Download Primes. - Elad Aigner

The prime numbers: fundamentals
Elad Aigner-Horev
§1. The fundamental theorem of arithmetics
Definition 1.1. A prime is a positive integer greater than 1 divisible by no positive integer other than
1 and itself.
Positive integers that are not prime are called composite. That is, a positive integer n is composite if
n = a · b for some 2 ≤ a, b < n. The following theorem asserts that the primes are the building blocks of
number theory.
Theorem 1.2. (The fundamental theorem of arithmetics)
Every integer greater than 1 can be uniquely written as a product of primes up to the order of the primes
involved.
Corollary 1.3. Any integer n > 1 can be written uniquely in a canonical form
n = pa11 · pa22 · · · pakk ,
where for each i ∈ [k] we have that ai > 0, each pi is prime, and p1 < p2 < · · · < pk .
The act of decomposing a number to prime numbers in such a way is called factorisation. We distinguish
between two types of factorisations.
Example 1.4.
240 = 2| · 2 · 2{z
· 2 · 3 · 5} =
prime factorisation
2| 4 ·{z3 · 5}
,
289 = 17 · 17 = 172 ,
1001 = 7 · 11 · 13.
prime−powerf actorisation
Definition 1.5. The distinct primes appearing in the prime-power factorisation of a number are called
its prime factors.
Prior to proving Theorem 1.2 let us prove something significantly more naive in order to be convinced
that primes and in fact Theorem 1.2 are of interest.
Lemma 1.6. Every positive integer greater than 1 has a prime divisor.
Proof. Assume towards a contradiction that the claim is false. Then there are positive integers that form
a counter example to this claim. Let n be the least counter example to the claim (we appeal here to the
WOP). We may assume that n is not a prime as then n is not a counter example. Hence, n is composite
and we may write n = ab for some two integers 2 ≤ a, b < n. By the choice of n the number a is not a
counter example to the claim. Hence, a has a prime divisor; and consequently so does n; contradiction. Actually, we can do much better than Lemma 1.6.
Theorem 1.7. If n is composite then it has a prime factor not exceeding
1
√
n.
Number Theory ArielU 2016
Elad Aigner-Horev
Proof. We may write n = ab where 1 < a ≤ b < n. We may also assume that a ≤
ab > n. By Lemma 1.6 a has a prime divisor and the claim follows.
√
n for otherwise
We commence our preparation towards a proof of Theorem 1.2.
Lemma 1.8. If a, b, c are integers such that (a, b) = 1 and a | bc then a | c.
Proof. There exist integers x and y such that 1 = (a, b) = ax + by. Then
c = c · 1 = c(ax + by) = acx + bcy.
Surely a | acx and by assumption we have that a | bcy. Consequently a | c.
An immediate corollary of Lemma 1.8 is the following.
Corollary 1.9. If p is a prime and p | ab then p | a or p | b.
Proof. If p | a we are done, so assume that p 6 |a implying that (p, a) = 1 (indeed, as p is prime
(p, a) ∈ {1, p}; having p 6 |a implies that (p, a) = 1). The claim then follows by Lemma 1.8.
Note that Corollary 1.9 is unique to prime numbers. For instance 4 · 6 = 24 and 12 | 4 · 6. However,
12 6 |4 and 12 6 |6. Corollary 1.9 extends to products of more than two integers.
Corollary 1.10. If a prime p satisfies p | a1 · · · an , where ai is a positive integer for every i ∈ [n], then
there exists a j ∈ [n] such that p | aj .
Proof. The proof is by induction on n. For n = 1 the claim is trivial (Corollary 1.9 handles n = 2) .
Suppose then that the claim is true for n ≥ 1 and consider a product of n + 1 numbers a1 · · · an an+1 that
is divisible by p. Either p | a1 · a2 · · · an or p 6 |a1 · a2 · · · an . In the former case the claim follows by the
induction hypothesis. In the latter case p | an+1 by Corollary 1.9.
Taking all of the ai , i ∈ [n], to be prime in Corollary 1.10 yields the following.
