The Circle Method on the Binary Goldbach Conjecture

Transcription

1 The Circle Method on the Binary Goldbach Conjecture Jeff Law April 3, 2005 Matheatics Departent Princeton University Princeton, J 08544

2 The Goldbach Conjecture The Goldbach Conjecture, appears to be very siple at first glance. It can be stated as thus: Every even nuber can be represented by the su of two prie nubers. Or in atheatical notation: 2n p + p 2.) n, and where p and p 2 depend on n. However, it has shown itself to be quite difficult to prove. This conjecture is soeties called the binary Goldbach proble because a siilar proble, soeties called the ternary Goldbach proble 2n + ) p + p 2 + p 3.2) for sufficiently large n or every odd nuber can be represented as the su of three pries was proven by I.M. Vinogradov in 937. We will approach the binary case by exaining the ethods eployed in the proof of the ternary proble. Let M be the odd nuber which we are trying to represent as the su of three pries. Consider the su F M x) p Mlog p)epx).3) where eα) exp2πiα) and p is prie. ote: We will be using this notation throughout the reainder of this paper. ow consider the integral R 3 M) 0 0 F 3 Mx)e Mx)dx p Mlog p)epx) 3 e M x)dx Observe that since our function eαx) exp2πiαx) 0 eαx)dx 0 for any value of α Z such that α 0. ote that the integral is if α 0. 2

3 Since p i and M, p + p 2 + p 3 M) Z, so the only ters that contribute to the integral R 3 M) are the ters where p + p 2 + p 3 M 0. Or So continuing on, we see that R 3 M) 0 log p log p 2 log p 3 )e p + p 2 + p 3 M)x) p,p 2,p 3 M p +p 2 +p 3 M p + p 2 + p 3 M log p )log p 2 )log p 3 )e p + p 2 + p 3 M)x) dx log p log p 2 log p 3.4) This integral R 3 M) is then a weighted su of the nuber of representations of our odd nuber M as the su of three pries. ote that CM) p,p 2,p 3 M p +p 2 +p 3 M.5) is siply the count of different ways that each M can be represented as the su of three pries. It is also evident that when R 3 M) > 0, by partial suation we have that CM) > 0 and hence we have a representation of M. Through siilar calculations we can derive R 2 ) where is an even nuber, and find it to be where R 2 ) 0 Fx)e x) 2 log p log p 2.6) p +p 2 F x) p log p)epx).7) In general, R s ) deterines the weighted count of the nuber of ways that is the su of s nuber of pries. 3

4 2 Decoposition into Major and Minor Arcs ow let us return to the ternary proble. To evaluate the integral R 3 ), we can eploy the circle ethod by decoposing the unit interval, over which the intgral is evaluated, into ajor and inor arcs. The ajor arcs are defined as the following: Let B > 0 and where [α] is the greatest integer less than α. Consider all pairs a, q) such that Q [log ) B ] 2.8) 0 a q, q Q, a, q) 2.9) Each ajor arc M a,q is defined as the interval of x R such that x a q Q and the set of ajor arcs M is M Q q q a,q) M q,a 2.0) ote that for sufficiently large, the ajor arcs becoe disjoint. The set of inor arcs is defined to be the copleent of the set of all ajor arcs in the unit interval. More clearly, [0, ] M 2.) Fro this definition of the ajor arcs, we can find an eleentary upper bound for the easure of the set of ajor arcs, M. 4

5 For sufficiently large that the ajor arcs are disjoint, we have M Q q q a,q) Q q q a,q) M a,q 2Q Q q Q q q a,q) 2Qq µm a,q ) Since 2 does not depend on q 2 q Q Qq Q2 Q + ) Q3 2.2) as grows large. Observe also that as, li M 0 So as grows large, the ass of the ajor arcs decreases and we see that the unit interval consists ostly of inor arcs. However, it has been shown that despite this density of inor arcs, the contribution of FMx)e Mx)dx 3 is negligible and the ain ter of 0 FMx)e Mx) 3 M FMx)e Mx)dx 3 + FMx)e Mx)dx 3 2.3) coes fro the integral over the ajor arcs. This is not apparent for the binary case. It becoes uch ore difficult to show that Fx)e x)dx 2 is neglible copared to the ter Fx)e x)dx 2 M 5

