Using quarternions instead of Gaussian primes when factoring a real number?
 

Do you think this would be efficient? It works. Here's a multiplication table and some factorizations:

[CODE] 1 i j k

1 1 i j k
i i -1 k -j
j j -k -1 i
k k j -i -1[/CODE]

911=(30+3i+j+k)*(30-3i-j-k)
569=(5+12i+12j+16k)*(5-12i-12j-16k)

Here's how I did it:
569=5^2+12^2+12^2+16^2=13^2+20^2
911=30^2+3^2+1^2+1^2

There is more than one way to factor these numbers:

(30+3i+3j+3k)*(30-3i-3j-3k)=911=(30-3i+3j+3k)*(30+3i-3j-3k)

Gaussian primes are the sum of 4 squares (with the exception of 3 which is 1^2+1^2+1^2), so they have 24 different quarternion factorizations each:
p=(a±bi±cj±dk)*(a±bi±cj±dk) where the signs are opposite (as in the example with 911). What do you think?

[QUOTE=Stargate38;295856]Do you think this would be efficient? It works. Here's a multiplication table and some factorizations:

[CODE] 1 i j k

1 1 i j k
i i -1 k -j
j j -k -1 i
k k j -i -1[/CODE]

911=(30+3i+j+k)*(30-3i-j-k)
569=(5+12i+12j+16k)*(5-12i-12j-16k)

Here's how I did it:
569=5^2+12^2+12^2+16^2=13^2+20^2
911=30^2+3^2+1^2+1^2

There is more than one way to factor these numbers:

(30+3i+3j+3k)*(30-3i-3j-3k)=911=(30-3i+3j+3k)*(30+3i-3j-3k)

Gaussian primes are the sum of 4 squares (with the exception of 3 which is 1^2+1^2+1^2), so they have 24 different quarternion factorizations each:
p=(a±bi±cj±dk)*(a±bi±cj±dk) where the signs are opposite (as in the example with 911). What do you think?[/QUOTE]

You are aware that the Quaternions (e.g. a Quaternion ring) are not commutative????

Get back to use after you get a degree in mathematics. Then your posts
may start to make sense. You bandy about words such as "efficient"
with no clue what the words mean.

Given away the values for 3[SUP]b[/SUP] (mod N), 3[SUP]a[/SUP] (mod N), b > a
we do not know the values for b, a directly at all

Can we calculate the value for 3[SUP]b mod a[/SUP] mod N efficiently?

[QUOTE=Raman;295910]Given away the values for 3[SUP]b[/SUP] (mod N), 3[SUP]a[/SUP] (mod N), b > a
we do not know the values for b, a directly at all

Can we calculate the value for 3[SUP]b mod a[/SUP] mod N efficiently?[/QUOTE]

If 3^a = 6 mod 101 and 3^b = 91 mod 101, is 3^(b mod a) 32 or 39 mod 101? It depends on what a and b are. So you can't compute this at all, let alone efficiently.

[QUOTE=Raman;295910]If we could be able to do that instantly, then we will be able
to factor N quite quickly.

Starting away right with the values for u = 3^Phi(N) = 1 (mod N)
v = 3^N mod N. Since that if GCD(Phi(N), N) = 1,
whether believing it, then we could be able to calculate the
step-by-step ladder for the values for 3^(u mod v) mod N,
3^(v mod u) mod N, alternatively, similar to the GCD calculation
process. And then reverse the direction starting away right directly
from the value for the 3^0 = 1 (mod N), 3^GCD(Phi(N), N) mod N, to get
the intermediate exponents, and then ultimately the value for Phi(N)[/QUOTE]

If instead of 3, some other generator element, then... ?
Oops! But that we do not know beforehand itself if whether an element is a generator, on over the ring Z[sub]n[/sub][sup]*[/sup] at all
But that it is being possible to randomly choose a base, there are being 50% chances, turns, for it to be the primitive root...

3[sup]b-a[/sup] mod N = (3b)(3a)[sup]-1[/sup] mod N
But that how do we know that how many times we have to carry out the above mentioned subtraction process before itself