Introduction to Number Theory

Introduction to Number Theory

By ynori7 | 10335 Reads |

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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 → 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