Theorem 2.6 There are infinitely many primes
Proof. The proof is done via the contradiction method.
Supposed only finitely primes, say 𝑝1 , 𝑝2 , ⋯ , 𝑝𝑘 . Let m be
defined by
𝑚 = 𝑝1 𝑝2 ⋯ 𝑝𝑘 + 1.
Since 𝑚 is an integer > 1 then FTA implies that m is
divisible by some prime p. By assumption the prime p
must be one of those primes 𝑝1 , 𝑝2 , ⋯ , 𝑝𝑘 . Thus,
p divides 𝑝1 𝑝2 ⋯ 𝑝𝑘 since p is one of them. Conseq.,
p also divides 𝑚 − 𝑝1 𝑝2 ⋯ 𝑝𝑘 = 1 which is impossible.
We deduce that assumption was false, so that there
must be infinite primes.
Let 𝑝𝑛 denote the 𝑛𝑡ℎ prime, so that 𝑝1 = 2, 𝑝2 =
3, 𝑝5 = 11, and so on.
2𝑛−1
Corollary 2.7. For all 𝑛 ≥ 1, 𝑝𝑛 ≤ 2
.
This estimation is too weak, for instance 𝑝4 = 7 but
24−1
2
= 28 = 256 → 7 << 256.
Proof: see on the textbook.
For any positive number 𝑥, let 𝜋(𝑥) denote the number
of primes 𝑝 ≤ 𝑥, for example 𝜋(10) = 4, 𝜋(100) = 25.
Corrolary 2.8. 𝜋(𝑥 ) ≥ ⌊log 2 (log 2 𝑥)⌋ + 1.
2𝑛−1
Proof. 𝑛 = ⌊log 2 (log 2 𝑥)⌋ + 1 ↔ 2
≤ 𝑥. Then, by
Corollary 2.7 there are at least n primes 𝑝1 , 𝑝2 , ⋯ , 𝑝𝑛 ≤
2
2𝑛−1
≤ 𝑥 → 𝜋(𝑥 ) ≥ 𝑛 = ⌊log 2 (log 2 𝑥)⌋ + 1.
This estimation is also too weak, for example:
𝑥 = 109 → 𝑛 = 5.
In fact, the the number of primes 𝑝 ≤ 109 is
approximately 5 × 107 much numerous than 5.
GAUSS CONJECTURE in 1793:
𝑥
𝑑𝑡
𝑥
𝜋 (𝑥 ) ≈ ∫
↔ 𝜋(𝑥) ≈
ln 𝑥
2 ln 𝑡
In the sense that
𝜋(𝑥)
𝑥/ ln 𝑥
→ 1 as 𝑥 → ∞. This conjecture
finally proved by Hadamark and de la Vallee Poussin in
1896.
Theorem 2.9. There are infinitely many primes of the
form 4𝑞 + 3.
Example: 7, 11, 19, 23, . . . all of has above type.
Theorem 2.10. If 𝑎 and 𝑏 are co-prime then there are
infinitely many primes of the form 𝑎𝑞 + 𝑏.
This is a general theorem proved by Dirichlet in 1837.
Theorem 2.9 correspond to 𝑎 = 4 and 𝑏 = 3.
Goldbach’s Conjecture: every even integer 𝑛 ≥ 4 is the
sum of two primes, e.g. 4=2+2, 6=3+3, 8=3+5, 12=5+7...
FERMAT AND MERSENNE PRIMES
Consider the integers of the form 2𝑚 ± 1; many small
primes 3, 5, 7, 17, 31, ... have this form.
Lemma 2.11. 2𝑚 + 1 is prime → 𝑚 = 2𝑛 for some 𝑛 ∈ 𝑍
Examples: 3 = 21 + 1 → 1 = 20 , 5 = 22 + 1 → 2 = 21 ,
17 = 24 + 1 → 4 = 22 , 31 = 25 − 1 ≠ 2𝑚 + 1 →? ? ?
Outline of proof. We prove the contraposition, that
𝑚 ≠ 2𝑛 → 2𝑚 + 1 is not a prime. Since 𝑚 ≠ 2𝑛 then it
must be hold 𝑚 = 2𝑛 𝑞 for some odd 𝑞 > 1. Then,
consider the polynomial 𝑃(𝑡) ≔ 𝑡 𝑞 + 1 has root 𝑡 = −1.
