Goldbach's problem with primes in arithmetic progressions and in short intervals

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1 Goldbach's problem with primes in arithmetic progressions and in short intervals Karin Halupczok Journées Arithmétiques 2011 in Vilnius, June 30, 2011 Abstract: We study the number of solutions in Goldbach's problem, where the primes are taken from arithmetic progressions as well as from short intervals. For this, we show mean value theorems in the style of Bombieri-Vinogradov's theorem and apply them in a sieve argument to get results for Goldbach's problem with primes from short intervals such that p i + 2, i = 1, 2, 3, are almost-primes.

2 Binary mean value theorems and ternary consequences Corollaries using Sieve Theory The limits when putting conditions on the primes

3 A binary theorem of Kawada As a special case of [Kawada 1993], the following is known: Theorem: Let X 2/3 1 L C < Y, Q YX 1/2 q Q max a (q) k [X 1,X 1 +Y ] p 2 +p 3 =2k p 2 a (q) p 2 [X 2,X 2 +Y ] 2 L B, X 2 + Y X 1. Then log p 2 log p 3 S(2k, q, a)y Y 2 L A.

4 A binary theorem of Kawada As a special case of [Kawada 1993], the following is known: Theorem: Let X 2/3 1 L C < Y, Q YX 1/2 q Q max a (q) k [X 1,X 1 +Y ] p 2 +p 3 =2k p 2 a (q) p 2 [X 2,X 2 +Y ] 2 L B, X 2 + Y X 1. Then log p 2 log p 3 S(2k, q, a)y Originally: X 1 = 2X 2, but the proof works also for any X 2 X 1 Y. Y 2 L A.

5 A binary theorem of Kawada As a special case of [Kawada 1993], the following is known: Theorem: Let X 2/3 1 L C < Y, Q YX 1/2 q Q max a (q) k [X 1,X 1 +Y ] p 2 +p 3 =2k p 2 a (q) p 2 [X 2,X 2 +Y ] 2 L B, X 2 + Y X 1. Then log p 2 log p 3 S(2k, q, a)y Originally: X 1 = 2X 2, but the proof works also for any X 2 X 1 Y. Y 2 L A. A similar variant is true if p 2 + p 3 = 2k is replaced by p 2 p 3 = 2k.

6 A binary theorem of Kawada As a special case of [Kawada 1993], the following is known: Theorem: Let X 2/3 1 L C < Y, Q YX 1/2 q Q max a (q) k [X 1,X 1 +Y ] p 2 +p 3 =2k p 2 a (q) p 2 [X 2,X 2 +Y ] 2 L B, X 2 + Y X 1. Then log p 2 log p 3 S(2k, q, a)y Originally: X 1 = 2X 2, but the proof works also for any X 2 X 1 Y. Y 2 L A. A similar variant is true if p 2 + p 3 = 2k is replaced by p 2 p 3 = 2k. We can put easily an additional condition on 2k, namely such that 2k lies in an arithmetic progression (AP):

7 Corollary with 2k in AP Theorem 1: Let X 2/3 1 L C < Y, Q 2 YX 1/2 2 L B and X 2 + Y X 1. Then, for xed a 1, a 2 N and X 1 > a 1, we have log p 2 log p 3 S(2k, q 2, a 2 )Y q 1 Q 1 q 2 Q 2 k [X 1,X 1 +Y ] 2k a 1 (q 1 ) p 2 +p 3 =2k p 2 a 2 (q 2 ) p 2 [X 2,X 2 +Y ] Y 2 L A.

8 Corollary with 2k in AP Theorem 1: Let X 2/3 1 L C < Y, Q 2 YX 1/2 2 L B and X 2 + Y X 1. Then, for xed a 1, a 2 N and X 1 > a 1, we have log p 2 log p 3 S(2k, q 2, a 2 )Y q 1 Q 1 q 2 Q 2 k [X 1,X 1 +Y ] 2k a 1 (q 1 ) p 2 +p 3 =2k p 2 a 2 (q 2 ) p 2 [X 2,X 2 +Y ] Y 2 L A. If in addition Q 1 Y 1/2 L D holds for some xed D > 0, then the estimate holds true with max a1 (q 1 ) inserted after q 1 Q 1. Also max a2 (q 2 ) can be inserted after q 2 Q 2.

