Brun's Theorem on Twin Primes. Mohamed Laradji. 1. Introduction. Primes have long been known to be infinite in number, for which there are several proofs, but ...
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Brun’s Theorem on Twin Primes Mohamed Laradji
1. Introduction Primes have long been known to be infinite in number, for which there are several proofs, but it is not yet known whether twin prime pairs, a pair of primes that differ by 2, are finite or infinite. However, Brun has proved that the sum of the reciprocals of twin primes actually converges. This, of course, does not answer the question about their number, but it does lend some support to the sum being finite. In any case, the convergence of the sum shows the relative scarcity of twin primes within primes. Brun’s theorem on twin primes, in fact, gives an upper bound on the number of twin prime pairs less than or equal to a certain number. Using another result, an immediate corollary of the theorem is the convergence of the sum. In this paper, a proof for Brun’s theorem on twin primes is detailed. In Section 2, the big ’O’ notation is introduced, along with several general results that will be used in the proof of the theorem. Section 3 is divided into two parts. The first part details the proof for the upper bound on the number of twin prime pairs, using a sieve method. The second part details the proof for the convergence of the sum. Reference [1] was relied on heavily in the writing of this paper, and the proof for Brun’s Theorem is nearly identical to that in [1]. However, all other results that were proved were proved by the author, and the proof for Brun’s Theorem has been presented here with much more detail. Results or definitions that have been fully obtained from [1] (with no or cosmetic changes) are marked with ’**’, and those with some differences from [1] are marked with ’*’. Please see the Correspondence with LeVeque’s Book section of this paper for more information.
Notation The following notation, adopted from [1], will be used throughout this paper: • p - prime • π(x) - the number of primes less than or equal to x • π2 (x) - the number of twin prime pairs in which the smaller prime of the pair is less than or equal to x
2 • τ(n) - the number of divisors of the integer n • ln x - logarithm to the base e
2. Background In this section, some notation and general results are introduced which will be used in proving Brun’s Theorem. Most of the results presented here are left unproved, and the interested reader can refer to [1] for proofs. 2.1. Big ’O’ Notation Big ’O’ notation will be used in the proof of Brun’s Theorem, and as such a definition and some rules are presented in Definition 1 and Theorem 2 below. Definition 1. ** Let S be an unbounded set of positive numbers, and let f , g be defined on S such that g(x) > 0 ∀x ∈ S. If ∃M > 0, y ∈ S, such that ∀x ∈ S ≥ y, | f (x)| < M · g(x), then we write that f (x) = O(g(x)), or f (x) g(x). Theorem 2. * Let f , g be functions defined on an unbounded set S of positive integers, such that f (x), g(x) > 0 ∀x ∈ S. Then the following rules apply: 1. O(O( f (x))) = O( f (x)), 2. O( f (x)) ± O(g(x)) = O(max( f (x), g(x))), 3. O( f (x)) · O(g(x)) = O( f (x) · g(x)), 4. If f is integrable on (a, x) ⊆ S, then
ˆx
ˆx O( f (t))d t = O( f (t)d t).
a
a
Proof. Assume the hypothesis. Now: 1. Let h be a function defined on S, and assume that f (x) = O(g(x)) and h(x) = O( f (x)).Then, ∃M1 , M2 ≥ 0 and y ∈ S such that ∀x ∈ S ≥ y, 0 < f (x) < M1 g(x) and |h(x)| < M2 f (x), from which it follows that |h(x)| < M1 M2 g(x), so that h(x) = O(g(x)). 2. Let h1 , h2 be functions defined on S such that h1 (x) = O( f (x)) and h2 (x) = O(g(x)). Then, ∃M1 , M2 ≥ 0 and y ∈ S such that ∀x ∈ S ≥ y, |h1 (x)| < M1 f (x) and |h2 (x)| < M2 g(x), from which it follows that |h1 (x) ± h2 (x)| ≤ |h1 (x)| +
2.2 Results on Primes
3
|h2 (x)| < M1 f (x) + M2 g(x) ≤ 2 · max(M1 , M2 ) · max( f (x), g(x)), so that h1 (x) ± h2 (x) = O(max( f (x), g(x)). 3. Let h1 , h2 be functions defined on S such that h1 (x) = O( f (x)) and h2 (x) = O(g(x)). Then, ∃M1 , M2 ≥ 0 and y ∈ S such that ∀x ∈ S ≥ y, |h1 (x)| < M1 f (x) and |h2 (x)| < M2 g(x), from which it follows that |h1 (x)·h2 (x)| < M1 M2 · f (x)· g(x), so that h1 (x) · h2 (x) = O( f (x) · g(x)). 4. Let h be a function defined on S and integrable on (a, x) ⊆ S such that h(x) = O( f (x)). Then, ∃M ≥ 0 and y ∈ S such that ∀x ∈ S ≥ y, |h(x)| < M · f (x). Now, for sufficiently large x, we have that, for some constant C,
ˆx
ˆy h(t)d t| = |
| a
ˆy
ˆx h(t)d t +
a
|h(t)|d t + M
a
ˆy f (t) d t < C
so that
a
ˆx
|h(t)|d t < y
ˆx f (t) d t + M
y
ˆx |h(t)|d t +
h(t)d t| ≤ y
ˆx
a
ˆy
ˆx f (t) d t < max(C, M ) ·
y
f (t) d t, a
ˆx h(t)d t = O( f (t)d t).