6 The contribution of the integral over the inor arcs is precisely the otivation for this investigation. We hope to see that the the contribution of F 2 x) fro the ajor arcs is large, while the contribution fro the inor arcs is sall. 3 Properties of the Integral over the Major Arcs The integral over the ajor arcs for the binary proble R M ) Fx)e x)dx 2 3.4) M is not trivial to evaluate so we will first find a function siilar to it which is easy to evaluate and forulate R M ) in ters of this function. This section will be devoted to deriving a ore accessible for of R M ). We begin by defining this function siilar to F x) by ow consider the function uβ) eβ) 3.5) J) Theore 3.. As, 0 u 2 β)e β)dβ 3.6) J) Proof: Since, as before, when integrated fro 0 to, our function eα) exp2πiα) gives a contribution of 0 for all α Z such that α 0, and a contribution of when α 0, we have that J) 0 0 u 2 β)e β)dβ e + 2 )β) dβ 2 since 2 < there are values of 2, and for each value of 2 there is a value of that satisfies

7 except when 2. Therefore there are pairs, 2 ) that give a non-zero contribution to J). Since each tie + 2, we conclude that and as grows large e + 2 )β) J) + O) J) 3.7) and we are finished. It is not iediately apparent why this result is significant but we shall use this it later on as a lea to a ore iportant theore approxiating the function F x) 2. Before continuing, let us first define a few useful functions.. Let φq) be the nuber of integers b < q such that b, q) 2. Let µq) be defined as such: if q µq) 0 if q is divisible by the square of a prie ) r if q is the product of r distinct pries 3. Let c q ), the Raanujan Su, be defined by 4. Let λ) be defined by λ) c q ) { log p q a a,q) e ) a q if p is prie 0 otherwise 3.8) 3.9) 3.20) 7

8 Theore 3.2. Siegel-Walfisz) If q log n) C, a, q), and n 2 then, for any C > B > 0, Theore 3.3. Let φn; a, q) p n p aq) log p n ) φq) + O n log n) C 3.2) F n x) p nlog p)epx) 3.22) Let B, C > 0 and B, C R. Then for n, and q Q, ) a F n q Proof: µq) φq) n + O Q ) log ) C 3.23) ) a F n p)epx) q p nlog q r p n p rq) log p)e pa q ) Let p rodq) Then p divides q if and only if r, q) >. To bound the contribution of ters where p q we have q r r,q)> p n p rq) log p)e pa q ) p n p q log p) epa q ) p q log p log q 3.24) A weaker, but ore intuitive bound can be found by considering that there are less than q integers r q such that r, q) >. Each one contributes a ter less than or equal to log q since e pa q ). So a very crude bound would be q r r,q)> p n p rq) log p)e pa ) q log q 3.25) q 8

9 We shall use the stronger bound but both are sufficient. Continuing on, F n a q ) q r r,q) q r r,q) p n p rq) e ra q log p)e ) p n p rq) ) ra + Olog q) q log p + Olog q) By the Siegel-Walfisz Theore: Since q r r,q) q r r,q) e e ) q r e ra q q, ) ra θn; q, r) + Olog q) q ) ra n q φq) + O n log n) C ) c q a) qn φq) n + O log n) C + Olog Q) Since c q a) µa) for a, q), q Q, and n, µq) ) Q φq) n + O log ) C and we are finished. )) + Olog q) Theore 3.4. Let B, C R such that B, C > 0 and C > 2B. If x M a,q and β x a q, then and F x) µq) Q 2 ) φq) uβ) + O log ) C Fx) 2 µ2 q) Q 2 2 ) φ 2 q) u2 β) + O log ) C ) 3.27)

10 Proof: If x M a,q) then x a q + β where β Q For n, F x) µq) φq) uβ) log p)epx) µq) φq) p eβ) λ)ex) µq) φq) λ)e a ) q + β) eβ) µq) φq) eβ) λ)e a q ) µq) ) eβ) 3.28) φq) We apply Partial Suation to this last ter. Let An) be defined by An) λ)e a q ) µq) ) 3.29) φq) Since µq) φq) n does not depend on n, and µq), An) n λ)e a q ) µq) φq) n Fro the definition of λ) we can see that the only ters which give a non-zero contribution are when p where p is a prie. So we have log p)e pa q ) µq) φq) n F n p n ) a µq) q φq) n By Theore 3.3, ) Q An) O log p) C 3.30) 0

