1. From the last time:
gcd(a, b) can be characterized in two different ways:
It is the least positive value of ax + by where x and y range over
integers.
It is the positive common divisor of a and b which is divisible by
every common divisor.

2. Consider now positive integers Z+ = {1, 2, 3, …}.
Any positive integer n >1 has at least two dividers, 1 and n .
An integer p >1, that does not have any other dividers except
1 and itself, is called a prime.
An integer n >1, that is not a prime, is composite.
Theorem. Any composite integer n Z+ has a prime factor.
Proof by contradiction. Assume there exists some positive integer,
that has no prime factors. Then the set of such integers S  and
we can find the smallest element n S  Z+.
Since n is composite, n = k  m, with 1< k, m < n,
so k, m S , so they are either primes or have prime factors.
In either case n has a prime factor.

3. The first primes: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, …
Does this sequence has an end?
This question is not as trivial as it seems!
Theorem (Euclid) There are infinitely many primes.
Proof. Suppose there were only a finite number of primes p1, p2, …pk. Then form a number n = p1 p2  …  pk +1=2357…  pk +1.
n is not divisible by 2, for then both n and 2357…  pk would be
divisible by 2, and therefore their difference would be divisible
by 2.
This difference is 1, and is not divisible by 2.
In the same way, n is not divisible by 3 or by 5 or … or pk.
But n is either a prime or has a prime factor.
In any case it is divisible by some prime p that is not among
the list 2, 3, 5,…pk. It implies that there is a prime distinct
from 2, 3, 5…pk, and so greater then pk.
Consequently, the list of primes can never end.

4. Fundamental Theorem of Arithmetic.
Any integer n > 1 can be written as a product of prime numbers.
Further, this product is unique except for rearrangement of factors.
For example, take number 666.
It is not a prime, because it has a factor 2, so we get 666=2333.
Now 333 has an obvious factor 3, so 333=2111.
Again 111 has a factor 3, and 111=337, hence:
666=23337 is a representation of the composite number 666
as a product of primes.
Other examples: 12=223=223; 120 = 22235=23 3 5;
But is there any another representation of 666 as a product of
primes (we don’t distinguish different orders of factors)?

5. Proof that a prime factorization exists for any integer n >1.
Prove by strong induction on n >1.
Basis. n =2 is prime itself, so the proposition is true.
Inductive Hypothesis. Assume that for some k >1 there exists
prime factorization for all integers 1<nk.
Inductive Step Consider n=k+1.
We can have two cases: either n is a prime, or n is composite.
In the first case we have nothing to prove.
In the second case n = m1 m2 and 1< m1 , m2 <n.
By IH both m1 , m2 have prime factorization, so n has a prime
factorization as well.

6. Lemma 1. If a prime p divides the product of two numbers,
p | ab, it must divide at least one of them.
Proof. Assume p | ab to prove that p | a or p | b.
If p | a we are done, so consider the case p | a
What can be implied about gcd(p, a)?
gcd(p, a)=1
Then the only common factor of p and a is 1.
It implies that there exist integers x0 and y0 such that p x0 +ay0=1
Then b =b(px0 +ay0) = p (b x0)+(ba) y0 is divisible by p
because both p(b x0) and (ba) y0 are divisible by p.
Suppose now that some number c divides the product ab, c | ab. Can we imply that c divides either a or b ?

7. Proof of the uniqueness of the prime factorization
Prove it by contradiction. For this assume that there exists some
integer that has non-unique prime factorization.
By Well-Ordering Principle we can find the smallest such integer,
let it be n. So we have
n =p1 p2…pk = q1 q2… qs ,
where all pi, qj are primes.

8. Note that p1 divides q1(q2… qs), so it either divides q1 or (q2… qs).
If p1| q1 then p1= q1 , both are primes.
If p1| (q2… qs), we repeat the argument, and ultimately reach
the conclusion, that p1 equals one of the primes q1, q2,… qs .
Then we can cancel the common prime from the two
representations and find another integer n/p1 <n that has
non-unique prime factorization in contradiction with assumption,
that n is the smallest one.

9. Now we can find another form for gcd(a, b).
Suppose that the prime factorizations of the integers a and b are:
(ai, bi 0 )
where all primes occurring in either factorization are included in
both factorizations with zero exponent if necessary.
Then
This integer does divide both a and b.
No larger integer can divide both a and b.
The least common multiple of two integers:

10. Lemma. For any two integers x and y we have:
x + y = max(x, y)+min(x, y)
Theorem. For any two integers a and b
ab= gcd (a, b) lcm(a, b)

11. Example. Find gcd(120, 500) using prime factorization.
We have 120=23 3 5 and 500=22 53 , then
gcd(120, 500)= 2min(3, 2) 3min(1, 0) 5min(1, 3) =22 30 51=20.
The same value can be found from Euclid Algorithm as follows:
500 = 1204 + 20
120 =206
So, the last nonzero remainder is 20.
What is the lcm(120, 500)?
120500 = 60,000 = gcd(120, 500) lcm(120, 500)
=20 lcm(120, 500)
lcm(120,500)= 3000

12. Given an integer n how can we decide is it a prime or not?
How many factors we need to check?
Obviously we don’t need to check factors above n.
But there exists better restriction.
Theorem. If n is a composite integer, it has a prime factor less
than or equal to
Proof. If n is composite, it has a factor a with 0<a<n.
Hence, n = ab, where both a and b are positive integers greater
then 1. So, either a or b , since otherwise ab >
Hence, n has a divisor less or equal It may be either prime
or composite, but in any case n has a prime factor less or equal .
The contrapositive of this theorem: If an integer n does not have
a prime factor less or equal to , then n is prime.

13. Sieve of Eratosthenes
We need to consider only primes less or equal 10: 2, 3, 5, 7

14. Example. Show that 101 is a prime.
The only primes not exceeding are : 2, 3, 5, 7.
So we check that 101 is not divisible by 2, not divisible by 3,
by 5 and by 7. So, 101 is prime.