PMATH 440/640 ANALYTIC NUMBER THEORY

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1 PMATH 440/640 ANALYTIC NUMBER THEORY. Lecture: Monday, Setember, 000 NO TEXT References: Introduction to Analytic Number Theory. Tom Aostol, Sringer-Verlag, 976 An Introduction to the Theory of Numbers. G. H. Hardy and E. M. Wright, Oxford University Press (5th ed, 979 Marks: Final Exam 65% Midterm 5% Assignments 0% Is there a formula for nth rime n? =, = 3,... Yes, but such a formula is comlicated. For examle, is there a olynomial f Z[x] for which f(n = n? f(x = a n x n + + a x + a 0 f(a 0 = a n a n a a 0 + a 0 so a 0 f. Suose q is rime and f(n = q. Then q f(n + kq for each k Z +. So, in articular, we see that if f(m is rime for each ositive integer m, then f is a constant. In articular, f(x = q for some rime q. The olynomial n + n + 4 is rime for n = 0,,..., 39. Further, (n 40 + (n is rime for 0 n 79. These examles are connected with the fact that the ring of algebraic integers of the field Q( 63 is a Unique Factorization Domain. (Note that 63 is the largest squarefree integer D such that Q( D is a U.F.D. By using ideas of Matijȧsevic which were used to rove Hilbert s 0th roblem, one can find a olynomial f Z[a, b,...,z] such that the set of ositive values assumed by f as the variables run over the non-negative integers is the set of rime numbers. Is n + a rime for infinitely many n? Almost surely yes. The best result in this direction is that n + is a P for infinitely many integers n. A P is an integer which is the roduct of at most rimes.

2 PMATH 440/640 ANALYTIC NUMBER THEORY There is no olynomial of degree > which is known to be rime infinitely often. On the other hand, we can deal with degree. Let k and l be corime ositive integers. Then kn + l is rime for infinitely many ositive integers n. This is Dirichlet s Theorem. Theorem (Euclid. There are infinitely many rime numbers. Proof. Assume that there are finitely many rimes,..., n say, and consider m = n + Then m can be written as a roduct of rimes, and so k m for some k with k n. Then k m n, so k, which is a contradiction. Definition. For any real number x, let π(x denote the number of rimes x. We can estimate π(x from below using Euclid s roof. In articular, we ll show that k, the kth rime, satisfies k k for k =,...,n We rove this by induction. The result holds for k =, since = = 4. Assume the result holds for j k. Then, by Euclid s argument, k k + and so by our inductive assumtion, k k + k + k. Lecture: Wednesday, Setember 3, 000 Thus, given x let s be the integer satisfying ( s x s+ Then, since k k for k =,,... then we have π(x s. By (, log x < log s+, hence log( log x log < s + log( and so π(x log log x. Another way of roving such a lower bound for π(x is the following. Suose x. Then π(x ( = ( x + n du x x n x u log x Hence π(x log log x log log log x

3 PMATH 440/640 ANALYTIC NUMBER THEORY 3 Fermat ( conjectured that the numbers of the form n + for n = 0,,,... are rime. He checked it for n = 0,,, 3, 4. These are known as the Fermat numbers and are denoted by F n, so F n = n + Euler in 73 roved that 64 F 5. It is known that F 6,...,F are comosite. Quite likely F,... are comosite. Almost certainly only finitely many Fermat numbers are rime. Theorem (Polya. If n and m are ositive integers with n < m then (F n, F m =. Proof. Let m = n + k with k. We first show that F n F m. We have F m = ( n+k + = n+k. Put x = n. Then F m = xk F n x + = x xk k + Z Thus we have F n F m. Suose that d F n and d F m. Then d but F n and so d = ±. The result follows. Thus we obtain another roof of Euclid s Theorem and we see that Theorem 3. For x, and for n, π(x n < n log x log( n < 4 n Proof. Let x with x Z. Let =,..., j be the rimes x. For each integer n with n x we can write n = n m where n is a ositive integer and m is squarefree, so not divisible by the square of a rime. Therefore, m = ǫ j ǫ j where ǫ i is in {0, } for each i =,...,j. Thus, there are at most j ossible values for m. There are at most x ossible values for n. Therefore, j x x and so ( j x Since j = π(x we see that π(x log log x

4 4 PMATH 440/640 ANALYTIC NUMBER THEORY amd the first art follows. Now take x = n so that π( n = n. By (, n n and so 4 n n. In 896 Hadamard and de la Vallée Poussin roved indeendently the Prime Number Theorem. That is, they roved that π(x lim x x = log x The result had been conjectured by Gauss. Remark. Let n be a ositive integer and let be a rime. Then the exact ower of dividing n! is log n log n n = k= k k= 3. Lecture: Friday, Setember 5, 000 k Theorem 4. For x, ( 3 log x 8 log x < π(x < (6 log x log x Proof. (Argument due to Erdős. Let us first rove the lower bound for π(x. Note that ( n n is an integer and that ( n ( = (n! n (n! r where r satisfies the inequality r n < r+. To see this note the exact ower of dividing (n! is r n <n and the exact ower dividing n! is k k= r n So the exact ower of dividing ( n n is r ( n In articular, by (, k= k k= k n r k ( n r (n π(n n n

5 Note that ( n n and so n (n π(n hence PMATH 440/640 ANALYTIC NUMBER THEORY 5 ( log π(n n log(n x Note that is increasing for x e and so for x 6, let n be such that n x n +, then log x ( ( log n log 3 π(x π(n log(n x 4 3 log x log( 3 > x 8 log x 4 We can then check that the result holds for 6 x n. Now the uer bound. Note that Thus On the other hand, Therefore, Take n = k. Then Therefore, and so n n n n n n ( n n < ( + n = n n π(n π(n π(n log(n π(n log(n/ < log(n + log(π(n < (3 log n π( k+ log( k π( k log( k < (3 log k π( k log( k π( k log( k < (3 log k. π(8 log(4 π(4 log( < (3 log 4 π( k+ log( k < (3 log ( k + k π(4 log( < (3 log k+ ( π( k+ k+ < (3 log log( k Thus, given x choose k so that k x k+. Then π(x π( k+. Therefore, π(x (3 log k+ log( k for x > e. We check for x e. ( ( k x (6 log (6 log log( k log x In 845 Bertrand showed that there is always a rime in the interval [n, n] for n Z + rovided that n < He conjectured this always holds. Chebyshev roved this in 850.

6 6 PMATH 440/640 ANALYTIC NUMBER THEORY Theorem 5. For all n Z +, 4. Lecture: Monday, Setember 8, 000 < 4 n n Proof. We ll rove the result by induction. The claim is certainly true for n = or n =. Suose the result is true for k n. We first remark that we can restrict our attention to the case where n is odd, since if n is even, and n >, then = n n and the result follows by induction. We write n = m +, and we consider ( m+ m. In articular, ( m + m m+< m+ Note that ( ( m+ m and m+ m+ occur in the binomial exansion of ( + m+, and ( ( m+ m = m+ m+. Thus, ( Now, m+ = ( ( m + m (m+ = 4 m m+ ( m+< m+ 4 m+ 4 m = 4 m+ by our inductive assumtion and (. The result now follows by induction. Theorem 6. If n 3 and is a rime with 3 n < n then ( n n. Proof. Since n 3 we see that if is in the range n < n then >. Then and are the 3 only multiles of with n and so (n! (Here b means 3 b and b. Further, n!, hence (n! since n < n. 3 Since ( n n = (n! we see that ( n (n! n. Theorem 7. For each ositive integer n, there is a rime with n < n.