9 Simple tools This can be proven quickly using the following Lemmas: Lemma 1: Let a N be xed, Q 1, N N with N > a and v N,..., v 2N 1 C. Then Q q<2q N n<2n n a (q) v n N 1/2 (log N) 3/2 ( N n<2n v n 2 ) 1/2.

10 Simple tools This can be proven quickly using the following Lemmas: Lemma 1: Let a N be xed, Q 1, N N with N > a and v N,..., v 2N 1 C. Then Q q<2q N n<2n n a (q) v n N 1/2 (log N) 3/2 ( N n<2n v n 2 ) 1/2. Lemma 2: Let Q 1, N N with N 1 and v N,..., v 2N 1 C. Then Q q<2q max a (q) N n<2n n a (q) v n (N log 2 Q+Q 2 ) 1/2 ( N n<2n v n 2 ) 1/2.

11 Simple tools This can be proven quickly using the following Lemmas: Lemma 1: Let a N be xed, Q 1, N N with N > a and v N,..., v 2N 1 C. Then Q q<2q N n<2n n a (q) v n N 1/2 (log N) 3/2 ( N n<2n v n 2 ) 1/2. Lemma 2: Let Q 1, N N with N 1 and v N,..., v 2N 1 C. Then Q q<2q max a (q) N n<2n n a (q) v n (N log 2 Q+Q 2 ) 1/2 ( N n<2n v n 2 ) 1/2. Both Lemmas are simple consequences of Halasz-Montgomery's inequality.

12 A ternary mean value theorem as corollary The following theorem can be deduced for the ternary Goldbach problem with extra conditions on the primes: Corollary 1: Let X 2/3 1 L C Y, X 3/5+ε 2 Y, and Q i YX 1/2 L B for i = 1, 2. i Let n X 1 + X 2 + 2Y be odd. Then max max a 1 (q 1 ) a 2 (q 2 ) q 1 Q 1 q 2 Q 2 log p 1 log p 2 log p 3 T(n, q 1, a 1, q 2, a 2 )Y 2 p 1 +p 2 +p 3 =n p i [X i,x i +Y ], i=1,2 p i a i (q i ), i=1,2 Y 2 L A.

13 A ternary mean value theorem as corollary The following theorem can be deduced for the ternary Goldbach problem with extra conditions on the primes: Corollary 1: Let X 2/3 1 L C Y, X 3/5+ε 2 Y, and Q i YX 1/2 L B for i = 1, 2. i Let n X 1 + X 2 + 2Y be odd. Then max max a 1 (q 1 ) a 2 (q 2 ) q 1 Q 1 q 2 Q 2 log p 1 log p 2 log p 3 T(n, q 1, a 1, q 2, a 2 )Y 2 p 1 +p 2 +p 3 =n p i [X i,x i +Y ], i=1,2 p i a i (q i ), i=1,2 Y 2 L A. Proof: Sum up over 2k = n p 1 instead of p 1 and insert the estimate of the previous theorem. The correct singular series is obtained.

14 Getting prime ranges of dierent length Disadvantage in above Theorems: Same magnitude Y for the interval length of p 1 and p 2 necessary in the proof.

15 Getting prime ranges of dierent length Disadvantage in above Theorems: Same magnitude Y for the interval length of p 1 and p 2 necessary in the proof. The well known Theorem of [Perelli/Pintz 1993] states: If X 1/3+ε 1 R X 1, then log p 2 log p 3 S(2k)2k 2 RX 2 1 L A. 2k [X 1,X 1 +R] p 2 +p 3 =2k

16 Getting prime ranges of dierent length Disadvantage in above Theorems: Same magnitude Y for the interval length of p 1 and p 2 necessary in the proof. The well known Theorem of [Perelli/Pintz 1993] states: If X 1/3+ε 1 R X 1, then log p 2 log p 3 S(2k)2k 2 RX 2 1 L A. 2k [X 1,X 1 +R] p 2 +p 3 =2k Can we put extra conditions on the primes? Especially a short interval condition?