2𝑛
Then, show that 2 + 1 is a proper factor of integer
2𝑚 + 1, which can not be prime.
DEFINITION:
2𝑛
1. Numbers of the form 𝐹𝑛 = 2 + 1 are called
Fermat numbers, and those which are primes are
called Fermat primes.
Examples: 𝑛 = 0, 1, 2, 3, 4 → 𝐹𝑛 = 3, 5, 17, 257, 65537
are prime, but Euler in 1732 that
𝐹5 = 232 = 4294967297 = 641 × 6700417 is composite.
No more Fermat prime have been found.
2. Integers of the form 2𝑝 − 1, where 𝑝 prime, are
called Mersenne numbers (in 1644); those which
are primes are called Mersenne primes.
Examples: for primes 𝑝 = 2, 3, 5, 7 the Mersenne
numbers 𝑀𝑝 = 3, 7, 31, 127 are primes, but 𝑀11 =
2047 = 23 × 89 is composite.
Primality-testing and factorization
(1) How do we decide that a given integer (big one) is
prime?
(2) How do we find the prime-power factorization of a
given large integer.
More specific,Is 8703585473 divisible by 3? Is it divisible
by 11?
Lemma 2.14. An integer 𝑛 > 1 is composite ↔ it is
divisible by some primes 𝑝 ≤ √𝑛.
Proof. (←) If 𝑛 is divisible by such 𝑝 where 1 < 𝑝 ≤
√𝑛 < 𝑛 it follows that 𝑛 is composite.
(→) 𝑛 composite ↔ 𝑛 = 𝑎𝑏, 1 < 𝑎 < 𝑛, 1 < 𝑏 < 𝑛; at
least one of 𝑎 and 𝑏 is less than or equal to √𝑛, (since if
not then 𝑎𝑏 > 𝑛) and this factor will be divisible by a
prime 𝑝 ≤ √𝑛, which then divides 𝑛.
Eratosthenes Sieve is a systematic way of compiling a list
of all the primes up to a given integer N.
1. List the integers 2, 3, . . . , N.
2. Underline or encircle 2 (which is prime) and cross
out all the proper multiples 4, 6, 8, . . . of 2.
3. The first integer which is neither encircle nor cross
out is 3; this is prime. We encircle 3 and cross out
all the proper multiples 6, 9, 12, . . . of 3.
4. Encircle 5 and cross out it multiples.
5. We continue like this till every number in the list is
either encircled or crossed out.
All encircled numbers are primes.
Remark: the process can be stopped when the proper
multiples of all the primes 𝑝 ≤ √𝑁 have been crossed
out.
Decimal form of Integer:
𝑁 = 𝑎𝑘 𝑎𝑘−1 ⋯ 𝑎1 𝑎0
≔ 𝑎𝑘 10𝑘 + 𝑎𝑘−1 10𝑘−1 + 𝑎1 10 + 𝑎0 .
where 𝑎𝑖 are integers in {0, 1, 2, ⋯ , 9} and 𝑎𝑘 ≠ 0.
1. N is divisible by 2 ↔ the last digit 𝑎0 is even.
2. N is divisible by 3 ↔ the sum of all the digits is
divisible by 3.
3. N is divisible by 4 ↔ the number formed by last two
digits is divisible by 4.
4. N is divisible by 5 ↔ the last digit is either 0 or 5.
5. N is divisible by 6 ↔ the sum of all digits is divisible
by 3 and the last digit is even.
6. N is divisible by 7 ↔ M is divisible by 7 where M is
the number derived from N by subtracting twice the
last digit.
7. N is divisible by 8 ↔ the numbers formed by three
last numbers is divisible by 8.
8. N is divisible by 9 ↔ the sum of all the digits is
divisible by 9.
9. N is divisible by 10 ↔ the last digit is zero.
10.N is divisible by 11 ↔ the difference between the
sum of digits on the odd places and the sum of
digits on the even places is divisible by 11.
11.N is divisible by 12 ↔ the number formed by the
last two digits is divisible by 4 and the sum of all
digits is divisible by 3.
12.N is divisible by 25 ↔ the number formed by the
last two digits is divisible by 25.
13.N is divisible by 125 ↔ the number formed by the
last three digits is divisible by 125.