11 By using An) and Partial Suation again, we can find an upper bound for F x) µq) φq) uβ) A)eβ) 2πiβ An)enβ)dn A) + β ax{an) : n } Q2 log ) C 3.3) It follows directly that F x) µq) Q 2 ) φq) uβ) + O log ) C Let us now derive 3.32) F 2 x) µq) Q 2 φq) uβ) + O log ) C ) µ2 q) µq) φ 2 q) u2 β) + 2 φq) uβ) As, we find that so the cross ter becoes negligible. Since Q log ) B and C > 2B, O )) 2 Q 2 log ) C Q 2 log ) C Q4 2 log ) 2C ) µq) Q 2 ) 2 φq) uβ) O log ) C log ) C log ) 2B ) Q 4 2 ) + O log ) 2C

12 so Q 4 2 ) log ) 2B Q 2 2 ) Q 2 2 ) O log ) 2C O log ) 2B log ) C O log ) C and we conclude that Fx) 2 µ2 q) Q 2 2 ) φ 2 q) u2 β) + O log ) C ote that the error ter Q 2 2 ) 2 ) O log ) C O log ) C 2B is uch saller than 2 for C > 2B. We shall now investigate the contribution fro the ajor arcs. 3.33) Theore 3.5. Let B, C R such that B, C > 0 and C > 2B. Let ɛ > 0. Then the weighted su R M ) over the ajor arcs M can be represented as M F 2 x)e x)dx S) 2 where S) is defined below. Proof: First consider this integral: M q Q 2 + O 2 log ) ɛ)b Fx) 2 µ2 q) φ 2 q) u2 x a )) e x)dx q q a0 a,q) M a,q) siply fro the definition of the ajor arcs M. ) + O 2 ) log ) C 5B 3.34) Fx) 2 µ2 q) φ 2 q) u2 x a )) e x)dx q 2

13 Then, since the function eαx), and Theore 3.4, q Q q a0 a,q) Ma,q) Q 2 2 log ) C dx Observe that when q, M a,q) Q, and when q 2, M a,q) 2Q. Then since C > 2B, q Q q a0 a,q) Q 5 log ) C M a,q) Q 3 log ) C dx log ) C 5B 3.35) Since M to deterine Fx) 2 µ2 q) φ 2 q) u2 x a )) e x)dx q it is sufficient to study M q Q q Q R M ) M F 2 x)e x)dx µ 2 q) φ 2 q) u2 x a ) e x)dx q q a a,q) q a a,q) M a,q) µ 2 q) φ 2 q) u2 µ 2 q) φ 2 q) a q + Q a q Q 3 log C 5B ) x a ) e x)dx q u 2 x a ) e x)dx q

14 Since x a q + β, and the integrand is periodic, q Q q Q µ 2 q) φ 2 q) µ 2 q) φ 2 q) q a a,q) q a a,q) Q Q Recalling the definition of c q ), the above is q Q µ 2 q)c q ) φ 2 q) u 2 β)e aq ) + β) e a ) Q u 2 β)e β) q Q Q Q u 2 β)e β)dβ where Q S, Q) u 2 β)e β)dβ 3.36) Q S, Q) q Q µ 2 q)c q ) φ 2 q) 3.37) Continuing with the proof, observe that if β 2, then uβ) β. Since eαx), 2 Q 2 Q 2 Q u 2 β)e β)dβ u 2 β)dβ β 2 Q 2 Q Ipleenting the sae calculation, we obtain that Q 2 u 2 β)e β)dβ Q 4

15 Fro these approxiations, it is easy to see that Q Q u 2 β)e β)dβ 2 2 u 2 β)e β)dβ up to a sall error. By Theore 3., we have that 2 2 u 2 β)e β)dβ + O ) Q 3.38) ow consider S, Q): Let S) be defined by S) q µ 2 q)c q ) φ 2 q) 3.39) This is the Singular Series for the Binary Goldbach Proble. We first will show that it converges. Then we will show that for Q, S, Q) converges to S) We know that c q ) φq) and that φq) > q ɛ for ɛ > 0 and sufficiently large q. We know that c q ) is ultiplicative. Let us, for the oent, use a clearer notation for this discussion. Since is fixed, we shall denote c q ) by c q). We see that since c q) is ultiplicative in q, that if q p p 2... p r, that c q) c p p 2... p r ) c p )c p 2 )... c p r ) ote that each p i ust be distinct since if q is divisible by p k i where k 2, then µq) 0. Also note that fro the definition of we have c q ) c q) c p) { p q a a,q) e ) a q if p divides otherwise 3.40) 5