7 PMATH 440/640 ANALYTIC NUMBER THEORY 7 Proof. (This roof due to Erdős. Note that the result holds for n =,, 3. Assume the result is false for some integer n with n 4. Then, by Theorem 6, every rime which divides ( n n is n. 3 Let be such a rime and suose that α ( n n. Then, as in the roof of Theorem 4, α r, so α r n. If α then α n, so n. Thus ( ( n (n π( n (n n n since π(x x. Further, by Theorem 5, ( n ( n 4 n/3 (n n n 3 n Note that ( n n is the largest of the n + terms in the binomial exansion of ( + n hence ( n (3 n n n + By ( and (, 4 n n + 4n/3 (n n 4 n/3 (n n (n + < (n n+ Taking logarithms, we find that n 3 log 4 < ( n + log(n Notice that if n = 5 the inquality is false. By calculus, the claim holds for n 5. Finally,, 3, 5, 7, 3, 3, 43, 83, 63, 37, 557 are all rimes and so the result is general. 5. Lecture: Wednesday, Setember 0, 000 What can we say about differences between consecutive rimes? Let ( =,,... be the sequence of rimes. By Theorem 7, n+ n or n+ n n. By robabalistic reasoning, Cramer was in 936 led to conjecture that ( n+ n lim su n (log n The best uer bound for n+ n is due to Baker and Harman. They roved that, for n sufficiently large, n+ n < n.

8 8 PMATH 440/640 ANALYTIC NUMBER THEORY What about small gas between consecutive rimes? We conjecture that n+ n = for infinitely many integers n. If we assume the rimes are randomly distributed and an integer x is rime with robability then we could exect x and x + to be rime with robability. log x (log x x Thus we exect about rimes with + rime and x. A more careful heuristic suggests (log x that there are about C x such rimes with C > 0 and C. In the 60 s, Chen roved that (log x x there are more than 0.6 rimes with x such that + is a P (log x, rovided that x is sufficiently large. Surrisingly, it is not known that n+ n lim inf = 0 n log n The best result in this direction is due to Maier who roved in 988 that lim inf n n+ n log n < 4 In 955 Ricci roved that the set of accumulation oints of { n+ n log n The only oint of accumulation that is actually known is. In the 930 s Erdős roved that for infinitely many integers n, log log n n+ n > c log n (log log log n for a ositive constant c. In 938 Rankin added a factor of log log log log n. : n Z + } has ositive measure. Conclusion: We have much work to do. We don t even know many rimes. In fact, < 4 is a known rime Primes of the form where is a rime are known as Mersenne rimes and they are easy to test for rimality. The largest rime known is the Mersenne rime 6,97,593 (Hajratwala Lecture: Friday, Setember, 000 Let a,...,a n be distinct ositive integers. For m Z + we define ν(m to be the number of distinct residue classes modulo m occuied by the integers a,...,a k. Thus ν(m = the cardinality of {a i + mz : i =,...,k} In 93, Hardy and Littlewood conjectured that if ν( < for all rimes then there exist infinitely many integers n such that the integers n + a, n + a,...,n + a k are all rimes. (e.g. a = 0, a gives the twin rime conjecture. The conjecture is known as the k-tule conjecture. Hardy and Littlewood also conjectured π(x + y π(y π(x for all x >, y >

9 PMATH 440/640 ANALYTIC NUMBER THEORY 9 Hensley and Richards roved, about 30 years ago that these conjectures are incomatible. At least one is false. Definition. We introduce the symbols O, o,. Let f and g be functions from Z + or R + to R, and suose g mas to R +. (i f = O(g means that there exist ositive numbers c and c such that for x > c, f(x c g(x. f(n (ii f = o(g means that lim n = 0. g(n f(n (iii f g means that lim n =. (read: f is asymtotic to g. g(n Observe 0x = O(x, sin(x = O(, x = O(x, x = o(x, sin(x = o(log x, x+ x, x+ x x. By the Prime Number Theorem, π(x x log x or equivalently, π(x = x log x + o( x Let ǫ > 0. Then the number of rimes in [x, ( + ǫx] is Note Therefore, ( + ǫx log(( + ǫx = π(( + ǫx π(x = ( + ǫx log x + log( + ǫ = log x ( + ǫx log(( + ǫx x log x + o( x log x ( + ǫx (log x( + log(+ǫ log x ( + ǫx = log x ( + ǫx π(( + ǫx π(x = log x x log x + o( x = ǫx log x log x + o( x Note that we can take ǫ =. Then π(x π(x = x + o( x. log x log x But π(x = x + o( x. (Should this be worrying? log x log x + o(( x log x log x Definition. For any integer n we define Λ(n by the rule { log if n = Λ(n = k for some k Z + 0 otherwise Also define, for x R, θ(x = ( log = log x x and ψ(x = log = Λ(n k x n x

10 0 PMATH 440/640 ANALYTIC NUMBER THEORY Notice that ψ(x = log x log log x Also observe that since x is equivalent to x, 3 x is equivalent to 3 x, then we have ψ(x = θ(x + θ( x + θ(x /3 + Note that θ(x m = 0 rovided m > log x. log Thus log x log ψ(x = θ(x k Observe that θ(x = O(x logx, hence Therefore, k k= θ(x k = O(x (log x ( ψ(x = θ(x + O(x (log x

11 PMATH 440/640 ANALYTIC NUMBER THEORY 7. Lecture: Monday, Setember 5, 000 By Theorem 4, θ(x < c x and so by (, ψ(x < c x. Further, by the roof of Theorem 4, ( n log( n log log n log ψ(n n log n so that for x, we have ψ(x > c 3 x, hence θ(x > c 4 x. Here c, c, c 3, c 4 are ositive constants. What is the link between θ(x, ψ(x, and π(x? Note that θ(x π(x log x, thus ( π(x θ(x log x > c x 4 log x Theorem 8. π(x θ(x log x ψ(x log x Proof. Since ψ(x = θ(x + O(x / (log x and θ(x > c 4 x, we see that θ(x ψ(x. In articular, θ(x ψ(x θ(x and so it suffices to show that π(x. log x log x log x By (, π(x θ(x log x, so π(x log x lim inf x θ(x We need an uer bound for π(x in terms of θ(x. Note that for any δ > 0, we have θ(x = log log(x ( δ x ( δ(log x(π(x π(x ( δ Thus, so Therefore, x δ x θ(x + ( δx ( δ log x ( δ(log xπ(x θ(x ( δ log x + x( δ π(x δ + x( δ log x π(x log x θ(x θ(x Given ǫ > 0, we can choose δ > 0 so that < + ǫ and then choose x δ 0 so that if x > x 0, x ( δ log x θ(x < ǫ

12 PMATH 440/640 ANALYTIC NUMBER THEORY since θ(x > c x for x. Then, for x > x 0, and so our result follows. π(x log x θ(x < + ǫ Lemma (Abel s summation formula. Let {a n } n= be a sequence of comlex numbers. Let f be a function from {x R : x } to C. For x R we introduce A(x = n x a n If f is continuous for x, a n f(n = A(xf(x n x x A(uf (udu Proof. Put N = x. Then a n f(n = A(f( + (A( A(f( + + (A(N A(N f(n n N = A((f( f( + A((f( f(3 + + A(N (f(n f(n + A(Nf(N Next observe that if i is a ositive integer and t is a real number with i t < i+ then A(t = A(i. Thus Accordingly, A(i(f(i f(i + = n N i+ i A(uf (udu N a n f(n = A(uf (udu + A(Nf(N This gives us the result when x is an integer. Finally note that A(t = A(N for x t N. Hence x N and so our result follows. A(uf (udu = A(x(f(x f(n = A(xf(x (A(Nf(N Definition. We define Euler s constant γ by Note γ = Lecture: Wednesday, Setember 7, 000 γ = t t dt t It is not known but it is conjectured that γ is irrational and indeed that γ is transcendental.