17 Getting prime ranges of dierent length Disadvantage in above Theorems: Same magnitude Y for the interval length of p 1 and p 2 necessary in the proof. The well known Theorem of [Perelli/Pintz 1993] states: If X 1/3+ε 1 R X 1, then log p 2 log p 3 S(2k)2k 2 RX 2 1 L A. 2k [X 1,X 1 +R] p 2 +p 3 =2k Can we put extra conditions on the primes? Especially a short interval condition? Using minor adaptions, we can deduce: Theorem 2: Let Y 1/3+ε R Y X 1 X 2 Y 0 and X 3/5+ε 2 Y X 2, then log p 2 log p 3 S(2k)Y RYL A. 2k [X 1,X 1 +R] p 2 +p 3 =2k p 2 [X 2,X 2 +Y ]

18 Putting an AP-condition on 2k Using both lemmas, we deduce as above: Theorem 3: Let Y 1/3+ε R Y X 1 X 2 Y 0 and X 3/5+ε 2 Y X 2. Fix a N. Then, for X 1 > a, log p 2 log p 3 S(2k)Y RYL A. q Q 2k [X 1,X 1 +R] 2k a (q) p 2 +p 3 =2k p 2 [X 2,X 2 +Y ]

19 Putting an AP-condition on 2k Using both lemmas, we deduce as above: Theorem 3: Let Y 1/3+ε R Y X 1 X 2 Y 0 and X 3/5+ε 2 Y X 2. Fix a N. Then, for X 1 > a, log p 2 log p 3 S(2k)Y RYL A. q Q 2k [X 1,X 1 +R] 2k a (q) p 2 +p 3 =2k p 2 [X 2,X 2 +Y ] If in addition Q R 1/2 L D, the same estimate holds with max a (q) inserted after q Q.

20 Another ternary corollary Again, we deduce for the ternary Goldbach problem in the same way as above (with Y 1 = R and Y 2 = Y ): Corollary 2: Let n be odd, Y 1/3+ε 2 Y 1 Y 2 n X 1 Y 1 X 2 Y 2 0 and X 3/5+ε Y i i for i = 1, 2. Let n X 1 + Y 1 + X 2 + Y 2 be odd. Then, for Q Y 1 X 1/2 1 L B, q Q max a (q) p 1 +p 2 +p 3 =n p i [X i,x i +Y i ], i=1,2 p 1 a (q) log p 1 log p 2 log p 3 T(n, q, a)y 1 Y 2 Y 1 Y 2 L A.

21 Binary mean value theorems and ternary consequences Corollaries using Sieve Theory The limits when putting conditions on the primes

22 Corollaries Corollary 3: For all but YL A even integers 2k 2 (6) with k [X 1, X 1 + Y ], the equation 2k = p 2 + p 3 is solvable in primes p 2, p 3 such that p = P 3, p 2 [X 2, X 2 + Y ], where X θ 2 = Y with θ 0.861, and X 2 + Y X 1, X 2/3+ε 1 Y.

23 Corollaries Corollary 3: For all but YL A even integers 2k 2 (6) with k [X 1, X 1 + Y ], the equation 2k = p 2 + p 3 is solvable in primes p 2, p 3 such that p = P 3, p 2 [X 2, X 2 + Y ], where X θ 2 = Y with θ 0.861, and X 2 + Y X 1, X 2/3+ε 1 Y. Proof: Theorem 9.3 in [Halberstam/Richert], using Theorem 1 above.

24 Corollaries Corollary 3: For all but YL A even integers 2k 2 (6) with k [X 1, X 1 + Y ], the equation 2k = p 2 + p 3 is solvable in primes p 2, p 3 such that p = P 3, p 2 [X 2, X 2 + Y ], where X θ 2 = Y with θ 0.861, and X 2 + Y X 1, X 2/3+ε 1 Y. Proof: Theorem 9.3 in [Halberstam/Richert], using Theorem 1 above. Corollary 4: Let Y be large and consider an odd integer n 1 (6) with n X 1 + X 2 + 2Y. Then the equation n = p 1 + p 2 + p 3 is solvable in primes p 1, p 2, p 3 such that (p 1 + 2)(p 2 + 2) = P 9, where p i [X i, X i + Y ], X θ i = Y, with θ i i 0.933, i = 1, 2.