Corollary 1.11. Let p, q1 , . . . , qn be prime. If p | q1 · q2 · · · qn , then there exists a k ∈ [n] such that
p = qk .
We are now in a position to prove Theorem 1.2.
Proof of Theorem 1.2. The argument consists of two parts. First we will show that for every positive
integer there exists at least one way to express the number as a product of primes. Second, we will show
that this expression is in fact unique.
Existence. Assume towards a contradiction that the claim is false and that there are positive integers
that cannot be written as a product of primes. Let n be the least1 integer amongst these counter examples
(which are all positive integers). We now analyse this counter example n. Surely, n is not a prime or
n is not a counter example to the statement. We may assume that n is composite then and thus may
write n = ab for some integers 2 ≤ a, b < n. Each of a and b do not constitute a counter example to
the statement, by the choice of n. Hence, both a and b can be expressed as a product of primes and
consequently n as well; a contradiction.
Uniqueness. Suppose that there exists an integer n for which two factorisations exist; that is
n = p1 · · · ps = q1 · · · q` .
Amongst all numbers with at least two representations choose n such that its two representations share
no common elements (put another way, reduce common primes in the original representations). Then
1
Here we appeal to the WOP.
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Number Theory ArielU 2016
Elad Aigner-Horev
p1 | q1 · · · q` (as p1 · (p2 · · · ps ) = q1 · · · q` ), and consequently there exists a j ∈ [`] such that p1 = qj , by
Corollary 1.11; contradiction.
In the proof of Existence we used the proof technique of minimal counter example. We mentioned
before that this is just another way to do induction. Let us translate this proof to the induction framework.
Proposition 1.12. Every integer n > 1 is either prime or can be written as a product of primes.
Proof. The proof is by induction on n. The claim is true for n = 2. Assume the claim holds true for all
integers < n. If n is prime we are done. Suppose then that n is composite, that is n = ab for some two
integers 2 ≤ a, b < n. By the induction hypothesis both a and b are either primes or can be written as a
product of primes. The claim follows.
Next we give an alternative proof of Theorem 1.2 using induction as follows.
Proof of Theorem 1.2 (Alternative proof). The proof is by induction on n. For n = 2 the claim holds.
Assume then that the claim is true for all integers 1 < k < n. We may assume that n is composite and
that n can be written as a product of primes, by Proposition 1.12. It remains to prove the uniqueness of
the factorisation for n. Assume towards a contradiction that n has two distinct prime factorisations, that
is
n = p1 · · · ps = q1 · · · qt .
We show that s = t and that these two factorisations are comprised of the same primes appearing the
same number of times.
As p1 · (p2 · · · ps ) = q1 · · · qt we have that p1 | q1 · · · qt . Then, by Corollary 1.10, there exists a qj
satisfying p1 = qj . Without loss of generality we may assume that j = 1. Cancelling p1 = q1 on both sides
we obtain
n
= p 2 · · · p s = q2 · · · qt
p1
As n is composite we have that s > 1 (and t > 1 as well) and thus 1 < n/p1 < n. The induction hypothesis
for n/p1 asserts that this number has a unique prime factorisation. Therefore s = t and the factorisation
of n is unique.
Message: d | n if and only if:
1. the prime factors of d are amongst the prime factors of n.
2. Every prime factor of d that is also a prime factor of n has its power in the factorisation of d bounded
by its power in the factorisation of n.
Example 1.13. List all divisors of 120. We start of by finding the prime factorisation of 120. Here
we have that 120 = 23 · 3 · 5. The number 1 we list automatically. To produce all the others we go over
all options to choose prime factors from this factorisation where 2 can be chosen at most 3 times in each
combination. We leave the details to the reader.