13 Theorem 9. PMATH 440/640 ANALYTIC NUMBER THEORY 3 n x Proof. Take a n = and f(t = t. Then A(x = n x a n = n x = x. By Abel s summation formula, n x n = log x + γ + O( x n = x x x + x u u du x (x x u (u u = + du x u x = + O( + du x x u u u du u ( = + O( + log x u u u u du du x u x u = log x + γ + O( + u u du x = log x + γ + O( x + O( x x u u du = log x + γ + O( x Theorem 0. n x Λ(n n = log x + O( Proof. Aly Abel s summation formula with a n = and f(n = log n. Then x u log n = x log x u du n x u (u u = (x (x x log x du u x u u = x log x + O(logx (x + du u ( = x log x x + O(log x Also we have log n = log( x! = ( x log k n x x k= = x log = x Λ(n n m m x n x

14 4 PMATH 440/640 ANALYTIC NUMBER THEORY = n x But n x Λ(n = ψ(x = O(x and so x n Λ(n ( x x n Λ(n n n x = x n x Λ(n n O( n x Λ(n By (, hence Thus log n = x n x n x Λ(n n x log x x + O(log x = x n x x n x n x Λ(n n O(x Λ(n n = x log x + O(x O(x Λ(n n = log x + O( Theorem. Proof. Note that But, m m x The result follows. log m x x log log = n x = log x + O( Λ(n n = log x + O( m m ( + + log 3 m x m x log m Theorem. There exists a real number β such that = log log x + β + O( log x x log m log ( log n n(n = O( n=

15 Proof. We aly Lemma with and with f(n = log n. PMATH 440/640 ANALYTIC NUMBER THEORY 5 a n = { log if is a rime and n = 0 otherwise Then on utting A(x = n x a n, we have = A(x log x + By Theorem, we have Therefore, where τ(u = O(. Thus, x x x A(x = x log x = + O( log x + x A(u u(log u du = log x + O( log u + τ(u du u(log u = + O( log x + log log x log log + x τ(u = log log x + log log + u(log u du x = log log x + β + O( log x In fact, β = γ + ( log( + = τ(u u(log u du τ(u u(log u du + O( log x 9. Lecture: Wednesday, Setember 9, 000 To rove the Prime Number Theorem we introduce the Riemann zeta function, ζ(s. Definition. For s C with Re(s >, we define ζ(s by ζ(s = n s We write s = σ + it with σ, t R. The series n= converges absolutely for Re(s > and we have the Euler roduct reresentation n s for Re(s > given by ( n= ( s = n= n s

16 6 PMATH 440/640 ANALYTIC NUMBER THEORY To see this, note that ( s = ( + + s + s and a tyical term is α s = α k k s ( α k α k s and by the Fundamental Theorem of Arithmetic ( holds. Notice that ( allows us to give another roof that there are infinitely many rimes. If there were only finitely many, then ( is bounded as we let s tend to from above on s the real line, whereas n= diverges. This argument goes back to Euler. n Theorem 3. ζ(s can be analytically continued to Re(s > 0 with s. It is analytic excet at the oint s = where it has a simle ole with residue. Proof. For Re(s > we have ζ(s = n=. n s By Lemma, with a n = and f(x =, we find that x s Letting x we find that for Re(s >. Note that n x n s = x x s + s ζ(s = 0 + s = s = s u du u s ( s+ u s = s s s = s s s x u u s+du u u s+du u (u u du u s+ u u du u s+ u u du u s+ u u du u s+ u u du u s+ converges for Re(s > 0 and reresents an analytic function. s Therefore, s u u du reresents an analytic function for Re(s > 0, s and so gives an s u s+ analytic continuation of ζ(s to the region. s Note that has a simle ole of residue at s =. s

17 PMATH 440/640 ANALYTIC NUMBER THEORY 7 Theorem 4. ζ(s has no zeroes in the region Re(s. Proof. If Re(s >, then since ( converges, then ζ(s 0. s Recall s = σ + it with σ, t R. Let σ >. For all t R, log (ζ(σ + it = ( ( log = ( s n s n= where log denotes the rincial branch of the logarithm and log denotes some branch of the logarithm. We now consider the real art of both sides of the above equality to get log ζ(σ + it = σn cos(nt log n n= since int = e intlog( = cos( nt log + i sin( nt log = cos(nt log + i sin(nt log So Re( int = cos(nt log. We aeal to the inequality We then deduce hence that In articular, 0 ( + cosθ = ( + cosθ + cos θ n= σn n = + 4 cosθ + cos θ = cosθ + ( cos θ = cosθ + cos(θ (3 + 4 cos(nt log + cos(nt log 0 log ζ(σ 3 + log ζ(σ + it 4 + log ζ(σ + it 0 ( ζ(σ 3 ζ(σ + it ζ(σ + it for σ > and t R. Suose that + it 0 is a zero of ζ(s. Then t 0 0 since ζ(s has a ole at s =. Note that as σ from the right, then ζ(s = O((σ since is a simle ole of ζ(s. Also since + it 0 is a zero of ζ(s, then ζ(σ + it 0 = O(σ as σ from the right.

18 8 PMATH 440/640 ANALYTIC NUMBER THEORY Finally, ζ(σ + it 0 = O( as σ from the right, since + it 0 is not a ole of ζ(s. Therefore, ζ(σ 3 ζ(σ + it ζ(σ + it = O((σ 3 O((σ 4 O( = O(σ Thus ζ(σ 3 ζ(σ + it ζ(σ + it tends to 0 as σ which contradicts (. Therefore,ζ(s has no zero on Re(s = and the result follows.

19 PMATH 440/640 ANALYTIC NUMBER THEORY 9 0. Lecture: Monday, October, 000. Lecture: Wednesday, October 4, 000 Theorem 5 (Donald J. Newman. Suose a n C with a n for n =,,.... Form the series a n n= n s. The series converges to the analytic function F(s for Re(s >. If F(s can be analytically continued to Re(s then a n n= n s converges to F(s for Re(s. Proof. Let w C with Re(w. Thus F(z + w is analytic for Re(z 0. Choose R and determine δ = δ(r > 0 so that F(z + w is analytic on the region {z : δ Re(z and z R}. Let M denote the maximum of F(z + w on the region. Let Γ denote the contour obtained by following the outside of the region in a counterclockwise ath. Let A be the art of Γ in Re(z 0 and B be the remainder of Γ. By Cauchy s residue theorem, for any N Z +, ( ( πif(w = F(z + wn z z + z dz R Γ Now on A, F(z + w is equal to its series and we may slit the series as S N (z + w = N n= and R N (z + w = F(z + w S N (z + w. Again by Cauchy s residue theorem, ( πis N (w = C S N (z + wn z ( z + z R dz where C is the contour given by the ath z = R taken in counterclockwise direction. (Note that S N (z + w is analytic for z C. Note that C = A ( A {ir, ir}. Thus πis N (w = A ( S N (z + wn z x + z dz + R A S N (z + wn z ( z + z R dz In the second integral change, change variables z z. We see that ( S N (z + wn z z + z ( dz S R N (w zn z z + z dz R A A a N n z+w