a
a
.
2.2. Results on Primes One important result which will be used is Chebyshev’s result on the order of magnitude of the prime-counting function. Theorem 3. ** ∃c1 , c2 > 0 such that for x ≥ 2, c1
x x < π(x) < c2 . ln x ln x
Another result which will be used is the bound on the sum of the reciprocal of primes. In proving Brun’s Theorem, we will instead use Corollary 5, proved below, which is an immediate consequence of Theorem 4. Theorem 4. ** For x ≥ 2, ∃C ∈ R such that X1 p≤x
1 = ln ln x + C + O . p ln x
2.3 Other Results
4
Corollary 5. For x ≥ 2, ∃c ∈ R such that X1 p≤x
p
< ln ln x + c.
Proof. By Theorem 5, ∃M > 0, C ∈ R such that for sufficiently large x ∈ R, X1 p≤x
p
< ln ln x + C + M ·
1 < ln ln x + C + M = ln ln x + c ln x
for some constant c ∈ R. Another theorem which will be used in the proof of Brun’s Theorem is the upper bound on the following product. Theorem 6. ** For x ≥ 2, we have that Y 1 1 1− < . p ln x p≤x
2.3. Other Results The following theorem, along with Brun’s Theorem, was used in proving the convergence of the sum of twin prime reciprocals. Theorem 7. ** Let λ1 , λ2 , . . . be an unbounded nondecreasing sequence of real numbers, and c1 , c2 , . . . be an arbitrary sequence of real numbers, and let f ∈ C 1 [λ1 , ∞). Define X C(x) = cn . n; λn ≤x
Then, ∀x ≥ λ1 , X n; λn ≤x
ˆ cn f (λn ) = C(x) f (x) −
x
λ1
C(t) f 0(t) d t.
Finally, this general result on the coefficient of the product of a number of terms in the power of a multinomial was also used in the proof of Brun’s Theorem. Lemma 8. * Let l, s ∈ N with l ≤ s, and let t 1 , t 2 , . . . , t s ∈ R. The coefficient in the expansion of (t 1 + t 2 + · · · + t s )l of the term t h1 t h2 · · · t hl , where 1 ≤ h1 < h2 < · · · < hl ≤ s, is equal to l!.
5
Proof. Suppose that we desire to find the coefficient of some product of l distinct terms of the t i ’s in the expansion. We rearrange the terms so that t 1 t 2 · · · t l is the product of concern. Using the binomial theorem, consider that (t 1 +t 2 +· · ·+t s−1 +t s )l = (t 1 +(t 2 +· · ·+t s−1 +t s ))l =
l X i=0
l X i=0
l−i
X (l − i)! l! j (t 3 + · · · + t s )l−i− j t 1i t 2 = (l − i)!i! j=0 (l − i − j)! j!
l X l−i X i=0 j=0
l X i1 =0
···
l−i1 −iX 2 −...−il−1 il =0
l! (t 2 +· · ·+t s )l−i t 1i = (l − i)!i!
l! j (t 3 + · · · + t s−1 )l−i− j t 1i t 2 = (l − i − j)!i! j!
l! il−1 il i (t l+1 +· · ·+t s )l−i1 −i2 −...−il−1 −il ·t 11 · · · t l−1 tl (l − i1 − . . . − il )!i1 ! · · · il !
Putting i1 = i2 = . . . = il−1 = il = 1, we have that i1 + i2 + . . . + il−1 + il = l, so that we obtain the coefficient as l! (t l+1 + t l+2 + · · · + t s )0 · t 1 · · · t l−1 t l = l! · t 1 · · · t l−1 t l . 0!1!1! · · · 1!
3. Brun’s Theorem on Twin Primes 3.1. An Upper Bound on the Number of Twin Primes We will first find an upper bound on π2 (x), the number of twin prime pairs less than or equal to x. This is done in the following theorem. Theorem 9. ** π2 (x) = O(
x(ln ln x)2 ln2 (x) ).