Q
Proposition 1.14. The set of divisors of n = ri=1 pai i is given by
( r
)
Y c
i
pi : 0 ≤ ci ≤ ai .
i=1
Example 1.15. We can use prime factorisation to find the gcd of numbers. For instance, what is
(720, 2100)? First let us note that 720 = 24 · 32 · 5 and that 2100 = 22 · 3 · 52 · 7. Recall the Message
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Elad Aigner-Horev
we had above. Any common divisor of 720 and 2100 will be comprised of prime factors that are common
to both these numbers. We get the gcd if we pick as many common factors as we can. Here the common
prime factors are 2, 3, 5. Then (720, 2100) = 22 · 3 · 5.
Let us make this last example more general.
Proposition 1.16. Let a and b be positive integers with factorisations
a = pa11 · · · pann and a = pb11 · · · pbnn ,
then
min{a1 ,b1 }
(a, b) = p1
· · · pnmin{an ,bn }
§2. The infinitude of primes
How many primes are there? For the infinitude of N we had to resort to the axiom of infinity asserting
that infinite sets exist. Once this was established we were able to prove the infinitude of Z and Q in a
fairly straightforward manner.
Theorem 2.1. (Euclid) There are infinitely many primes.
Proof. Assume towards a contradiction that there are finitely many primes. Let p1 , . . . , pn denote these
primes. Put
Q := p1 p2 · · · pn + 1.
By Lemma 1.6 the number Q has at least one prime divisor, namely q. We show that q is not amongst
the primes p1 , . . . , pn and thus attain a contradiction. Suppose that q = pj for some j ∈ [n]. Then
q | Q − p1 p2 · · · pn . As Q − p1 p2 · · · pn = 1 this implies that q | 1 which is impossible as q ≥ 2. It follows
that q 6= pj for every j ∈ [n]; contradiction.
Remarks.
1. The proof above uses Lemma 1.6 to deduce the existence of q. We could in fact have used Theorem 1.2 to that end. Notice that Lemma 1.6 is much more naive than Theorem 1.2.
2. The proof we have seen here discovers a new prime q. The proof however does not reveal a method
to finding q. It just shows it exists. In math we distinguish between non-constructive or existential
proofs (like the one we have here) and constructive proofs.
§3. The distribution of primes
Let pn denote the nth prime number. In the proof of Theorem 2.1 we have seen that p1 ·p2 · · · pn +1 has
at least one prime divisor that is not amongst {p1 , . . . , pn }. If there are several then pn+1 cannot exceed
the least of these. In particular, through the proof of Theorem 2.1 we have established the following.
Proposition 3.1. For every n ≥ 1
pn+1 < p1 · p2 · · · pn + 1
holds.
Proposition 3.1 forms a rather crude estimate; indeed
p5 = 11 ≤ 2 · 3 · 5 · 7 + 1 = 210.
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Number Theory ArielU 2016
Elad Aigner-Horev
A better estimate is that of Bonse.
Theorem 3.2.
p2n+1 < p1 · p2 · · · pn , whenever n ≥ 5.
Proof of Theorem 3.2 is left to the practical session. Suffices here to note that
√
√
p5 = 11 < 2 · 3 · 5 · 7 = 210 ≈ 14.491.
n−1
Theorem 3.3. pn ≤ 22
.
Proof. The proof is by induction on n. For n = 1 the claim is trivial. Let then n > 1 and assume the
claim holds true for all integers ≤ n. We prove the claim for n + 1. By Proposition 3.1 we have that
pn+1 < p1 · p2 · · · pn + 1.
Applying the induction hypothesis to each pi we arrive at
n−1
≤ 2 · 22 · · · 22
+1
1+2+22 +···+2n−1
=2
+ 1.
P
In the induction lecture we proved that ki=1 2i ≤ 2k+1 , for every k ≥ 1. In fact, one can prove (by
P
induction) that ki=1 2i ≤ 2k+1 − 1. Applying this inequality to the exponent here we arrive at
n −1
≤ 22
+ 1.
n −1
To reach the form we want let us note that 1 ≤ 22
n −1
pn+1 ≤ 22
n −1
+ 22
for all n ≥ 1; then
n −1
= 2 · 22
n
= 22
and the claim follows.
n
Corollary 3.4. For n ≥ 1 there are at least n + 1 primes less than 22 .
n
Proof. By Theorem 3.3 the primes p1 , p2 . . . , pn+1 all do not exceed 22 .