20 0 PMATH 440/640 ANALYTIC NUMBER THEORY Therefore, (3 πis N (w = From ( and (3, (4 πi(f(w S N (w = A A ( (S N (z + wn z + S N (w zn z ( (R N (z+wn z S N (w zn z We want to show that S N (w F(w as N. So let s get down to it. n=n+ z + z R z + z R Observe that if we write z = x + iy with x, y, R then for z A, (5 z + z R = x R (6 R N (z + w n x+ N u x+du = xn x So S N (w z N n x N x + n= N (7 S N (w z N x ( N + x dz ( dz+ F(z+wN z B z + z dz R u x du N x + Nx x Therefore from (5, (6, (7, ( (R N (z + wn z S N (w zn z A z + z dz R ( ( N x A x N + x Nx N + x N x x R dz ( A x + ( dx 4 N R dz = A R + x dz NR A ( 4 = πr R + = 4π NR R + π N We now estimate the integral over B. Observe that z + z R = z z z + zz ( R + z δ R δ for Re(z = δ, z R. ( 4 R + NR dz

21 PMATH 440/640 ANALYTIC NUMBER THEORY Therefore since F(z + w is at most M we see that ( F(z + wn z B z + z R dz R M R δ N δ dz + 4MR + 8M 0 δn δ R xn x dx δ 4MR + 8Mδ ( δn δ R log N N δ log N 4RM + 8Mδ δn δ R log N Thus 0 πi(f(w S N (w 4π R + π N + 4RM + 8Mδ δn δ R log N F(w S N (w R + N + RM δn δ + δ Mδ R log N MN x x R dx Given ǫ > 0, choose R = 3 ǫ. Then for N sufficiently large, F(w S N(w < ǫ. Thue S N (w F(w as N. The result follows.. Lecture: Friday, October 6, 000 Definition. We now introduce the Möbius function µ : Z + {, 0, } under the rule that µ( = and µ(n = ( r if n is the roduct of r distinct rimes and µ(n = 0 otherwise. E.g. µ( = 0, µ(5 =, µ(30 =. Notice that for Re(s >, ζ(s = ( = s n= µ(n n s Theorem 6. (i Let n Z +, µ(k = k n { if n = 0 otherwise (ii Let f : R + C and define F : R + C by F(x = n x f( x n. Then f(x = n x µ(nf( x n. (iii Let f : Z + C and define F : Z + C by F(n = d n f(d. Then f(n = d n µ(df(n d.

22 PMATH 440/640 ANALYTIC NUMBER THEORY Proof. (i If n = the result is obvious. l l If n > we can write n = r r with,..., r distinct rimes and l,..., l r ositive integers. Then k n µ(k = k m µ(k where m = r. But notice that (ii By (i, (iii Again by (i, Theorem 7. µ(k = k m f(x = n x = k x µ(k f(n = c n = d n ( r + ( ( r r + + ( r = ( r = 0 r ( x µ(k f = n ( x k n kl xµ(kf kl l x k d µ(d c n c ( x f lk = ( x µ(kf k k x µ(d f(c = µ(df(c n d n= cd n f(c = ( n d d nµ(df µ(n n = 0 Proof. For Re(s > we have = µ(s ζ(s n=. n s It follows from Theorems 3 and 4 that (s ζ(s is analytic and non-zero in Re(s, hence ζ(s is analytic in Re(s. By Theorem 5, µ(n n= converges to n s ζ(s ζ(s has a ole at s = so = 0. ζ( for Re(s. In articular, it converges at s =. But Theorem 8. µ(n = o(x n x

23 PMATH 440/640 ANALYTIC NUMBER THEORY 3 Proof. Aly Lemma with a n = µ(n and f(x = x. Then n x µ(n = A(xx A(udu where n x ( A(t = n t µ(n n By Theorem 7, A(t = o(, so ( A(xx = o(x and (3 The result follows from (,(,(3. x A(udu = o(x

24 4 PMATH 440/640 ANALYTIC NUMBER THEORY 3. Lecture: Wednesday, October, 000 For any ositive integer n, let d(n denote the number of ositive integers which divide n. Theorem 9. n d(m = m= where γ is Euler s constant. n n = n log n + (γ n + O(n m / m= Proof. Let D n be the region in the uer right hand quadrant not containing the x or y axes and under and including the hyerbola xy = n. That is, D n = {(x, y R : x > 0, y > 0, xy n}. Every lattice oint in D n, in other words every oint with integer coordinates in D n, is contained in a hyerbola xy = s for some s with s n. Thus n s= d(s is the number of lattice oints in D n. Notice that this is equivalent to n n m= m since we may count the number of lattice oints with x coordinates,,..., n. Observe that to estimate the number of lattice oints in D n we first remark that the number above the line x = y is the same as the number below. By Theorem 9, ( = n n m m= n n = m m= n = n x= n x= n x= ( n x x + n ( n x x + O( + n ( n ( n + + O( n x = n(log n + γ + O( n (n + O( n + O( n Since n = n { n} where {x} denotes the fractional art of x for any x R, we have log( n = log( n { ( ( n n} = log { n} n = log( ( n + log { n} = log( n + O( n n

25 PMATH 440/640 ANALYTIC NUMBER THEORY 5 Thus, from (, we have as required. n n = n log n + (γ n + O( m n m= 4. Lecture: Friday, October 3, 000 Theorem 0 (Prime Number Theorem. π(x x log x Proof. By Theorem 8 it suffices to rove that ψ(x x. Put F(x = ( ( x x ψ + γ n n n x By Möbius inversion, Theorem 6 (ii, we have ψ(x x + γ = ( x µ(nf n n x It remains to show that n x µ(nf ( x n = o(x. To do this we first estimate F(x. We have F(x = ( x ψ n x + γ x n n x n x also ( x ψ = n Λ(m = Λ(n n x n x m x n x m x n n = x Λ(n = m x log k n x k x = ( x x + + log = log( x! = log n x n x and as in the roof of Theorem 0, log n = x log x x + O(log x hence n x n x ( x ψ = x log x x + O(log x n

26 6 PMATH 440/640 ANALYTIC NUMBER THEORY Further, by Theorem 9, Note that Thus Therefore, x n= x = x log x + (γ x + O(x / n x n= x n= x x x n n n= x + n= x + x = x log x + (γ x + O(x n / F(x = (x log x x + O(log x (x log x + (γ x + O(x / + (γx + O( = O(x / Thus there is a ositive constant c such that F(x < cx / for x Let t be an integer larger than. Then ( x µ(nf n n x n x t t x cx ( / du + u / n ( x F n n x t c x n / cx ( / + u / x t ( x / ( cx ( / + cx t t / Observe that F is a ste function. In articular, if a is an integer and a x < a + then F(x = F(a. Therefore, ( x µ(nf = F( x n µ(n + F( µ(n + + F(t t <n x x <n x x 3 <n x Thus so But x t <n x t ( x µ(nf x n t <n x F( µ(n + + F(t µ(n x <n x x t <n x t ( x µ(nf ( F( + + F(t max x n t <n x i t µ(n x t <n x t x i <n x i µ(n = n x i µ(n x i <n µ(n = o(x µ(n

27 PMATH 440/640 ANALYTIC NUMBER THEORY 7 Therefore, ( x µ(nf n x t <n x = o(x Thus given ǫ > 0 choose t so that c < ǫ. Then for x sufficiently large, t / ( x µ(nf n x t <n x < ǫ x And so by (, ( x µ(nf < ǫ n x + ǫ x = ǫx n x In articular, n x µ(nf( x n = o(x and so ψ(x x.

