numbertheory.htm

A Bit of Number Theory

Fundamental Lemma of Number Theory: Let be two integers. There exits unique integers $\QTR{Large}{q}$ and $\QTR{Large}{r}$ with and

Note: Uniqueness follows from the fact that implies

Definition: Let be two integers. The greatest common divisor of $\QTR{Large}{m}$ and $\QTR{Large}{n}$ is defined to be largest integer $\QTR{Large}{d}$ such that

$\QTR{Large}{d}$ divides both $\QTR{Large}{m}$ and $\QTR{Large}{m}$ .

Notation .

Since $\QTR{Large}{1}$ divides both $\QTR{Large}{m}$ and $\QTR{Large}{n}$ , we can use induction to show that exists. For that matter, since one can compute $\QTR{Large}{d}$ (very inefficiently!) by

trying all values, starting at $\QTR{Large}{1}$ and ending at the least of $\QTR{Large}{|m|}$ and $\QTR{Large}{|n|}.$

Examples: .

Lemma: Suppose then .

$\qquad$ PROOF: In fact all of their common divisors are in common!

The Euclidean Algorithm for computing : (See the proof below)

Theorem( Extended Euclidean Algorithm): Let There exists integers $\QTR{Large}{a}$ and $\QTR{Large}{b}$ , computable using the Euclidean Algorithm such that

$\vspace{1pt}$

$\qquad$ PROOF: (By induction)

If $\QTR{Large}{m=n}$ the theorem is trivial so for simplicity assume that $\QTR{Large}{m>n>0}$ . Again if $\QTR{Large}{n}$ divides $\QTR{Large}{m}$ the theorem is trivial so assume

$\ \QTR{bf}{\ \ }$ hence

If $\QTR{Large}{r}_{1}$ divides $\QTR{Large}{n}$ then and .

Otherwise

$\ \QTR{bf}{\ \ }$ hence and

or again if $\QTR{Large}{r}_{2}$ divides $\QTR{Large}{r}_{1}$ we are done else......

Example- Compute

Computing the simple-minded way requires long divisions. Divide $\QTR{Large}{746}$ and $\QTR{Large}{83}$ by $\QTR{Large}{2}$ through $\QTR{Large}{41}$ .

Now by the euclidean algorithm,

This admittedly is an easy example but we see that EA requires long divisions to find .

Lemma: Suppose $\QTR{Large}{p}$ is prime (the only divisors are $\QTR{Large}{p}$ and $\QTR{Large}{1}$ ). Suppose $\QTR{Large}{p}$ divides $\QTR{Large}{cd}$ then $\QTR{Large}{p}$ divides $\QTR{Large}{c}$ or $\QTR{Large}{p}$ divides $\QTR{Large}{d}$ .

$\qquad$ PROOF: Suppose $\QTR{Large}{p}$ does not divide $\QTR{Large}{c}$ , then . Thus there exists integers $\QTR{Large}{a}$ and $\QTR{Large}{b}$ , such that

Multiplying both sides of the equation by $\QTR{Large}{d}$ gives $\left( {}\right)$

Since both terms on the left hand side are divisible by $\QTR{Large}{p}$ so is the right hand side.

Simple Observation: Let $\QTR{Large}{p}$ and $\QTR{Large}{q}$ be primes. Let $\QTR{Large}{m=pq}$ . The only divisors of $\QTR{Large}{m}$ are $\QTR{Large}{p}$ , $\QTR{Large}{q}$ , and $\QTR{Large}{1}$ .

A Big Question: Suppose I know that $\QTR{Large}{m}$ is the product of two primes $\QTR{Large}{p}$ and $\QTR{Large}{q}$ . How hard is it to compute this primes? Again there is a simple-minded answer. . Hence we only have to do at most long divisions.

Suppose $\QTR{Large}{m}$ has $\QTR{Large}{500}$ digits, how long might that take?