25 Corollaries Corollary 3: For all but YL A even integers 2k 2 (6) with k [X 1, X 1 + Y ], the equation 2k = p 2 + p 3 is solvable in primes p 2, p 3 such that p = P 3, p 2 [X 2, X 2 + Y ], where X θ 2 = Y with θ 0.861, and X 2 + Y X 1, X 2/3+ε 1 Y. Proof: Theorem 9.3 in [Halberstam/Richert], using Theorem 1 above. Corollary 4: Let Y be large and consider an odd integer n 1 (6) with n X 1 + X 2 + 2Y. Then the equation n = p 1 + p 2 + p 3 is solvable in primes p 1, p 2, p 3 such that (p 1 + 2)(p 2 + 2) = P 9, where p i [X i, X i + Y ], X θ i = Y, with θ i i 0.933, i = 1, 2. Proof: Theorem 10.3 in [Halberstam/Richert], using Corollary 1 (beginning with the result of [Kawada]) above.

26 A similar result using a counting argument Corollary 5: Let X θ 1 1 = Y = X θ 2 2 be large, where θ and θ Let n be an odd integer n 1 (6) with X 1 + X 2 + 2Y n < Y 3/2 ε. Then the equation n = p 1 + p 2 + p 3 is solvable in primes p 1, p 2, p 3 such that p = P 2, p = P 3 and p i [X i, X i + Y ], i = 1, 2.

27 A similar result using a counting argument Corollary 5: Let X θ 1 1 = Y = X θ 2 2 be large, where θ and θ Let n be an odd integer n 1 (6) with X 1 + X 2 + 2Y n < Y 3/2 ε. Then the equation n = p 1 + p 2 + p 3 is solvable in primes p 1, p 2, p 3 such that p = P 2, p = P 3 and p i [X i, X i + Y ], i = 1, 2. Proof: Counting argument using Corollary 3 and the result of [Wu 2004] on the number of Chen primes (= primes p with p + 2 = P 2 ) in short intervals [X, X + X θ ] with θ

28 A similar result using a counting argument Corollary 5: Let X θ 1 1 = Y = X θ 2 2 be large, where θ and θ Let n be an odd integer n 1 (6) with X 1 + X 2 + 2Y n < Y 3/2 ε. Then the equation n = p 1 + p 2 + p 3 is solvable in primes p 1, p 2, p 3 such that p = P 2, p = P 3 and p i [X i, X i + Y ], i = 1, 2. Proof: Counting argument using Corollary 3 and the result of [Wu 2004] on the number of Chen primes (= primes p with p + 2 = P 2 ) in short intervals [X, X + X θ ] with θ Corollary 5 cannot be deduced from Corollary 4 since >

29 Another similar result with dierent interval lengths Corollary 6: Let X θ 1 1 = Y 1 be large with θ , let X θ 2 2 = Y 2 with θ 2 > 3/5, and let Y 1/3+ε 2 Y 1 Y 2. Let n be an odd integer n 1 (6) with Y 2 n X 1 Y 1 X 2 Y 2 > 0. Then the equation n = p 1 + p 2 + p 3 is solvable in primes p 1, p 2, p 3 such that p = P 3 and p i [X i, X i + Y i ], i = 1, 2.

30 Another similar result with dierent interval lengths Corollary 6: Let X θ 1 1 = Y 1 be large with θ , let X θ 2 2 = Y 2 with θ 2 > 3/5, and let Y 1/3+ε 2 Y 1 Y 2. Let n be an odd integer n 1 (6) with Y 2 n X 1 Y 1 X 2 Y 2 > 0. Then the equation n = p 1 + p 2 + p 3 is solvable in primes p 1, p 2, p 3 such that p = P 3 and p i [X i, X i + Y i ], i = 1, 2. Proof: Theorem 9.3 in [Halberstam/Richert], using Corollary 2 (beginning with the result of [Perelli/Pintz]) above.