16 and since p, it follows that µ 2 q)c q) φ 2 q) )s φ 2 q) + )r s φ 2 q) < )s q 2 ɛ) + )r s q 2 ɛ) 3.4) where s is the nuber of prie factors coon to q and. Since the nuerators do not depend on q, it is obvious that as q, the Singular Series S) q µ 2 q)c q ) φ 2 q) converges. Convergence of the Singular Series is also apparent when we consider the bound Then we have c q ) φ) 3.42) µ 2 q)c q) φ 2 q) φ 2 q) < which tends to 0 as q. ow, for fixed, and q square-free, we have q 2 ɛ) 3.43) S) S, Q) q>q q>q Q ɛ µ 2 q)c q ) φ 2 q) φ 2 q) q>q q 2 ɛ therefore ) S, Q) S) + O Q ɛ 3.44) 6

17 but this approxiation is of no use since the error ter is of the sae order as the ain ter. With theore 3.5 as otivation, we have now discovered a proble which has not yet been solved. It has not yet been shown that can be approxiated by S, Q) S) with an error ter saller than the ain ter. We now digress into a discussion on this proble. We see that the proble arises in bounding c q ) when the coon factors between q and, q, ) is very large. We will consider the special case where has few factors relative to its size. In particular, fix an integer D > 0. We consider such that the nuber of factors of is at ost log D ). Let us represent q as the product of two functions, xq)yq), where xq) is the product of factors coon to q and, and yq) is the product of what reains. Observe that for each yq), there are at ost τ) log D ) ref: HW) choices of xq) that have the sae yq). ow let s consider the su S xq),yq) xq)yq)>q C xq)yq)) φ 2 xq)yq)) There are two cases to consider. The first is when xq) > Q /2 and the second is when yq) > Q /2. We can split the su into two sus with these conditions as follows: S xq),yq) xq)>q /2 C xq)yq)) φ 2 xq)yq)) + xq),yq) yq)>q /2 C xq)yq)) φ 2 xq)yq)) Let us first exaine the first su, when xq) > Q /2. Call it L. Since C and φ are both ultiplicative functions, we can separate the ters. We have, 7

18 L xq),yq) xq)>q /2 C xq)yq)) φ 2 xq)yq)) xq),yq) xq)>q /2 xq) xq)>q /2 C xq)) φ 2 xq)) C yq)) φ 2 yq)) C xq)) φ 2 xq)) C yq)) φ 2 yq)) yq) We observe that, by definition, and by the fact that xq), C xq)) And since yq), C yq)) We then have that, L < xq) xq)>q /2 xq) a a,xq)) yq) a a,yq)) φxq)) yq) < Q ɛ)/2 xq) yq) logd ) Q ɛ)/2 yq) ) a e xq) ) a e < yq) φyq)) < φyq)) φyq)) O xq) a a,q) yq) a a,q) xq) xq)>q /2 τ) Q ɛ)/2 yq) log D ) ) Q ɛ)/2 Let us now exaine the second su. We shall call it R. φxq)) φyq)) xq) ɛ yq) φyq)) φyq)) R xq),yq) yq)>q /2 C xq)yq)) φ 2 xq)yq)) xq),yq) yq)>q /2 yq) q)>q /2 C xq)) φ 2 xq)) C yq)) φ 2 yq)) C xq)) φ 2 xq)) C yq)) φ 2 yq)) xq) 8

19 By siilar reasoning as with L, we find that R < yq) yq)>q /2 φyq)) xq) Q ɛ)/2 φxq)) logd ) Q ɛ)/2 xq) φxq)) Since D was arbitrary, we see that we can choose D so that for such that τ) log D ), S, Q) is a good approxiation to S. We end our digression here. Returning to the proof, we have R M ) M F 2 x)e x)dx Q S, Q) Q u 2 β)e β)dβ + O S) + O + O Q ɛ ) log ) C 5B S) + O S) + O and this is the desired result. Q ɛ )) + O Q ) + O log ) C 5B log ) ɛ )B ) + O ) log ) C 5B )) ) log ) C 5B ) 9