28 8 PMATH 440/640 ANALYTIC NUMBER THEORY 5. Lecture: Monday, October 6, 000 Definition. For any ositive integer n let Ω(n denote the number of rime factors of n counted with multilicity and let ω(n denote the number of distinct rime factors of n. Thus if n = , then Ω(n = = 8 and ω(n = 3. Definition. Let k Z + and for each real number x let τ k (x be the number of ositive integers n with n x and Ω(n = k. Further, let π k (x be the number of ositive integers n with n x and ω(n = Ω(n = k. In other words, π k counts the n s u to x which are squarefree and have k rime factors. Note that π(x = π (x = τ (x. Theorem (Landau,900. Let k be a ositive integer. Then x π k (x τ k (x (log log xk (k! log x Proof. We introduce the following functions L k (x = Θ k (x = Π k (x = k x k x k k x log( k where * signifies that the sum is taken over all k-tules of rimes (,..., k with k x. Note that different k-tules may corresond to the same roduct k. For each ositive integer n, let c n = (c n (k denote the number of k-tules (,..., k for which k = n. Note c n = 0 if n is not the roduct of k rimes and is equal to k! if n is squarefree and ω(n = k. Thus we have ( k!π k (x Π k (x k!τ k (x Note also that Π k (x = n x c n and Θ k (x = n x c n log n. For k note that the number of ositive integers u to x with k rimes factors and divisible by the square of a rime is τ k (x = π k (x.

29 Thus ( τ k (x π k (x PMATH 440/640 ANALYTIC NUMBER THEORY 9 k x i = j for some i j ( k = k x We shall rove that x(log log xk Π k (x k log x and then our result follows from ( and (. ( k Π k (x From Lemma with a n = and f(x = log x, we have Observe that Thus Π k (u = O(u and so Therefore, it suffices to rove that We ll rove this by induction on k. Θ k (x = n x c n log n = Π k (x log x + Π k (x k!τ k (x k!x x Θ k (x = Π k (x log x + O(x Θ k (x kx(log log x k for k Z + Π k (u u du Note that Θ (x = θ(x x by the Prime Number Theorem. Assume now that Θ k (x kx(log log x k for k with k. We ll rove it for k +. First note that for k, L k (x (log log x k, since k ( k L k (x and by Theorem, and x k x k ( x k x k (log log(x k k (log log x k

30 30 PMATH 440/640 ANALYTIC NUMBER THEORY Further, (log log(x k k = ( log It suffices to show that Θ k (x k(log log x k. ( k log x = (log log x log k k (log log x k k The result holds for k = by the Prime Number Theorem. We now make the inductive hyothesis that Θ k (x k(log log x k for k and we ll deduce it for Θ k+ (x. Recall also that L k (x (log log x k. Therefore Θ k+ (x (k + (log log x k = Θ k+ (x (k + L k (x + o(x(log log x k kθ k+ (x = (log( k+ + log( 3 k+ + + log( k k+ x Next, we ut L 0 (x = and note Thus, By inductive hyothesis, L k (x = = (k + k+ x = (k + x k x log k+ Θ k ( x k = x Θ k+ (x (k + L k (x = (k + x ( x L k ( ( ( x k Θ k x L k x Θ k (y kyl k (y = o(y(loglog y k Thus given ǫ > 0, there exists x 0 = x 0 (ǫ, k such that for y > x 0, Θ k(y ky L k (y ǫy(log log yk Further, there exists a ositive number c = c(ǫ, k such that for y x, Θ k(y ky L k (y c

31 PMATH 440/640 ANALYTIC NUMBER THEORY 3 Therefore, for x sufficiently large, Θ k+ (x (k + L k (x Thus, x x 0 < x c + x x 0 ǫ x k (log log x cx + ǫx(log log x k x x 0 cx + 4ǫx(log log x k < 5ǫx(log log x k for x suffiently large Θ k+ (x (k + L k (x = o(x(log log x k and this comletes the induction. The result follows. Theorem. and 6. Lecture: Wednesday, October 8, 000 ω(n = x log log x + β x + o(x n x Ω(n = x log log x + β x + o(x n x where β is the constant in Theorem and β = β + Proof. Put S = S(x = n x ω(n. Then Thus by Theorem, S = x x S = = n x n x By the Prime Number Theorem, or Theorem 6, Put S = S (x = n x Ω(n. Then S S =. ( x + O(π(x = x(log log x + β + o( + O(π(x S = x log log x + xβ + o(x m x,m x m = m x,m x m + O(x log x (The x arises since m so is at most x. The log x arises since m goes u to log x.

32 3 PMATH 440/640 ANALYTIC NUMBER THEORY Continuing, Thus, S S = x S S = x and the result follows. ( ( ( O(x / log x 3 m m x ( + o( + O(x / log x = x ( + o(x 7. Lecture: Friday, October 0, 000 Definition. Let A be a subset of Z +. For any n Z + ut A(n = {,,..., n} A. Define the uer density d(a of A by d(a = lim su n A(n n d(a is also known as the asymtotic uer density of A. Similarly, we define d(a the lower asymtotic density of A by A(n d(a = lim inf n n We say A has an asymtotic density d(a if d(a = d(a in which case we ut d(a = d(a. Examles: ( A set of rimes, d(a = d(a = 0. ( A = {n Z + : n 0 (mod 5} then d(a = d(a = d(a = 5. (3 A = {n Z + : n not of the form k + for k Z }, d(a = d(a = d(a =. (4 Take A = {a Z : (k! < a < (k +! for k Z}. Then d(a = 0 and d(a =. Definition. Let f(n and F(n be functions from Z + R. We say that f(n has normal order F(n if for each ǫ > 0 the set A(ǫ = {n Z + : ( ǫf(n < f(n < ( + ǫf(n} has the roerty that d(a(ǫ =. (Equivalently, if B(ǫ = Z + \A(ǫ then d(b(ǫ = 0. We say that f has average order F if Examles: n f(j j= n F(j j=

33 PMATH 440/640 ANALYTIC NUMBER THEORY 33 ( Take { if n k! for k Z + f(n = n if n = k! Then f has normal order but not average order. ( Put { for n (mod f(n = 0 for n 0 (mod Then f has average order but does not have normal order. (3 Put { log n + (log n for n (mod f(n = log n (log n for n 0 (mod Then f has both normal and average order. We have already roven that ω(n and Ω(n have average order log log n. We ll now rove that they have normal order log log n.

34 34 PMATH 440/640 ANALYTIC NUMBER THEORY 8. Lecture: Monday, October 3, 000 Theorem 3. Let δ > 0. The number of ositive integers n x with f(n log log n < (log log n +δ is o(x where f(n = ω(n or f(n = Ω(n. Proof. The first remark we make is that it is enough to rove that the number of ositive integers n with n x and f(n log log x < (log log x +δ is o(x since for x e n x, ( log x log log x log log n log = log log x e Secondly, we may restrict our attention to f(n = ω(n, since by Theorem, (Ω(n ω(n = O(x Thus the number of n x for which Ω(n ω(n > (log log n / is o(x. n x For each n x, we consider the ordered airs (, q where and q are distinct rime factors of n. There are ω(n choices for and then ω(n choices for q so ω(n(ω(n = = Thus q n, q n x ω(n n x ω(n = n x ω(n(ω(n = n x Observe that and Thus ( Next note that q n n x x = O(x x x x x q q + O(x q x x n n x ω(n n x ω(n = n x ω(n(ω(n = q x x ( x q x = x x q q n q x x ( q x x q + O(x

35 PMATH 440/640 ANALYTIC NUMBER THEORY 35 Further, ( x = (log log x + O(log log x and Thus ( x By Theorem, (3 Thus by (, (, and (3, Therefore, = (log log x + O( = (log log x log + O( = (log log x + O(log log x q x q = (log log x + O(log log x ω(n = O(x log log x n x ω(n = x(log log x + O(x loglog x n x n x(ω(n log log x = n x ω(n n x ω(n loglog x + n x (log log x = ((x log log x + O(x log log x log log x( ω(n + x (log log x n x = x(log log x + O(x log log x x(log log x + O(log log x + x(log log x + O(loglog x (4 = O(x log log x Let ǫ > 0. If there are more than ǫx integers n with n x for which ω(n log log x > (log log x +δ then (ω(n log log x > ǫx(log log x +δ n x This cannot hold for x sufficiently large, by (4, and so the result follows. Recall, for n Z +, d(n denotes the number ositive divisors of n. If a a n = r r where a,...,a r are ositive integers and,..., r are distinct rimes, then ω(n = r, Ω(n = a + + a r, and d(n = (a + (a r +

