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#1 2013-07-10 12:43:09

Registered: 2013-07-10
Posts: 1

Modular Arithmetic: Generators for set Zp*

Hi guys, I'm taking a course about cryptography, and Im currently confused on a topic I learned about. Basically my task is to find 5 generators by testing g=2,3,4,5,6,7. I dont quite understand what generators are but what I understand ( or think I understand) is that if I take p, which is the number of the modulus and observe p-1's factors, I then raise g to those powers and mod by p and see if I get one as an answer. If I get one, then I know the number is not a generator. Please help me understand this topic- the specific question is test g=2,3,4,5,6,7... for being generators mod 103 until you can find one of them. Since 102's factors are 2,3,17,51 and 6, I first tried testing 2 out. I did 2^2,2^3,2^17,2^51 and 2^6 and modded them all by 103 and didnt get a single 1. Does this mean that 2 is a generator? (I already know it is, but I want to know why). Thanks for any help, its appreciated, thanks.

EDIT: this link may help explain

Last edited by napalmgrenade (2013-07-10 13:13:11)


#2 2013-07-11 00:21:01

bob bundy
Registered: 2010-06-20
Posts: 8,337

Re: Modular Arithmetic: Generators for set Zp*

hi napalmgrenade

Welcome to the forum.

This topic is not my area of expertise (if indeed I have one at all smile ) but I can get you started I think.

The underlying mathematics for this is group theory.  There's a bit on this page

If the operation of a group is multiplication, then 1 is the identity element because

x.1 = x for all x in the group.

A member of the group is called a generator if you can make all the other elements of the group from it.

Let's say the group is G = {1,2,3,4,5,6} and the operation is multiplication, mod 7

3 is a generator because

3^1 = 3 = 3mod7
3^2 = 9 = 2mod7;   
3^3 =27 = 6mod7; 
3^4 = 81 = 4mod7;
3^5 = 243 = 5mod7;
3^6 = 729 = 1mod7

Once you get to the identity like this, higher powers of 3 will just generate the same elements again (eg 3^7 = 3mod7)

Note: You can save some effort by using the previous answer to get the next, taking advantage of the mod

eg 3^3 = 3^2 x 3 = 2 x3mod 7 = 6mod7

On the other hand, 2 isn't a generator because

2^1 = 2;
2^2 = 4;
2^3 = 8 = 1mod7

I've reached the identity before generating all the group elements.  Higher powers will just keep producing 2,4 and 1 over and over.


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei


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