20 4 The Singular Series Earlier, we had defined the Singular Series for the Binary Goldbach Proble to be S) q µ 2 q)c q ) φ 2 q) This function has any interesting properties. The first that we will discuss is that it can be represented as an Euler Product. Theore 4.. The Singular Series has the Euler Product S) p p ) 2 ) p + ) p Proof: Recall the calculation done in Section 3 that showed that 4.45) µ 2 q)c q ) φ 2 q) )r q 2 ɛ) Hence the Singular Series converges. Since the functions c q ), µq), and φq) are ultiplicative in q, and the product of ultiplicative functions is ultiplicative, so is the Singular Series. Therefore it has the Euler Product S) p + j µ 2 p j )c p j) φ 3 p j ) Returning to the definition of the function µq), we see that for j 2, µq) 0. So we can write the Singular Series as S) µ 2 p)c p ) + φ 3 p) p Recall that c p ) { p j if p divides otherwise 20

21 It follows that + p p p j + c p) φ 2 p) p ) 2 µ 2 p)c p ) φ 3 p) p ) + c ) p) φ 2 p) p ) p + ) p + c ) p) φ 2 p) The Singular Series is very significant to the Circle Method. It basically encodes all of the inforation that restricts us to the proble we are investigating. For exaple: If we try to prove that every odd nuber is the su of two pries, we observe that in order for this to happen, one of the pries has to be 2. This is because 2 is the only even prie and the su of two odd nubers is even. ow, looking back at the Euler Product for our Singular Series for two pries, since is odd, our first prie 2 does not divide, and so we have 0 p p>2 ) 2 ) 2 p p>2 ) p ) 2 p p ) 2 ) p + ) 0 p + ) p and so the weighted su of the nuber of ways our odd can be represented as the su of two pries, R) is 0. This reflects that it is not true that all odd nubers can be represented as the su of two pries. We can find a siple counterexaple by considering the nuber 27. 2

22 For the sake of interest, let us now turn to the Singular Series for the ternary Goldbach Proble. It follows fro siilar analysis that SM) q p M µq)c q M) φ 3 q) + ) p ) 3 p M ) p ) ) If we violate the condition that M is odd and set M as even, we see iediately that the Singular Series forces the weighted su R 3 M) to equal zero since 2 divides M p M + ) p ) 3 p M ) 2 ) 2 + p M ) p ) This is very interesting since if the Singular Series allowed us to show that every even nuber is the su of three pries then we would be able to prove the binary Goldbach proble fro that result. Let us see why. Let it be true that 2n p + p 2 + p 3 n where p, p 2, p 3 depend on n. ow, there are three pries p, p 2, p 3 ) and each can be even or odd. This gives us eight different cobinations. The four such that p + p 2 + p 3 is even are. if all of the are even or 2. if one is even while the other two are odd. If all of the are even, they ust all be 2 since 2 is the only even prie. The su is then 6 and we are liited to this specific case. If we consider one of the pries, say p, to be even, and hence 2, and the other two to be odd, then we have Furtherore, we have 2n 2 + p 2 + p 3 2n ) p 2 + p 3 22

23 and for n 2 we have proven that every even nuber greater than 2 can be represented as the su of two pries. This deonstrates the power of S) and why the rigidity of the Singular Series is so iportant. It is a very precise expression. 5 The Integral Over the Minor Arcs ow that we have established the value of the integral over the Major Arcs within a certain error ter, we ust investigate the contribution of the integral over the Minor Arcs. Recall that our weighted su, R 2 ), of the nuber of different ways the nuber can be the su of two pries can be represented by R) 0 M Fx)e x)dx 2 Fx)e x)dx 2 + F 2 x)e x)dx Vinogradov showed that in the case of the ternary Goldbach proble, the integral over the inor arcs is bounded by F 3 Mx)e x)dx Fro this he showed that for A > 0 M 2 log M) B/2) ) RM) 0 FMx)e Mx)dx 3 SM) M 2 ) M O log M) A which is bounded away fro 0 since as M, log M and so 5.48) Therefore the function RM) > 0. ote that the function M 2 log M) A SM)M2 2 RM) p,p 2 M p +p 2 M log pepx) 23