31 Binary mean value theorems and ternary consequences Corollaries using Sieve Theory The limits when putting conditions on the primes

32 A conjectured generalization

33 A conjectured generalization All theorems so far can be deduced by the following single estimate, what we state as Conjecture: 2r [X 1,X 1 +R] q Q max a(q) p 2 [X 2,X 2 +Y ] p 2 a (q) p 2 +p 3 =2r log p 2 log p 3 S(2r, q, a)y for Q Y 1/2 L B, Y θ R Y and some 0 < θ < 1. (Also for a xed a N with weaker restriction on Q.) RY L A

34 A conjectured generalization All theorems so far can be deduced by the following single estimate, what we state as Conjecture: 2r [X 1,X 1 +R] q Q max a(q) p 2 [X 2,X 2 +Y ] p 2 a (q) p 2 +p 3 =2r log p 2 log p 3 S(2r, q, a)y for Q Y 1/2 L B, Y θ R Y and some 0 < θ < 1. (Also for a xed a N with weaker restriction on Q.) The same should be true in the variant p 2 p 3 = 2r. RY L A

35 A conjectured generalization All theorems so far can be deduced by the following single estimate, what we state as Conjecture: 2r [X 1,X 1 +R] q Q max a(q) p 2 [X 2,X 2 +Y ] p 2 a (q) p 2 +p 3 =2r log p 2 log p 3 S(2r, q, a)y for Q Y 1/2 L B, Y θ R Y and some 0 < θ < 1. (Also for a xed a N with weaker restriction on Q.) The same should be true in the variant p 2 p 3 = 2r. This is an estimate in the style of Bombieri-Vinogradov's theorem. RY L A

36 A BDH-variant of the conjecture is true Balog, Cojocaru, David (2010): Koblitz' Conjecture is true on average.

37 A BDH-variant of the conjecture is true Balog, Cojocaru, David (2010): Koblitz' Conjecture is true on average. They prove and apply a Barban-Davenport-Halberstam-variant of the stated conjecture (where max a is replaced by a ).

38 A BDH-variant of the conjecture is true Balog, Cojocaru, David (2010): Koblitz' Conjecture is true on average. They prove and apply a Barban-Davenport-Halberstam-variant of the stated conjecture (where max a is replaced by a ). But the stated conjecture is open!

39 A BDH-variant of the conjecture is true Balog, Cojocaru, David (2010): Koblitz' Conjecture is true on average. They prove and apply a Barban-Davenport-Halberstam-variant of the stated conjecture (where max a is replaced by a ). But the stated conjecture is open! The major arc contribution can be shown to be admissible using standard techniques.

40 A BDH-variant of the conjecture is true Balog, Cojocaru, David (2010): Koblitz' Conjecture is true on average. They prove and apply a Barban-Davenport-Halberstam-variant of the stated conjecture (where max a is replaced by a ). But the stated conjecture is open! The major arc contribution can be shown to be admissible using standard techniques. But the minor arc estimate seems to be dicult!

41 A BDH-variant of the conjecture is true Balog, Cojocaru, David (2010): Koblitz' Conjecture is true on average. They prove and apply a Barban-Davenport-Halberstam-variant of the stated conjecture (where max a is replaced by a ). But the stated conjecture is open! The major arc contribution can be shown to be admissible using standard techniques. But the minor arc estimate seems to be dicult! Further open: What about extra conditions on the second prime in binary theorems? Or the third prime in ternary theorems?

42 Thank you!

ON SUMS OF PRIMES FROM BEATTY SEQUENCES. Angel V. Kumchev 1 Department of Mathematics, Towson University, Towson, MD , U.S.A.

INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 8 (2008), #A08 ON SUMS OF PRIMES FROM BEATTY SEQUENCES Angel V. Kumchev 1 Department of Mathematics, Towson University, Towson, MD 21252-0001,