36 36 PMATH 440/640 ANALYTIC NUMBER THEORY Theorem 4. Let ǫ > 0. We have ( ǫ log log n (+ǫlog log n < d(n < for a set of ositive integers n with asymtotic density. Proof. Since for any a Z, The result now follows from Theorem 3. Recall that the average order of d(n is log(n. By Theorem 4, normally d(n satisfies for any ǫ > 0. ( + a a (log n log ǫ log +ǫ < d(n < (log n Definition. For any n Z +, let ϕ(n be the number of integers m with gcd(m, n =. ϕ(n is known as Euler s ϕ-function. Theorem (Euler s Theorem. Let a and b be ositive integers with gcd(a, n =. Then a ϕ(n (mod n Proof. Let c,...,c ϕ(n be a reduced residue system modulo n. Then ac,...,ac ϕ(n is also a reduced residue system modulo n. Thus c c ϕ(n (ac ac ϕ(n (mod n hence c c ϕ(n a ϕ(n c c ϕ(n (mod n and so a ϕ(n (mod n ( Fermat s Theorem follows from setting n =. Theorem (Wilson s Theorem. If is a rime then (! (mod. Proof. Consider x in Z Z [x]. It factors by Fermat s Theorem as x = (x (x (x ( in Z Z [x] since,,..., are roots. Considering the constant coefficient we find that so ( (! (mod. ( ( ( ( (mod If = the result holds since = ; otherwise is odd, so (! (mod as required.

37 PMATH 440/640 ANALYTIC NUMBER THEORY Lecture: Wednesday, October 5, 000 Wilson s Theorem was conjectured by Wilson ( He communicated the conjecture to Waring ( who ublished it in 770. Shortly after, Lagrange gave the first roof. In fact, Leibniz had conjectured the result in 68. Here is another roof due to Stern from 860. We have, for x <, ( log( x = log = x + x x + x3 3 + and so ex (x + x + x3 3 + = x = + x + x + But ex (x + x + x3 3 + = ex(x ex( x ex(x3 3 = ( ( ( + x + x! + x3 3! + + x + (x + + x3! 3 + (x3 3 +! ( = + x + + ( x +! 3! + + ( x ! + + x + In articular, the coefficient of x is of the form + r + where r and s are ositive integers with! s gcd(r, s = and gcd(s, =. But comaring coefficients in the ower series, we see that Therefore r s =! + s((!+. hence s r s =! + r s + =! +. Thus s r = s! + s, so (s r(! = s + s(! = Since (s r(! is an integer, we see that s((! +. But gcd(, s =. Therefore, (! +, as required. Definition. ( Let be a rime, and let a be an integer corime with. We define the Legendre a symbol by the rule: ( { a if x = a (mod has a solution if it has no solution ( a If = we say that a is a quadratic residue mod and otherwise a is a quadratic nonresidue mod.

38 38 PMATH 440/640 ANALYTIC NUMBER THEORY Theorem 5 (Euler s Criterion. Let be an odd rime and let a be and integer corime with. Then ( a a (mod Proof. The congruence ( x a (mod has at most solutions modulo. Suose that it has a solution (so that = and that b a (mod. a From b a (mod, we get Thus a ( a as required. a (b b (mod And suose that x a (mod has no solution. Slit the integers into airs (r, s with rs a (mod. Thus (! a (mod Note (! (mod by Wilson s Theorem, so a ( a (mod.

39 PMATH 440/640 ANALYTIC NUMBER THEORY Lecture: Monday, October 30, 000 ( a Extend the definition of the Legendre symbol to where is a rime and a by utting ( = 0 in this case. a Theorem 6. Let be an odd rime and let a and b be integers. Then ( ( ( a b ab = ( Also = ( (. Proof. The first art lainly holds if ab. So we may assume that a and b. By Euler s Criterion, (ab ( ( (ab (a a b (b (mod ( ( ( ab a b Since and are in {, } and is and odd rime, we see that ( ( ( a b ab = Similarly, by Euler s Criterion, ( and since is an odd rime ( ( ( (mod = ( (.. Lecture: Wednesday, November, 000 Theorem 7 (Gauss Lemma. Let be an odd rime and let a be an integer corime with. Let µ be the number of integers from {a, a,..., ( a} whose residues modulo of least absolute value are negative. Then ( a = ( µ E.g. take = 5 and a =. The set {, 4}. The residues mod 5 of least absolute value are,-. Thus µ = hence ( 5 = ( =. Proof. Relace the integers a,...,( a by their residues of least absolute value. Denote the ositive ones by r,...,r µ and the negative ones by s,..., s µ. Plainly no two r i s are equal and no two s j s are equal.

40 40 PMATH 440/640 ANALYTIC NUMBER THEORY Observe that if m a r i (mod and m a s j (mod with r i = s j then (m + m a 0 (mod, hence m + m +, which is not ossible if m t for t =,. Therefore, no r i is equal to an s j. In articular, r,..., r µ, s,...,s µ is a rearrangement of,...,. Thus ( ( a(a ( µ (mod and so a ( µ (mod. The result now follows from Euler s Criterion. Corollary. If is an odd rime, then ( = ( 8 Proof. We aly Gauss Lemma. We consider the number of integers from {,, 4,..., ( } whose residues of least absolute value are negative. ( µ = 4 If = 8k + then µ = 4k 8k+ 4 = 4k k 0 (mod. If = 8k + 3 then µ = 4k + k = k + (mod. If = 8k + 5 then µ = 4k + (k + = k + (mod. If = 8k + 7 then µ = 4k + 3 (k + = k + 0 (mod. Therefore, is a square modulo if ± (mod 8 and is a quadratic nonresidue if ±3 (mod 8.. Lecture: Friday, November 3, 000 Theorem 8 (Law of Quadratic Recirocity. If and q are distinct odd rimes, then ( ( q ( ( q ( q Euler had stated the law. Legendre attemted to rove it. Gauss gave 8 roofs. ( ( Proof. By Gauss Lemma, = ( µ and = ( ν, where µ is the number of integers from q q {q, q,..., ( q} whose residue mod of least absolute value is negative, and ν is the number of integers from {,,..., ( q } whose residue mod q of least absolute value is negative. It suffices to show that µ + ν = ( (q (mod. Given x with x we determine y such that < qx y <. Note that q x < y < q x so y is uniquely determined and that qx y is the residue mod of least absolute value of qx.