24 is greater than 0 if and only if the function CM) p,p 2 M p +p 2 M is greater than 0. Therefore, since RM) > 0, the nuber of ways M can be represented as the su of three pries is greater than 0. Since the function CM) only takes integer values, CM), and the ternary proble is proven. Let us investigate soe eleentary bounds for the integral over the inor arcs. For the ternary case, it was possible to deterine bounds for the integral over the inor arcs. If we siply take F 3 Mx)e x)dx F M x) 3 dx ax{ F M x) : x } F M x) 2 x)dx then it is possible to find a bound for the axiu of the function F M on the inor arcs and the integral 0 F M x) 2 p M log p) 2 log M log p p M and by Chebyshev s Theore, M log M And Vinogradov gives us a bound for the function F M x). F M x) ) q /2 + 4/5 + /2 q /2 log ) 4 log ) B/2) ) 24

25 Fro above, we have ax{ F M x) : x } log ) B/2) 4 2 log ) B/2) 5 0 F M x) 2 dx F M x) 2 x)dx When we try to apply this ethod to the binary Goldbach proble, we find that we cannot bound R) away fro 0 since we cannot deterine the contribution fro the inor arcs to be sall enough that were it to be negative, it still would not contribute enough to diinish the value of R) to less than 0. We have R) S) + O + and we cannot yet show that log ) ɛ )B ) + O log ) C 5B F 2 x)e x)dx 5.50) ) ) O + log ) ɛ )B O + log ) C 5B ) Fx)e x)dx 2 < S) If we try to bound the integral over the inor arcs in the sae way that we did for the ternary case we find that F 2 x)e x)dx F x) 2 dx ax{ F x) } F x) dx 25

26 but it is not easy to bound enough. As we saw earlier, F x) dx 0 F x) dx log but looking back at the contribution fro the ajor arcs we see that the ter fro the inor arcs is of larger order. S) + O log ) ɛ )B ) + O log ) C 5B ) + O log ) is not necessarily larger than 0. Let us try another ethod to bound the contribution fro the inor arcs. We eploy the Cauchy-Schwartz inequality. Again, we have ax{ F x) } F x) dx ax{ F x) } F x) dx ax{ F x) } ) /2 ) /2 F x) 2 2 log )/2 log ) B/2) 4 3/2 log ) B/2) 7/2) 5.5) which is still larger than the ter contributed by the ajor arcs. We now turn to coputational ethods to deterine the behavior of the integral over the inor arcs for the binary proble. 6 Minor Arc Behavior As we saw in the previous sections, it is not a siple task to theoretically bound the contribution fro the inor arcs. Consider the weighted function F x) log pe2πipx) cos 2πpx + i sin 2πpx) 6.52) 26

27 For the sake of siplicity we only investigate the behavior of the real part cos 2πpx) 6.53) of F x) in this paper. The ain questions we have to ask in bounding the contribution fro the inor arcs are: : What is the axiu value attained by the function? 2: How often is this axiu value attained, and what is the easure of the set that it attains this axiu value on? 3: How sall is the function away fro the peaks? By observing the graphs, we can get a good sense for how the function looks and the liklihood that Goldbach s Conjecture is true. Before exaining the behavior of the function on the inor arcs, let us first take a look at a plot of the ajor arcs theselves. ote that this is not a plot of the function on the ajor arcs, but siply an illustration of where the ajor arcs lie and their spacings. This particular case is the ajor arc intervals for the first two hundred pries

28 For the first five hundred pries, we have For the first thousand pries, we have For the first ten thousand pries, we have

29 For the first one hundred thousand pries, we have For the first illion pries, we have ote how thin and sparse the ajor arcs becoe as the nuber of pries increase. ow let us investigate the behavior of the function on the inor arcs. ote also that we are only investigating the cases where Q log B ) where B. This is because for larger values of B, the ajor arcs are so thick that they do not becoe distinct until is sufficiently high. We have shown behavior up to the first one hundred thousand pries, but for to take on any higher values would require an extreely long tie to copute with the software we have available. 29

