15. Bruns Theorem Definition Primes p and p < q are called twin primes if q = p + 2.

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1 15 Bruns Theorem Definition 151 Primes and < q are caed twin rimes if q = π ) is the number of airs of twin rimes u to Conjecture 15 There are infinitey many twin rimes Theorem 153 π ) P ) = og og ) og In articuar, the sum of the recirocas of a twin rimes ) ) ) 13 is convergent 153) indicates why the twin rime conjecture is so hard Indeed, even if there are infinitey many twin rimes the sum of the recirocas converges and so we cannot rove the eistence of infinitey many twin rimes a a Dirichet Proof of 153) The ast statement foows from the first by an easy aication of the artia summation formua Consider integers of the form a n = nn ), 1 n Remove those divisibe by a rime y, for some integer y Let A, y) denote the number of remaining a n We have π ) πy) A, y) Our goa is to show that A, y) P ) for some y = y) such that πy) P ) Put R = y We cacuate using incusion-ecusion A, y) = 1 R n, a n = S 0 S 1 S 1 <, 1 R n, 1 a n 1 One ossibiity is to roceed as we did in Lecture 5 However this introduces error terms which are too arge We introduce a variation on this theme: 1

2 Caim 154 There is an inequaity A, y) S 0 S 1 S S k, which is vaid for every even inde 0 < k πy) Proof of 154) Suose that a n is divisibe by eacty m rime factors of R If m = 0 then a n is counted once on the RHS and this is correct If m > 0 it is counted Ca n ) = 1 ) m 1 ) m ) m k times on the RHS and not at a on the LHS and so we just need to check that Ca n ) 0 There are three cases If k 1/m 1) then we use the fact that the binomia coefficients ) m increase unti the midde term or air of midde terms): { ) )} { ) )} m m m m Ca n ) = 1 > If k m then Ca n ) = 1 1) m = 0 Finay if 1/m 1) < k < m then we use the fact that the binomia coefficients decrease after the midde term: { ) ) } m m Ca n ) = 1 1) n k 1 k { ) )} m m = k 1 k > 0 Now we have 154) we sti have to give a good aroimation of the individua terms of the sum For this we need to estimate the inner sums in the doube summation for S m, T d = n,d a n 1 Note that d = 1 R, and µd) 0 First suose that d is odd Then nn ) 0 mod d if and ony if there is a factorisation d = d 1 d such that n 0 mod d 1 and n 0 mod d

3 Caim 155 n is determined, moduo d, by the factorisation Proof of 155) Suose aso that n 0 mod e 1 and n 0 mod e, where d = e 1 e It foows that the equations mod d 1 0 mod d mod e 1 0 mod e have a simutaneous soution, = n Then by a resut in 104A concerning simutaneous soutions to systems of inear congruences) we must have d 1, e ) and d, e 1 ) As d is odd this imies that both d 1 and e, and d and e 1 are corime As d 1 divides d = d 1 d it divides the roduct e 1 e = d As d 1 is corime to e, d 1 divides e 1 By symmetry, it foows that d 1 = e 1 and d = e, so that the factorisation is unique It foows that there are eacty τd) soutions, moduo d, of nn ) 0 mod d Hence the number of n satisfying the congruence is T d = d τd) θ 1τd) = τd) θτd) d where 0 θ 1 1 so that θ 1 Now suose that d is even Then n = m is even In this case we just have to count the number of m 1 such that By the same argument, mm 1) 0 mod 1 d T d = / τd/) θτd/), d/ where θ 1 If we define an auiiary function { τ τd) if d is odd d) = τd/) if d is even then we see that T d = d τ d) θτ d) where θ 1 3

4 It foows that A, y) 1 R τ ) 1 R 1 τ ) τ 1 ) R 1 R τ 1 ) 1 1 k R 1 k R τ 1 k ) We adot the convention that the sums run over increasing rimes, 1 < < If we fi then ) πy) τ 1 ) 1 R πy)πy) 1) πy) 1) =! = π y)! Thus the second term above, the error term, is at most τ ) τ 1 ) R 1 R 1 k R τ 1 k ) k =1 < π k y) τ 1 k ) 1 k π y)! k =1 < π k y)e Now we turn to estimating the main term, the term invoving As τ n) is mutiicative if we mutiy out 1 τ ) ) = 1 1 ), R R,> then we get the coefficient of if we incude V k = 1 k1 R τ 1 k1 ) 1 k1 1 k R! τ 1 k ) 1 k Thus A, y) 1 R,> 1 ) π k y) V k 4

5 Caim 156 There is a constant c 0 such that ) e og og y ec V k < >k Proof of 156) Note that for πy) we have τ 1 ) R 1 R ) 1! To see the ast inequaity, note that if we eand the mutinomia y t 1 t t s ) each of the ) s roducts t h1 t h t h, 1 h 1 < h < < t h s, occurs with coefficient! once can aso rove the inequaity directy by induction) Therefore, by 8), we have V k < ) 1! k πy) y < 1! og og y c), for some c > 0 As k πy) e = u=0 >!, it foows that )! > e Putting a of this together, we get ) e og og y ec u u! V k < >k 5

6 Let k = 6 og og y If y is sufficienty arge then and then k > e og og y ec, V k < >k = k 1 og og y < < og y) 8 On the other hand if we use 51) then we get 1 ) = { 1 1 ) } 1 < y < y < 1 1 ) < y 1 og y Putting a of this together we get π X) og y πk y) y og y og y og 8 y ) k Finay, if we ut 1/1 og og ) y = then by 71) we have ) og og ) 1/1 og og 1 og og π ) og og /1 og og og og ) og 6