20120409, 01:38  #1 
"Daniel Jackson"
May 2011
14285714285714285714
1011110100_{2} Posts 
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 1 569=(5+12i+12j+16k)*(512i12j16k) 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)*(303i3j3k)=911=(303i+3j+3k)*(30+3i3j3k) 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? 
20120409, 11:46  #2  
"Bob Silverman"
Nov 2003
North of Boston
2^{2}·1,877 Posts 
Quote:
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. 

20120409, 14:16  #3 
Noodles
"Mr. Tuch"
Dec 2007
Chennai, India
3×419 Posts 
Given away the values for 3^{b} (mod N), 3^{a} (mod N), b > a
we do not know the values for b, a directly at all Can we calculate the value for 3^{b mod a} mod N efficiently? Last fiddled with by Raman on 20120409 at 14:21 
20120410, 01:24  #4 
Aug 2006
5,987 Posts 
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.

20120410, 04:20  #5  
Noodles
"Mr. Tuch"
Dec 2007
Chennai, India
3×419 Posts 
Quote:
Oops! But that we do not know beforehand itself if whether an element is a generator, on over the ring Z_{n}^{*} 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^{ba} mod N = (3b)(3a)^{1} mod N But that how do we know that how many times we have to carry out the above mentioned subtraction process before itself Last fiddled with by Raman on 20120410 at 05:08 

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