30 We begin by observing the graph of the real part of the function over the inor arcs in the unit interval for the first five hundred pries For the first five hundred pries, the ajor arcs are still apparent and we can see the here as the intervals where the function over the inor arcs vanishes. We now focus in on a saller interval around one of the peaks to observe its behavior in ore detail. The interval is 0.098, 0.02) where the peak appears to be the highest. The constant plots are at 900 and 900, just as visual aid for deterining exactly where the axius of the peaks lie We can see that the axiu of the function over the inor arcs is slightly less than

31 For the first thousand pries, we have: And in the interval 0.689, 0.70), with the constants at agnitude 2000, And we see that the axiu is approxiately

32 For the first ten thousand pries, we have: And in the interval , ), with the constants at agnitude 9000, And we see that the axiu is approxiately

33 For the first one hundred thousand pries, we have: And in the interval , ), with the constants at 0000 and 38000, And we see that the axiu is approxiately

34 It is clear fro the graphs that although the peaks are still relatively high, they are extreely thin and rare. It is also proising to observe that aside fro the peaks, the function is otherwise extreely sall on the inor arcs. Vinogradov only gave a bound for the axiu of the function, which we now see to be significantly larger than the rest of the function. Vinogradov s crude bound using the axiu of the function was not sufficient to prove the binary case, but now that we see that the axiu of the function ost likely is far larger than the average of the function over the inor arcs, the binary case could very well be true. 34

35 7 Corrections that have yet to be incorporated 7. First reeber, yq) is relatively prie with THIS is what allows you to conclude that c yq)). for the yq) su, there are two pieces: Case : yq) sqrtq): then clearly the su over yq) will be sall Case 2: xq) sqrtq): in this case, you are correct and all one can say is the yq) su is O); however, there will be enorous savings fro the xq) su. ote case and 2 are OT utually exclusive. 7.2 Second the proble with your bound for L is that, by using the week bound, you now have a y-su of /phiyq)). This su is OT O). USIG THE BOUD THAT yq) is yq) epsilon, you are now suing soething like /yq) epsilon, and this DOES OT COVERGE!!! This is why your L bound is now WROG you CAOT use the weak bound,you MUST use c yq)). Your bound for the xq)-part of L is okay: xq) is large so you have soe cancellation, and you have at ost τ) log D ters. But your yq) su is no longer O) you ve ade it infinite! then see y coents for the R bound. 7.3 Third page 7: it is S L + R, not S L + R. Reeber, you are double counting the ters where both xq) and yq) sqrtq). This is inor. on page 8, your bound for L is wrong. The proble is you are using C yq)) < phiyq)) You should instead use C yq)), so it is bounded by. Reeber, since xq) sqrtq), it is possible for yq) to be sall yq) can start at. If you use a bound of phiyq)),then you do OT get a suable series for y you ll have /y epsilon and this blows up. Putting in the correct bound for C yq)), the su over y is bounded by a constant depending on epsilon since we have soething that is su y /y 2 2epsilon Then, for the x-su, note that there are at ost τ) > log D choices for x; as each is at least sqrtq), then /phixq)) /Q epsilon)/2. 35

36 7.4 Fourth Botto of 8 / top of 9: Again, you have copletely destroyed the savings on C yq)). You were so concerned before about using instead of or, that I find this surprising. We know C yq) why aren t you using this bound? To finish the proof, all you need reark is that for each y, there are at ost τ) log D choices of xq). Each gives a contribution of /phixq)). Trivially saying this is suffices for the proof. Thus, in this case, the su over xq) gives at ost log D. The su over yq) is bounded by /phiyq)) 2 /y 2 2epsilon Q 2epsilon)/2 y Q References y Q Goldbach, C. Letter to L. Euler, June 7, 742. Hardy, G. H. and Littlewood, J. E. Soe Probles of Partitio ueroru. III. On the Expression of a uber as a Su of Pries. Acta Math. 44, -70, 922. Hardy, G. H. and Littlewood, J. E. Soe Probles of Partitio ueroru V): A Further Contribution to the Study of Goldbach s Proble. Proc. London Math. Soc. Ser. 2 22, 46-56, 924. athanson, Melvyn B., Additive uber Theory: The Classical Bases, Springer,