We will organize the theorem into several claims. First, we will start with some definitons. We define the sequence an = n(n + 2). Let x, y ∈ R such that x Q ≥ 1 and p 1 ≤ y ≤ x. We define A(x, y) = {an : 1 ≤ n ≤ x; p - an ∀p ≤ y}, and R = p≤ y p. Throughout the proof of the theorem, we adopt the convention that p1 < p2 < . . .. We define S0 = [x], and for j ≥ 1, X X Sj = 1. p1 p2 ··· p j |R p1 p2 ··· p j |an ;n≤x
3.1 An Upper Bound on the Number of Twin Primes
6
Claim 10. π2 (x) ≤ π( y) + A(x, y). Proof. The proof of this claim is immediate. For let p, p + 2 be a prime twin pair such that p ≤ x. Then, either p ≤ y, or q - p(p + 2) ∀q ≤ y, q prime, and the claim follows. Now, let k ∈ N with 0 ≤ 2k ≤ π( y). Claim 11. A(x, y) ≤ S0 − S1 + S2 − · · · + S2k . Proof. To show this, suppose that, for n ≤ x, an is divisible by exactly m primes less than or equal to y. Let C(an ) denote the contribution of an to the sum S0 −S1 +S2 − · · · + S2k . Since the contribution of each an to A(x, y) is 1 if m = 0 and 0 if m > 0, we only need to show that C(an ) ≥ 1 for m = 0 and C(an ) ≥ 0 for m > 0. If m = 0, then S j = 0 ∀ j ≥ 1, and S0 = 1, so that the inequality is satisfied. If 0 < m ≤ 2k, then m m X X m C(an ) = Sj = (−1) j = (1 − 1)m = 0. j j=0
j=0
If m+1 2 ≥ 2k, then, C(an ) =
2k X
Sj =
j=0
§
1+
m 2
−
m 1
ª
2k X m (−1) j = j j=0
+ ··· +
§
m 2k
−
m 2k − 1
ª
> 0,
since the factorial is increasing in the first half of the binomial coefficient sequence. Similarly, if m+1 2 < 2k < m, C(an ) =
2k X j=0
(1 − 1)m +
2k m m X X X m m m j j Sj = (−1) = (−1) − (−1) j = j j j
§
j=0
m 2k + 1
j=0
−
m 2k + 2
ª
j=2k+1
+ ··· +
§
m m
−
m m−1
ª
> 0,
since the factorial is decreasing in the second half of the binomial coefficient sequence. Thus we have shown that A(x, y) ≤ S0 − S1 + S2 − · · · + S2k . Now we wish to estimate the term to the sum
P
p1 p2 ··· p j |an ;n≤x
1. Consider that this is equivalent
X
n(n+2)≡0 (mod p1 p2 ··· p j );n≤x
1,
3.1 An Upper Bound on the Number of Twin Primes
7
which we will denote by Td , where d = p1 p2 · · · p j . Clearly, Td is equal to the number of unique solutions n ≤ x of the congruence n(n + 2) ≡ 0 (mod d). We find this number in the following claim. Claim 12. For some −1 ≤ θ ≤ 1, Td = x d d d τ( 2 ) + θ τ( 2 ) if d is even.