We can do much better than Theorem 3.3.
Theorem 3.5. (Bertrand’s postulate) For n ≥ 2 the interval (n, 2n) contains at least one prime.
Bertrand conjectured Theorem 3.5 in 1845. It was then proved by Tchebycheff in 1852 and much later
by Erdős who gave rise to the limerick: Tchebycheff proved it, I proved it again; there is always a prime
between 2n and n. A proof of Bertrand’s postulate is a bit beyond us right now. Instead we shall prove
the following using this postulate.
Theorem 3.6. For n ≥ 2 we have that pn < 2n ; in particular pn+1 < 2pn .
Proof. The proof is by induction on n. For n = 2 we have that p2 = 3 ≤ 22 . Assume then that the claim
holds for n, i.e., that pn < 2n . We prove the assertion for n + 1. By Bertrand’s postulate there exists
a prime p satisfying 2n < p < 2n+1 . In particular, pn < p, by the induction hypothesis. Consequently,
pn+1 < p as well and the first part of the claim follows.
For the second part, note that the interval (pn , 2pn ) contains a prime number by Bertrand’s postulate
implying that pn+1 < 2pn .
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Number Theory ArielU 2016
Elad Aigner-Horev
Definition 3.7. For a positive integer n let π(n) denote the number of primes not exceeding n.
Theorem 3.8. (The prime number theorem)
lim
n→∞
π(n)
n
log n
= 1.
Corollary 3.9. pn ≈ n log n.
§4. Primes in arithmetic progressions
We have seen that there are infinitely many primes of which only 2 is even the rest are clearly odd.
The odd integers are of the form 4n + 1 or 4n + 3. The infinitude of primes does not imply directly that
we have infinitely many primes of the form 4n + 1 and of the form 4n + 3; it only implies that at least one
of these forms sees infinitely many primes. Could it be for instance that there are infinitely many primes
of the form 4n + 1 and only finitely many of the form 4n + 3? Perhaps the other way around? In this
section we answer these questions. We start with an observation.
Observation 4.1. The product of two numbers of the form 4k + 1 is also of the form 4k + 1.
Proof.
(4n + 1)(4m + 1) = 16mn + 4n + 4m + 1
= 4 · 4mn + 4(m + n) + 1
= 4 · 4 · (mn + m + n) +1
|
{z
}
Integer
Proposition 4.2. There are infinitely many primes of the form 4n + 3.
Prior to proving Proposition 4.2 let us consider the following failed attempt to prove this proposition.
Assume towards a contradiction that there are only finitely many primes of the form 4n + 3 and let p be
the largest such prime. Put
N := 4 · (3 · 7 · · · p) + 3,
where 3 · 7 · · · p contains all primes of the form 4n + 3 (which we now assume a finite number of them).
The advantage of such a definition is that N cannot be prime as it has the form 4n + 3 (and we have used
all other such primes to define N in particular N > p). By Lemma 1.6 N has a prime divisor q. We wish
to say that q is of the form 4n + 3 in order to replicate our argument from Theorem 2.1 in that q is not
amongst the list of primes of this form that we assumed is finite.
Other than the prime 2 all primes are odd and consequently of the form 4n + 1 or of the form 4n + 3.
As N is odd 2 6 |N . In addition, it cannot be that all prime factors of N are of the form 4n + 1, by
Observation 4.1. It follows then that N has a prime factor q of the form 4n + 3. At this point we would
have liked to reach a contradiction but we fail. Indeed, q could be equal to 3. This attempt failed but we
are not far.
