Download Elementary Number Theory Fall 2014 Lecture 3: Prime numbers and

Elementary Number Theory
Fall 2014
Lecture 3: Prime numbers and unique factorization
Definition: A natural number p is prime if it is only divisible by 1 and p itself.
Theorem: Every natural number is either a prime number or a product of prime numbers.
Proof: n = 2, 3 are prime numbers and n = 4 = 2 × 2 is a product of prime numbers.
Suppose that the statement is true for all natural numbers less than equal to n. Consider
n + 1. If it is not a prime number then ∃d > 1 such that d|n + 1 ⇒ n + 1 = cd. Both c, d
are less than n + 1 and are therefore primes or product of primes. Thus n + 1 is a product
of primes.
Theorem: There are infinitely many prime numbers.
Proof: Suppose that there are only a finite number of primes, p1 , p2 , p3 , · · · pr . Then
consider
N = p1 p2 · · · pr + 1.
(1)
N is not divisible by any of the pi , pi - N i = 1, 2, 3, · · · r. But by previous theorem N is either
a prime or a product of primes. If N is a prime then N is not in the above list of primes
hence given r primes we generated r + 1 primes. But if N is not a prime it is product of
primes and therefore there exists primes other than the ones in the list given above. Hence
our assumption that there are only r primes is wrong and there are infinitely many primes.
Theorem: (Fundamental theorem of arithmetic) Every natural number greater than one
can be represented in a unique way as a product of prime numbers.
Proof: Let n ∈ N and n > 1. The statement is certainly true for n = 2, 3, 4 so suppose
the statement is true for all n ≤ k and consider n = k + 1. If k + 1 is a prime number we
are done. So suppose that k + 1 is not a prime number then since a composite number can
be written as product of primes therefore
k + 1 = p1 p2 · · · ps
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Elementary Number Theory
Fall 2014
where p1,2,··· ,s are prime numbers. The above equation implies
k+1
= p 2 · · · ps
p1
Since
k+1
p1
< k + 1 therefore
k+1
p1
has a unique representation as product of primes therefore
k + 1 = p1 k+1
has a unique representation as product of primes.
p1
If n ∈ Z, then n = ±
Q
i
pai i . ai ≥ 0 and only a finite number of ai are non-zero.
i
pai i then the set of divisors of n is given by {
i
pαi i , b =
If n =
Q
If a =
Q
Q
i
pβi i then gcd(a, b) =
Q
min{αi ,βi }
i
pi
Q
i
pci i | ai ≥ ci ≥ 0}.
.
If p is an odd prime then either p = 4k + 1 or p = 4k − 1 for some k ∈ N.
Theorem: There are infinitely many prime of the form 4k − 1. (In the set {4k − 1 | k ∈ N}
there are infinitely many primes but not every number is prime.)
Proof: Suppose that there are only finitely many primes, {p1 , · · · pr } of the form 4k − 1.
Then consider the number
N = 4p1 p2 · · · pr − 1.
(2)
It is clear that pi - N for i = 1, 2, · · · , r. But since every natural number can be written as
product of primes therefore there exist primes p̃1,2,··· ,m such that N = p̃1 · · · p̃m . Now if all
of p̃i are of the form 4k + 1 then N must itself be of the form 4A + 1 for some A ∈ N. But
since N is not of this form therefore one of the primes, say, p̃1 is of the form 4x − 1. And
since p̃1 is not one of the primes p1,··· ,r therefore we get a contradiction that there are only
r primes of the form 4x − 1. Thus there must of infinitely many primes of the form 4x − 1.
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