x d τ(d)
+ θ τ(d) if d is odd, and Td =
Proof. Assume first that d is odd. Consider that n(n + 2) ≡ 0 (mod d) only if each pi divides n or n + 2. Since (n, n + 2) ≤ 2, and d is odd, the congruence holds if and only if for some factorization d = d1 d2 , n ≡ 0 (mod d1 ) and n + 2 ≡ 0 (mod d2 ). Since d is a product of distinct primes, (d1 , d2 ) = 1, and by the Chinese Remainder Theorem, for each factorization,n is a unique solution modulo d. Now, assume that 0 0 d = d1 d2 is another factorization and that n is a solution for both factorizations; 0 0 that is, n ≡ 0 (mod d1 ),n + 2 ≡ 0 (mod d2 ), n ≡ 0 (mod d1 ) andn + 2 ≡ 0 (mod d2 ). 0 0 By the Generalized Chinese Remainder Theorem, (d1 , d2 ), (d2 , d1 ) | (n + 2) − n = 2, 0 0 and so (d1 , d2 ), (d2 , d1 ) = 1. Then any prime divisor of d1 cannot be a prime divisor 0 0 0 of d2 , and so it is a prime divisor of d1 , and likewise any prime divisor of d1 is a 0 0 prime divisor of d1 . Thus, d1 = d1 and d2 = d2 . Hence, for each factorization of d, there is a unique solution of the congruence modulo d. Since d is the product of distinct primes, the number of solutions modulo d is then equal to τ(d). Now, let 1 ≤ n ≤ d be a solution of the congruence. Then, for 1 ≤ a ≤ x, the congruence is satisfied ∀a such that a = n + bd for some b ∈ Z. There are at most dx + 1 and at least dx such a. Hence, for some −1 ≤ θ ≤ 1, the total number of solutions is Td = dx τ(d) + θ τ(d). Assume now that d is even. Then, n is a solution of n(n + 2) ≡ 0 (mod d) only if n = 2m for some m ∈ Z. The previous congruence is then equivalent to the congruence m(m + 1) ≡ 0 (mod d2 ), which holds if and only if for some factorization d d d = 2d1 d2 , m ≡ 0 (mod 21 ) and m + 1 ≡ 0 (mod 22 ). By the Chinese Remainder 0 0 Theorem, for each factorization,m is a unique solution modulo d2 . If d = 2d1 d2 is another factorization and m is a solution for both factorizations, then, by the d
d
0
d
d
0
0
Generalized Chinese Remainder Theorem, ( 21 , 22 ), ( 22 , 21 ) = 1. Thus, d1 = d1 0 and d2 = d2 . Hence, for each factorization of d2 , there is a unique solution of the congruence modulo d2 . Thus, for some −1 ≤ θ ≤ 1, the total number of solutions a x/2 such that1 ≤ a ≤ x is Td = d/2 τ( d2 ) + θ τ( d2 ) = dx τ( d2 ) + θ τ( d2 ). For d = p1 p2 · · · p j , where p1 , p2 , · · · , p j are distinct primes, we define τ0 by τ0(d) = τ(d) if d is odd, and τ0(d) = τ( d2 ) if d is even, so that, in Claim 12, Td = dx τ0(d) + θ τ0(d) for some −1 ≤ θ ≤ 1. Then, with the aid of Claim 11, we have that, for some −1 ≤ θ ≤ 1,
A(x, y) ≤ S0 − S1 + S2 − · · · + S2k = [x] +
2k X (−1) j j=1
X
X
p1 p2 ··· p j |R p1 p2 ··· p j |an ;n≤x
1=
3.1 An Upper Bound on the Number of Twin Primes
2k X [x] + (−1) j
X
j=1
p1 p2 ··· p j |R
2k X x 1+ (−1) j j=1
x·
τ0(p1 p2 · · · p j ) p1 p2 · · · p j
τ0(p1 p2 · · · p j )
X p1 p2 ··· p j |R
! +
p1 p2 · · · p j
8
+ θ τ0(p1 p2 · · · p j ) ≤
2k X
X
τ0(p1 p2 · · · p j ).
j=1 p1 p2 ··· p j |R
To find an upper bound for A(x, y), we will find upper bounds on the sums. P2k P 2 2k Claim 13. p1 p2 ··· p j |R τ0(p1 p2 · · · p j ) < e π ( y). j=1 Proof. Let d = p1 p2 ...p j be the product of j distinct primes. If d is odd, then τ0(d) = τ(d) = 2 j . If d is even, then τ0(d) = τ( d2 ) = 2 j−1 < 2 j . Now, the number of products of j distinct prime divisors of R is the number of possible combinations of j distinct primes less than or equal to y. Hence, X π( y) τ0(p1 p2 . . . p j ) ≤ 2 j = j p1 p2 ... p j |R
2j
π( y)(π( y) − 1) · · · (π( y) − j + 1) 2 j π j ( y) < j! j!
so that 2k X
X
τ0(p1 p2 · · · p j )
, j! j! i=0
j
so that j! > ( e ) j . Let k = [6 ln ln y]. For sufficiently large y, by Theorem 3,2k ≤ y c1 ln y < π( y), so that our choice of k is valid, and, for sufficiently large y, k > 2e ln ln y + ec. Then, X
|Vk | ≤ |
p1 p2 ... p2k+1
τ0(p1 p2 . . . p2k+1 ) | + ... + | p1 p2 . . . p2k+1 |R p
X
τ0(p1 p2 . . . pπ( y) ) p1 p2 . . . pπ( y)
1 p2 ... pπ( y) |R
|
2k j>2k 2k< j2k Since −12 ln ln y = log2 (ln y)−12 ln 2, 2−12 ln ln y = (2log2 (ln y) (ln y)−8 . Thus, |Vk | < (ln y)−8 .
−12
)ln 2 = (ln y)−12 ln 2