41 PMATH 440/640 ANALYTIC NUMBER THEORY 4 Note that y is nonnegative. If y = 0 there is no contribution to µ since qx 0. Further, if x =, then q x = q( = q ( < q hence y q (since y is an integer. Thus the number µ corresonds to the number of combinations of x and y from the sequences (A :,,..., (B :,,..., q resectively, such that < qx y < 0, or equivalently that 0 < y qx <. Similarly, ν is the number of combinations of x and y from the sequences (A and (B resectively, for which q > y qx < 0. For any other air (x, y with x from (A and y from (B, either y qx < q or y qx >. Let ρ be the number of airs (x, y for which the first ossibility holds, and λ the number of airs for which the second ossibility holds. Then ( ( q = µ + ν + ρ + λ As x and y run through (A and (B resectively, x = + x and y = q+ y run through (A and (B resectively, but in reverse order. And note y qx > if and only if y qx = ( q+ y q( + x = q (y qx < q. Further, y qx < y if and only if y qx = q (y qx >. Then λ = ρ. So ( Examles: ( q = µ + ν + λ µ + ν (mod ( What is ( 3 7? By the Law of Quadratic Recirocity, ( 3 7 But ( ( 7 3 = 4 ( 3 = so 3 7 =. ( What is ( ? 009 is rime and 73 = 3 3. So ( ( = By the Law of Quadratic Recirocity, ( ( = Now, ( ( 0 3 = 4 ( 5 ( 3 3 = 5 ( 3 and 5 ( 3 = 3 ( 5 = 3 5 Further ( ( 7 3 = 3 ( 7 = 4 ( 7 = ( ( 7 3 = ( ( 7 ( 3 =. ( ( = 0 3 =. and ( ( = 009 ( 3 = 7 3.

42 4 PMATH 440/640 ANALYTIC NUMBER THEORY Alying the result on the quadratic character of from our corollary to Gauss Lemma, we have ( ( 7 =, so 7 ( 3 = 7 ( 7 = 3 7 =. Thus ( =. (3 5 is a quadratic residue for all rimes of the form 0k ± and is a quadratic( nonresidue 5 for all rimes of the form 0k ± 3 since by the Law of Quadratic Recirocity, = ( 5, and and - are quadratic residues mod 5 while 3 and -3 are not. (4 The equation x 4 7y 4 = w has no solution in the integers x, y, w. Assume we have a solution. We may suose that x and y are corime, hence that x and w are corime. ( 7 If is an odd rime which divides w then = since x 4 7y 4 (mod. ( ( 7 ( By the Law of Quadratic Recirocity, ( 7 = ( 8 =. 7. Further, ( 7 7 = so is a quadratic residue of Thus, w t (mod 7 for some t. Therefore, x 4 t 4 (mod 7 so there exists an integer r with r 4 (mod 7, which is a contradiction, since no such r exists. (5 Is the congruence 3x 7x 4 0 (mod 39 solvable? Note 39 = 7 3. Multily by : 36x + 84x 56 0 (mod 39 thus we have (6x (mod 39. This is equivalent to solving x 74 (mod 39. Note x 74 (mod 7 is x 4 (mod 7, which has a solution. Also, x 74 (mod 3 if and only if x 3 (mod 3. And ( ( 3 3 = 3 ( 3 = 0 ( 3 = 5 ( 3( 3 = 5 ( 3 = 3 ( 5 = 3 5 =. Hence by the Chinese Remainder Theorem, the original congruence has a solution. Review of congruences 3. Lecture: Monday, November 6, 000 Recall: a, b Z imlies there exist x, y Z such that ax + by = gcd(a, b. x, y are found using the Euclidean Algorithm. Theorem 9 (Chinese Remainder Theorem. Let m,..., m t Z + with gcd(m i, m j for i j. Set m = m m t. Let b,...,b t Z +. Then the simultaneous congruences x b (mod m x b (mod m. x b t (mod m t

43 have a unique solution modulo m. PMATH 440/640 ANALYTIC NUMBER THEORY 43 Proof. Let n i = M m i for i =,...,t. Then gcd(m i, n i = so there exist r i, s i Z such that r i n i + s i m i = for i =,...,t. Let e i = r i n i so that e i (mod m i. Then take x 0 = t i= b ie i. Notice that x 0 b i (mod m i for i =,..., t. Suose also x b i (mod M i for i =,..., t. Then m i x x 0 for i =,...,t and since gcd(m i, m j for i j then m = t i= m idvx x 0. Thus there is a unique solution modulo m. For any ositive integer n, (Z/nZ is the set of invertible elements in Z/nZ. In articular, it is the set of congruence classes r + nz for which there exists s + nz with (r + nz(s + nz = + nz. Theorem 30. Let m,...,m t be ositive integers with gcd(m i, m j = for i j and ut m = m m t. Then the ring Z/mZ is isomorhic to the ring Z/m Z Z/m k Z and the grou (Z/mZ is isomorhic to the grou (Z/m Z (Z/m k Z. Proof. Let ψ : Z Z/m Z Z/m k Z be defined by ψ(n = (n + m Z,...,n + m t Z. We check that ψ is a ring morhism. ψ is surjective by the Chinese Remainder Theorem. Also by the Chinese Remainder Theorem, kerψ = mz. Thus by the First Isomorhism Theorem for Rings, Z/mZ = Z/m Z Z/m k Z Let λ : (Z/mZ (Z/m Z (Z/m k Z be defined by λ(n+mz = (n+m Z,...,n+m t Z. We see that λ is a grou morhism. It is bijective by the Chinese Remainder Theorem. Corollary. Let m,...,m t be ositive integers which are airwise corime. Put m = m m t. Then ϕ(m = ϕ(m ϕ(m t Proof. ϕ(m = (Z/mZ and ϕ(m ϕ(m t = (Z/m Z (Z/m Z = (Z/m Z (Z/m Z and so the result follows from Theorem 30. a a Corollary. Let m = t t where, t are distinct rimes, a,...,a t ositive integers. Then t ϕ(m = m ( i i= Proof. Take m i = i a i for i =,..., t in Corollary. Since ϕ( i a i = i a i i a i = i ( i, then ϕ(m = ϕ( a ϕ( t a t = ai t a t ( ( t = m t i= ( i

44 44 PMATH 440/640 ANALYTIC NUMBER THEORY Proosition. Let be a rime. If d then x d (mod has exactly d solutions modulo. Proof. Suose = dk with k Z +. Then x x d = (xd k x d = (xd k + + = g(x Z/Z[x] By Fermat s Theorem, x has distinct roots in Z/Z so (x d g(x factors into linear factors in Z/Z[x] and our result follows. Theorem 3. (Z/Z is a cyclic grou. Proof. For each divisor d of we let λ(d denote the number of elements of (Z/Z of order d. By Proosition, there are exactly d elements of (Z/Z whose order divides d hence d = c d λ(c Now by Möbius inversion, we then have λ(d = c d µ(c d c = d c d µ(c c = d d ( = ϕ(d Thus there are ϕ( elements of (Z/Z of order. In articular, (Z/Z is cyclic. Definition. Let n be a ositive integer and let a be an integer. a is said to be a rimitive root modulo n if a + nz generates (Z/nZ. 4. Lecture: Wednesday, November 8, 000 Note: For any rime (Z/Z is cyclic and so there exists a rimitive root modulo. In fact, there are ϕ( rimitive roots modulo. Artin conjectured that if a is a ositive integer and a is not a erfect square, then a is a rimitive root modulo for infinitely many rimes. This is still oen. It can be deduced from the generalized Riemann Hyothesis. (Why do we require a not a erfect square? is even. We want a to have order. Well, if a = k then no ower of a is congruent to k, for if a i k then a i a so a i so i, but even cannot divide odd! Notice is a rimitive root mod 5 (, 4, 3 5, 4 but is not a rimitive root modulo 7 since 3. In general, (Z/nZ is not cyclic so rimitive roots do not exist modulo n. (E.g. (Z/8Z = {[], [3], [5], [7]} and, 3, 5, 7. Therefore this grou is not cyclic! Proosition. Let be a rime. If l and a b (mod l, then a b (mod l+.