Let us try yet again to prove Proposition 4.2. Assume towards a contradiction that there are only
finitely many primes of the form 4n + 3. Let S denote the (finite) set of such primes and let p be the
largest prime in S. Put
N := 4 · (3 · 7 · · · p − 1) + 3,
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Number Theory ArielU 2016
Elad Aigner-Horev
where 3 · 7 · · · p contains all primes of the form 4n + 3. As in the previous attempt N cannot be prime and
it must have a prime divisor of the form 4n + 3. We wish to argue that q ∈
/ S but it seems we are stuck; in
particular the "trick" we used in Euclid’s result (i.e., Theorem 2.1) does not seem applicable here. So at
this point it is not clear how to continue. This does not mean that this approach cannot lead to a proof
it just means that our choice of N makes it difficult to see how to end it gracefully.
At this point we make the following observation
{4n − 1 : n ∈ Z} = {4n + 3 : n ∈ Z}.
So instead of proving Proposition 4.2 we could instead prove the following.
Proposition 4.3. There are infinitely many primes of the form 4n − 1.
Proof. Assume towards a contradiction that there are only finitely many primes of the form 4n − 1. Let
S denote the (finite) set of such primes and let p be the largest prime in S. Put
N := 4 · (3 · 7 · · · p) − 1,
where 3 · 7 · · · p contains all primes of the form 4n − 1. Note that N cannot be prime as it has the form
4n − 1. It cannot be that all of its prime factors are of the form 4n + 1, by Observation 4.1, and as N is
odd we have that 2 6 |N . It follows then that N has a prime factor q of the form 4n − 1. If q ∈ S then
q | 3 · 7 · · · p implying that q | N − 4 · (3 · 7 · · · p). As N − 4 · (3 · 7 · · · p) = 1 and q ≥ 3 (recall that q is odd)
this is impossible; contradiction.
Returning to our original question, we are still left with the problem of determining whether there
are infinitely many primes of the form 4n + 1. While the proof of Proposition 4.2 was fairly easy, we are
currently not equipped to prove the same assertion of 4n + 1.
Proposition 4.4. There are infinitely many primes of the form 4n + 1.
It was Dirichlet that proved the following general result. Proof of which is beyond the scope of our
course.
Theorem 4.5. (Dirichlet’s theorem) If a and b are co-prime integers then the arithmetic progression
an + b contains infinitely many primes.
§4.1 Progressions in the primes. In the previous section we considered the presence of prime
numbers along arithmetic progressions. The following result by Green and Tao establishes something even
more astonishing.
Theorem 4.6. (Green-Tao theorem, abridged) The primes contain arbitrarily long arithmetic progressions.
Let us make this last theorem precise. For n ∈ Z+ , let Pn denote the first n prime numbers.
Theorem 4.7. (Green-Tao theorem, abridged) For every integer k ≥ 3 there exists an n0 = n0 (k) such
that for all n ≥ n0 the set Pn contains a k-term arithmetic progression.
Theorem 4.7 is best possible in the sense that there cannot be an infinite arithmetic progression in the
primes. For if there were, that would imply that the gap between every two consecutive primes is bounded
by some constant. This is easily negated.
Theorem 4.8. The gap between two consecutive primes can be arbitrarily large.
Proof. We show that for every n ∈ Z+ there exist two consecutive primes separated by at least n − 1
consecutive composite numbers. Indeed, given n the numbers
n! + 2, n! + 3, . . . , n! + n
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is such an interval of composite numbers.
§5. Conjectures regrading the primes
Definition 5.1. Two prime numbers that differ by two are called twin primes.
For instance, the pairs (3, 5), (5, 7), (11, 13), (101, 103), and (4967, 4969) are all twin primes.
Conjecture 5.2. (The twin prime conjecture) There are infinitely many twin primes.
Consider the following examples
10 = 3 + 7 = 5 + 5
24 = 5 + 19 = 7 + 17 = 11 + 13
100 = 3 + 97 = 11 + 89 = 17 + 83
= 29 + 71 = 41 + 59 = 47 + 53.
Conjecture 5.3. (The Goldbach conjecture) Every positive even number can be written as a sum
of two odd primes.
Conjecture 5.4. (The n2 + 1 conjecture) There are infinitely many primes of the form n2 + 1.
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