Math 597e Primes, Spring 2008, Solutions 4 Throughout we suppose that for each prime p, 0 (p) < p and that F (Q) = 1.
m qQ
(q)2
(p) p|q p(p) . r s(r)Q ph r (p) p h
(a) By considering
r,(r)=m s(r)Q ph
MATH 597E PRIMES, SPRING 2008, PROBLEMS 6
Due Tuesday 4th March Let S(x) =
1 nx (n) , 1 (n)
U (x) =
1 n
px
d(p + 1) and suppose that x 2.
1. (i) Prove that
=
(ii) Prove that S(x) =
2
(m)2 m|n (m) . (m
Math 597e Primes, Spring 2008, Problems 7 Due Tuesday 18th March
1 This and the next homework prove that lim inf pn+1 pn 2 (Bombieri & Davenport, 1965). log pn Let f () be as in the lectures, let Q =
MATH 597E PRIMES, SPRING 2008, PROBLEMS 5
Due Tuesday 26th February Let A be xed and positive, N be large, Q = (log N )B with B = B(A), and M(q, a), M, m, f , S(h, Q), S(h) be as in the lectures. Assu
Math 597e Primes, Spring 2008, Problems 4 Due Tuesday 19th February Throughout we suppose that for each prime p, 0 (p) < p and that F (Q) = 1. (Vaughan 1973). (a) By considering
(p) p(p) qQ
(q)2
(p) p
MATH 597E PRIMES, SPRING 2008, PROBLEMS 2
Due Tuesday 5th February 1. Let Q be a set of pairwise coprime positive integers not exceeding Q, suppose M +N M +N that T (x) = n=M +1 cn e(nx), and that Z(q
MATH 597E PRIMES, SPRING 2008, PROBLEMS 3
Due Tuesday 12th February Let S() = 1. Recall that
M +N n=M +1
an (n), T () =
q
M +N n=M +1
an e(n).
(x)e(xn/q) = (n) ()
x=1
holds for all modulo q when (n, q
MATH 597E PRIMES, SPRING 2008, SOLUTION 1
1. (The Larger Sieve of Gallagher [1971]) Let N be a subset of Z of the integers in an interval [M + 1, M + N ], and let Z(q, h) denote the number of n N such
MATH 597E PRIMES, SPRING 2008, SOLUTIONS 2
1. Let Q be a set of pairwise coprime positive integers not exceeding Q, suppose M +N M +N that T (x) = n=M +1 cn e(nx), and that Z(q, h) = n=M +1,nh (mod q)
Math 597e Primes, Spring 2008, Solutions 7 Let f () be as in the lectures, let Q = N 1/2 (log N )B and P = (log N )D where B and D will be xed later. H Further let K() = | h=1 e(h)|2 = |h|H (H |h|)e(h
Math 597e Primes, Spring 2008, Solutions 8 This is the previous homework continued. Let f () be as in the lectures, let Q = N 1/2 (log N )B and P = (log N )D H where B and D will be xed later. Further
MATH 597E PRIMES, SPRING 2008, SOLUTIONS 6
Let S(x) =
1 nx (n) , 1 (n)
U (x) = =
1 n
px
d(p + 1) and suppose that x 2. (ii) Prove that S(x) =
(m)2 1 rx/m r . mx m(m) 2 (m)2 + m>x (m) log m and mx (m)
MATH 597E PRIMES, SPRING 2008, SOLUTIONS 5
Let A be xed and positive, N be large, Q = (log N )B with B = B(A), and M(q, a), M, m, f , S(h, Q), S(h) be as in the lectures. Assume any result from the le
Math 597e Primes, Spring 2008, Solutions 3 1. Recall that x=1 (x)e(xn/q) = (n) () holds for all modulo q when (n, q) = 1. (a) Show that if an = 0 whenever (n, q) > 1 then | ()|2 |S()|2 = q 2 (q) a=1 |
MATH 597E PRIMES, SPRING 2008, PROBLEMS 1
Due 29th January 1. (The Larger Sieve of Gallagher [1971]) Let N be a subset of Z of the integers in an interval [M + 1, M + N ], and let Z(q, h) denote the n