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# Introduction to Number Theory

## A simple intro to basic number theory concepts and their applications to the field of encryption.

This article is intended to show a simple introduction into number theory, and give readers some insight into the applications of number theory in encryption.

---------------------------------------Division---------------------------------------
The first piece of knowledge you need to make sense of this is division. Note that this is not just 10/2=5, this is a general definition of the concept which will be used later on in the article.

Definition:
�a� (such that a is not equal to 0) divides �b� if there exists �k� such that b = ak
This is denoted as: a|b (read as: a divides b)

Example:
- 3 | 7 � false
- 3 | 12 � true

Let �a�, �b�, �c� be integers. Then these are a few tautologies (something that is always true):
-If a|b and a|c , then a|(b + c)
-If a|b then a | bc
-If a|b and b|c then a|c

I�ll prove the first tautology mathematically: a|b and a|c &#8594; a|(b + c)
- Assume a|b and a|c.
- b=ak1, c=ak2 where k1 and k2 are constants
- b+c=ak1 + ak2
- b+c=a(k1+k2)
- Since k1+k2 is an integer, by the definition of division a|b+c

---------------------------------------Modulus---------------------------------------
Definition:
�a� and �b� are integers, and �m� is a positive integer. �a� is congruent to b modulo m if m|(a-b).
-This is written as: a=b mod m
-a=b mod m if a%m=b%m where x%y is the remainder of x/y

Example:
-17%5=2, 12%5=2, therefore 17=12 mod 5
-3%10=7, 17%10=7, therefore -3 =17 mod 10

Here are a few tautologies (the first one being the most important in this case):
-a=b mod m if there exists �k� such that a=b+km. Note that this is equivalent to the definition shown above: m|(a-b)
-(a+b)mod m =((a mod m)+(b mod m)) mod m
-ac mod m=((a mod m)(b mod m)) mod m
-Let �m� be a positive integer. If a=b mod m and c=d mod m, then: a+c=b+d(mod m) and ac=bd(mod m)

Here�s an example: Prove that if �n� is odd, then n^2=1 mod 8
There exists a �k� such that n=2k+1 (note that this is the definition of an odd number)
So, n^2=4k^2+4k+1
n^2-1=4k^2+4k
n^2-1=4k(k+1)
n^2=1 mod 8 is the same as 8|n2-1 which is the same as 8|4k(k+1)
4k(k+1)=8c reduces to k(k+1)=2c when you divide both sides by 4
Note that both k(k+1) and 2c are always even, and since k and c are just arbitrary constants, this is true.

Modulus is used in the definition for the Caesar Cipher:
E(x)=(x+k) mod 26 (note that the original Caesar Cipher used k=3)
Where E(x) is the encrypted text and x is the original text.

---------------------------------------Greatest Common Divisor---------------------------------------
The greatest common divisor of two integers �a� and �b� is the largest integer �d� such that d|a and d|b. This is denoted by gcd(a, b).
Two numbers are considered relatively prime if they don�t have any common factors (besides 1).

Here is the link to the Euclidean algorithm for finding the gcd of two numbers: http://www.hellboundhackers.org/code/gcd-1218_python.html
This is written in python, but it�s a fairly simple algorithm that should be easy to convert into whatever language you�re comfortable with.

---------------------------------------Multiplicative Inverse---------------------------------------
Given a=b mod m, there exists �A� such that aA=1 mod m
The inverse only exists if the gcd(a, m)=1

So, for example, 3A=1 mod 7
�A� must be equal to -2 since 3*(-2) is -6, and -6=1 mod 7 (since by definition, mod is a-b=mk, so -7=7k)

The inverse can be used in this type of situation: 3x=2 mod 5
To solve for x, you need to isolate it (i.e. get rid of the 3 in front)
The inverse of 3=2 mod 5 is 2, so multiply both sides by the inverse:
3xA=2A mod 5, so we get x=4 mod 5

If gcd(a, m) is not equal to one, you have to solve it differently. For example: 2x=2 mod 4
Break the problem down to the definition of modulus.
2x=2+4k Notice that everything is divisible by 2.
x=1+2k Now you can use the definition of modulus to change it back.
x=1 mod 2, therefore x=1 is a solution (note that there are many solutions)

-----------------------------------------------------------------------------------------------------
This can all be applied to when creating simple encryption algorithms and reversing them. For example, suppose that messages are encrypted using this formula: F(p)=(ap+b) mod 26 such that gcd(a, 26)=1 where p is the original data, a and b are integers, and F(p) is the encrypted data.

Here�s how you can find the decryption algorithm:
First break the problem down using the definition of modulus: F(p)=26k+ap+b
Isolate �p�: F(p)-b+26k=ap
Convert it back to modulus form: ap=(F(p)-b) mod 26
Find the multiplicative inverse to solve for p: p=A(F(p)-b) mod 26

I hope you found this article to be interesting, and if so I can write another on this (or a similar) subject.
-Ynori7

Comments

 on January 05 2009 - 01:22:20Before the critics make it here, I'll go ahead and say that I was left wanting more at the end of this article. That could both be an indication of a very interesting article and an indication of an article that is a bit short. Number theory is a fascinating necessity when it comes to algorithmic comprehension, which can be applied to encryption, encoding, coding logic, and optimizing code. It's also enjoyable for closet mathematicians. Finally, this article is a prime example of how someone must think before they do... this applies to coding as well: Many people like to just jump into the code when, in fact, they should sit back and rationalize how the code will work first. So, though I felt like I hit an abrupt and unexpected halt at the end of the article, I quite enjoyed it. It was well-written and it gave information on a topic that gets very little attention here. Good job, man. on January 05 2009 - 01:32:12Very nice. This makes a good start on coding and encrypting. Nice format and nice detail. I am with zephyr in wanting more, but all in all its is a good article. on January 05 2009 - 10:54:32Format and grammar was good. The content was interesting, and I agree with previous comments that there could have been a bit more. But I guess that leaves more for the next one. Keep it up. on January 10 2009 - 03:29:31Nice article kept me interested like to see more. This is what we need on HBH. on January 10 2009 - 07:45:29I thought it was good. I understood some of it but after a while I get confused with just reading. Makes me want to get back into math again. Keep up the good work. on January 22 2009 - 03:00:32I greatly enjoyed this. If I could guess I would assume you are taking/have taken a course in discrete mathematics? This article reminded me of mine thats for sure. I will agree with both Zephyr in that you should do a follow up/continuation of this article, and include more topics and content and with korg in that this is what we need here. Keep up the good work mate.
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