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Primes
Definition 1.1 (Prime) An integer n > 1 is called a prime if it’s only divisors
are 1 and n
There are a lot of ways of proving that there are infinitely many primes. The
most well known is Euclid’s proof, but I will give a slightly more elegant version
of the proof which was created by the German mathematician Ernst Kummer.
Theorem 1.1 There are infinitely many primes
Proof : We assume there are only finitely many primes. p1 , p2 , . . . , pr . Let
N = p1 · p2 · · · pr . For some pi , 1 ≤ i ≤ r: pi |N − 1. Now we have pi |N ,
pi |N − 1 ⇒ pi |N − (N − 1) = 1 which is a contradiction.
Another way of proving the existence of an infinite amount of primes is by
proving that the sum of the reciprocals of primes diverges. This was proven by
James A. Clarkson.
Theorem 1.2
X
The series
p:p
is prime
1
p
diverges
Proof : (contradiction) We assume that the series converges.
If the series converges, we can choose k such that
∞
X
m=k+1
1
1
<
pm
2
Now let Q = p1 · p2 · · · pk , and consider the number 1 + nQ, for n = 1, 2, · · ·.
Now we know that 1 + nQ is not divisible by any of the primes, p1 , p2 . . . pk ,
since pi |Q ⇒ pi |nQ ∀n = 1, · · · , k ⇒ pi - 1 + nQ. So the prime factors of 1 + nQ
must occur among the primes pk+1 , pk+2 , · · ·. Now we can conclude that:
!t
r
∞
∞
X
X
X
1
1
≤
1 + nQ
p
n=1
t=1 m=1 m+k
The above inequality can be proven formally, but you can just observe that all
the terms on the left side are included in the right side! But because of the way
t
P∞
we chose k, the right side of the inequality is dominated by t=1 12 which is
a convergent geometric
We conclude that under the assumption we made
P∞ series.
1
at the beginning n=1 1+nQ
is a convergent series, but this is a contradiction,
since
∞
∞
∞
X
X
X
1
1
1
≥
= (1 + Q)
1 + nQ n=1 n(1 + Q)
n
n=1
n=1
P∞ 1
and we know that n=1 n is a divergent series.
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Twin Primes
Definition 2.1 (Twin primes) If integers n and n + 2 are prime, n and n + 2
are called twin primes.
The twin prime conjecture states that there are infinitely many twin primes!
Now, if we could prove that the series
X
p:p
is twin prime
1
p
diverges, we would have proven that there are infinitely many twin primes!
However Viggo Brun proved that
X
p:p
is twin prime
1
p
is in fact convergent. Of course this does not contradict the theory that there
are infinitely many twin primes! Now I will state a couple of interesting results
on twin primes.
Theorem 2.1 Let π2 (x) = |{p ≤ x : p is twin prime}| denote the number of
twin primes less than or equal to x.
We have:
∃c ∈ R such that π2 (x) < c
x(log(log(x)))2
(log(x))2
∀x ∈ N
This implies Brun’s theorem:
Theorem 2.2 (Brun’s theorem)
X
p:p
3
is twin prime
1
<∞
p
The Twin Prime Conjecture
Now we will look at the asymptotic distribution of primes. Let π(x) = |{p ≤
x : p is prime}| denote the number of twin primes less than or equal to x. The
prime number theorem states the following:
Theorem 3.1 (Prime number theorem)
lim x→∞
π(x)
x
log(x)
2
=1
Or in other words, the distribution of the number of primes less than or equal
x
to x is approximately log(x)
. We denote this as π(x) and write:
π(x) ≈
x
log(x)
If we look at this result in terms of probability, for a random number between
1 and x we have
1
P (x is prime) =
log(x)
So if for any random number the probability that n is prime is independent of
the probability that n + 2 is prime, we would have
P (n is prime, n + 2 is prime) = P (n is prime)P (n + 2 is prime) =
1
1
1
≈
log(n)
log(n + 2)
log(n)2
So if we use the prime number theorem, and we assume that n + 2 being prime
is independent of whether or not n is prime, we will expect that
1
π2 (x) ≈
log(x)2
However we can not assume that the distribution of primes is random, because
obviously if for any number n > 2, if n is prime, then it is also odd and therefor
n + 2 is also odd. Since n + 2, for n > 2 has to be odd to be prime, the probability that n + 2 is prime under the assumption that n is prime would be twice
is big as the probability that a random number is prime.
We can approach the problem in a different way which gives a more accurate
result. Obviously a prime, p, is an integer which is not divisible by any of the
primes q : q < p. The probability that two random numbers are not divisible
2
by p > 2 is 1 − p1 . Now we look at the probability that n and n + 2 are not
divisible by some p < n. We have n ≡ k(mod p) for some k ∈ {0, · · · , p − 1}.
Now neither n or n + 2 are divisibleby p ifand only if k 6= 0 and k 6= −2, and
the probability for this is obviously 1 − p2 .
We will now correct our estimation of π2 (x) by multiplying the result by
P (n + 2 is prime|n is prime)
P (n + 2 is prime|n is random)
We calculate:
P (n + 2 is prime|n is prime)
=
P (n + 2 is prime|n is random)
3
2
1−
p
−2
1
1−
=
p
1
1−
p
2
1
− 2
p
!
1
1−
p
−2
=
1−
!
1
(p − 1)
2
Now we have a better estimate of π2 , and we are able to write down the asymptotic result called the twin prime conjecture.
Theorem 3.2 (The twin prime conjecture)
!
Y
x
1
π2 (x) ≈
1−
2
(log(x))2
(p − 1)
p>2
Obviously since
π2 (x) → ∞ for x → ∞
this implies that the are infinitely many twin primes.
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