# Lecture 023

## Number Theory Revisit

### Division

a|b \land a | c \implies a| (bx+cy)

### Euclidean Algorithm

Continuously divide remainder gives gcd(a,b)
TODO: Proof

### Bezout's Lemma

- gcd(a, b) can be written in gcd(a, b) = ax + by
- gcd(a,b) is the smallest positive integer written in multiples of a and b xa + yb
TODO: Proof

### Corollary to Bezout's Lemma

- t|a \land t|d \implies t|gcd(a,b) greatest divisor divides common divisor
- a number divisible by gcd(a,b) can be written as ax+by
- gcd(a, b)=1 \implies ax+by=1 if a,b coprime, then we can find some number of multiple that gets close but not 0. (special case for Bezout's Lemma)
- gcd(ma, mb) = m \times gcd(a,b)

### Least Common Multiple

lcm[a,b]

- only multiples of lcm[a,b] are common multiple

### Corollary to LCM

## Number Theory on Friday

### Euclid's Lemma (coprime)

gcd(a,b)=1 \land a|bc \implies a|c
TODO: proof

When a is prime: p|bc \implies p|b \lor p|c

### Fundamental Theorem of Arithmetic (FTA)

Every integer n > 1 can be written as a product of primes in a unique way.

### The Infinitude of Primes

The next prime is a prime for P = (\prod_{i=1}^n p_i)+1

### Divisors

ord_p(n) = max\{a\in \mathbb{N} | p^a|n\}

- order(p, n) is how many times n can be divide by prime p.

so n can be expressed as
n=p_1^{ord_{p_1}(n)}p_2^{ord_{p_2}(n)}...

so a > b means that

- the power of b's factorization is smaller equal to the power of a's factorization. ((\forall p \in \mathbb{P})(ord_p(a) \leq ord_p(n)))
- we can calculate how many possible divisor of a and b

### GCD-LCM Theorem

We can express
a = \prod_{i=1}^n p_q^{a_i}
b = \prod_{i=1}^n p_q^{b_i}
Then
gcd(a,b) = \prod_{i=1}^n p_i^{min\{a_i, b_i\}}
lcm(a,b) = \prod_{i=1}^n p_i^{max\{a_i, b_i\}}
TODO: proof

### Corollary to GCD-LCM Theorem

ab = gcd(a,b) \times lcm[a,b]

## Additional Things

gcd(a,b)|c iff a(kx)+b(ky)=c (ax+by=gcd(a,b))
d|a, d|b -> d|gcd(a,b)