and X is a divisor

> of zero when x = \pm y.

>

> I rather suspect that the P-1 analog in this ring is just the P+1 algorithm, but haven't yet

> proved that.

If it's 'real' you stay real, and end up doing P-1.

If it's 'imaginary', you become real upon the first squaring, and P-1 ensues

So I assume you're 'complex'.

Do you agree with these power tables for (1+2i), which I chose arbitrarily?

(15:31) gp > v=init(7)

%5 = [Mod(1, 7), Mod(2, 7)]

(15:38) gp > for(i=0,15,print(i" "pow(v,i)))

0 [1, 0]

1 [Mod(1, 7), Mod(2, 7)]

2 [Mod(5, 7), Mod(4, 7)]

3 [Mod(6, 7), Mod(0, 7)]

4 [Mod(6, 7), Mod(5, 7)]

5 [Mod(2, 7), Mod(3, 7)]

6 [Mod(1, 7), Mod(0, 7)]

7 [Mod(1, 7), Mod(2, 7)]

O(v mod 7) = 6

(15:38) gp > v=init(11)

%6 = [Mod(1, 11), Mod(2, 11)]

(15:41) gp > for(i=0,15,print(i" "pow(v,i)))

0 [1, 0]

1 [Mod(1, 11), Mod(2, 11)]

2 [Mod(5, 11), Mod(4, 11)]

3 [Mod(2, 11), Mod(3, 11)]

4 [Mod(8, 11), Mod(7, 11)]

5 [Mod(0, 11), Mod(1, 11)]

6 [Mod(2, 11), Mod(1, 11)]

7 [Mod(4, 11), Mod(5, 11)]

8 [Mod(3, 11), Mod(2, 11)]

9 [Mod(7, 11), Mod(8, 11)]

10 [Mod(1, 11), Mod(0, 11)]

11 [Mod(1, 11), Mod(2, 11)]

O(v mod 11) = 10

If so, then:

O(v mod 3)=2

O(v mod 5)=4

O(v mod 7)=6

O(v mod 11)=10

O(v mod 13)=6

O(v mod 17)=16

O(v mod 19)=18

O(v mod 23)=22

O(v mod 29)=28

O(v mod 31)=30

O(v mod 37)=18

O(v mod 41)=8

O(v mod 43)=42

O(v mod 47)=46

O(v mod 53)=52

O(v mod 59)=58

O(v mod 61)=10

O(v mod 67)=22

O(v mod 71)=70

O(v mod 73)=12

O(v mod 79)=78

O(v mod 83)=82

O(v mod 89)=88

O(v mod 97)=48

Which looks like p-1.

I tried with other initial values (1+3i, 1+4i, etc.) and they all came out

with orders dividing p-1.

I also tried bases 2p with p=prime, and ditto. Likewise 3p.

Phil

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