45 PMATH 440/640 ANALYTIC NUMBER THEORY 45 Proof. We may write a = b + c l for some c Z. Then ( a = (b + c l = b + b c l + So a b (mod l+ since l l +. ( b (c l + Proosition 3. If l and is an odd rime, then for any integer a, ( + a l + a l (mod l Proof. We rove the result by induction on l. The result is immediate for l =. Assume the result holds for some integer l with l and we shall rove it for l +. By Proosition, and our inductive hyothesis, ( + a l ( + a l (mod l+ ( ( + a l + (a l + (mod l+ Note that (l + divides (a l k for k = 3,..., since l imlies that (l + 3(l k(l. Further, (l + divides ( is an integer since is odd. So (l + divides the sum ( (a l + (a l also, since ( (a l = ( ( (a l (a l = (al. Note ( (a l (mod l+ Thus since l imlies (l + l + since is odd, then ( ( ( + a l + (a l + (mod l+ + a l + a l (mod l+ The result follows by induction. Proosition 4. If is an odd rime, l a ositive integer and a an integer corime with then + a has order l in (Z/ l Z. Proof. By Proosition 3, Hence, since a is corime with, But again by Proosition 3, hence ( + a l + a l (mod l ( + a l (mod l ( + a l + a l (mod l+ ( + a l (mod l

46 46 PMATH 440/640 ANALYTIC NUMBER THEORY Thus + a has order l in (Z/ l Z. 5. Lecture: Friday, November 0, 000 Theorem 3. Let be an odd rime and let l be a ositive integer. Then (Z/ l Z is a cyclic grou. Proof. By Theorem 3, there is a rimitive root g modulo. If g (mod, then ( ( (g + = g + g + g 3 + so ( (g + + g (mod hence (g + (mod. Therefore at least one of g and (g + is not congruent to modulo. Without loss of generality, we may suose that g (mod. We claim that in this case, g is a rimitive root modulo l. Suose that g has order m. By Euler s Theorem, g ϕ(l (mod so m l l = ( l. Write m = d s where d and 0 s l. By Fermat s Theorem, g g (mod. Hence g s g (mod rovided s 0. But g m (mod l so g m (mod, so g d (mod. Since g is a rimitive root, d. Thus d =, so m = ( l. Since g (mod and g (mod there exists an integer a corime with such that g + a (mod. And by roosition 4, +a has order l in (Z/ l Z and so g has order ( l hence (Z/ l Z is cyclic. Theorem 33. Let l be a ositive integer. (Z/ l Z is cyclic for l =,. For l 3, In articular, (Z/ l Z = Z/Z Z/Z (Z/ l Z = {( a 5 b + l Z : a {0, }, b {0...., l }} Proof. (Z/Z and (Z/4Z are lainly cyclic. Suose l 3. We claim that ( 5 l 3 + l (mod l We ll rove it by induction. For l = 3 we have 5 = + (mod 3 as required.

47 Suose ( holds for some l for l 3. PMATH 440/640 ANALYTIC NUMBER THEORY 47 Note that ( + l = + l + (l and (l l + for l 3. Now, 5 l 3 = + l + k l for some k Z. Hence 5 l = ( + l + k l = + l + k l + l + l + k l + k l + k l = + l + k l+ + l + k l + k l so 5 l + l (mod l+ since l l + for l 3. The result follows by induction. Thus by 5 l 3 (mod l and 5 l (mod l, hence 5 has order l in Z/ l Z. We now show that the numbers are distinct modulo, for l 3. Suose that ( a 5 b with a {0, }, b {0,..., l } ( a 5 b ( a 5 b (mod l with 0 a i and 0 b i < l for i =,,.... Then ( a 5 b ( a 5 b (mod 4 So ( a ( a (mod 4 hence a = a. Therefore 5 b 5 b (mod l and since 5 has order l, we see that b = b. Our result follows. Theorem 34. The only ositive integers having rimitive roots are,, 4, a and a with a a ositive integer and an odd rime. Proof. Let n = l 0 l r l r with l 0, l....,l r, non-negative integers and,..., r distinct odd rimes. Then by Theorem 30, (Z/nZ is isomorhic to (Z/ l 0 Z (Z/ l Z (Z/ r l r Z. By Theorem 3, (Z/ i l i Z is cyclic for i =,..., r. By Theorem 33, (Z/ l 0 Z is cyclic for 0 l 0 and is isomorhic to Z/Z Z/ l 0 Z for l 3. Therefore, the order of any element of Z/nZ is a divisor of λ(n = lcm(b, ϕ( l,...,ϕ( r l r where { ϕ( l 0 for 0 l 0 b = ϕ( l 0 l 0 3 Plainly, λ(n < ϕ( l 0 l ϕ( l ϕ( r r excet in the cases,, 4, a and a. Definition. λ(n is known as the universal exonent of n.

48 48 PMATH 440/640 ANALYTIC NUMBER THEORY Theorem 35. Let n be a ositive integer and define λ(n as before. Then for any integer a corime with n, a λ(n (mod n Proof. The roof follows from the roof of Theorem 34. Theorem 35 gives a strengthening of Euler s Theorem. 6. Lecture: Monday, November 3, 000 Given a rime one can ask for an uer bound for the smallest ositive integer a which is a rimitive root modulo. Hua roved a < ω( + Theorem 36. If is a rime of the form 4q + with q an odd rime then is a rimitive root modulo. Proof. Let t be the order of modulo. By Fermat s Theorem, t so t 4q. Thus t is one of,, 4, q, q and 4q. Note that = 3 or > 0 so t is not,,4. Further, by Euler s Criterion, q ( ( But = ( 8 = ( (4q +8q 8 = ( q =. Thus t is not q or q hence t = 4q = as required. (mod Let k and l be ositive integers with k and l corime. Dirichlet s Theorem asserts that kn + l is rime for infinitely many integers n. For many rimes (k, l, we can rove Dirichlet s Theorem by elementary means. Consider (4,3. Suose that there are only finitely many rimes,,..., k say, of the form 4n+3. Consider 4 k + 3. This must be divisible by a rime of the form 4m + 3 (since a roduct of rimes congruent to modulo 4 can only yield numbers congruent to modulo 4, and this rime 4n + 3 cannot be any of the,..., k. Theorem 37. Let n be a ositive integer. There are infinitely many rimes congruent to modulo n. Proof. (Due to Birkhoff and Vandiver, 904

49 PMATH 440/640 ANALYTIC NUMBER THEORY 49 Let a be a ositive integer with a >, and consider Φ n (a, the nth cyclotomic olynomial evaluated at a. Φ n (a = (a ζ j n j=, (j,n= where ζ n = e πi n, Φ n (x Z[x], and x n = d n Φ n(x. We claim that if is a rime dividing Φ n (a, then n or (mod n. To see this, note that a n n hence a. If a d for d a roer divisor of n, then the order of a (mod is n. By Fermat s Theorem, n, hence (mod n. Suose next that a d a for some roer divisor d of n. Then since Φ n (a, then n. a d We have hence a n = ( + (a d n d = + n d (ad + a n a d = n d + ( n/d (a d + a Since n and a d ad n, we conclude that d ( n (a d + ( n (a d ( n/d (a d + 3 hence n as required. Assume that there only finitely many rimes,..., k which are congruent to modulo n. Φ n (x = x ϕ(n + ± so Φ(n k m is not divisible by i for i =,...,k and is corime with n. Letting m we see that for m sufficiently large, Φ(n k m and so has a rime divisor congruent to modulo n which is not one of {,..., k }. This is a contradiction. 7. Lecture: Wednesday, November 5, 000 Definition. Let G be a finite abelian grou. A character of G is a homomorhism χ : G C. The set of characters of a grou form a grou under (χ χ (g = χ (gχ (g. This grou is called the dual grou of G and is denoted Ĝ. The identity of Ĝ is the character χ 0 where χ 0 (g = for all g G. χ 0 is known as the rincial character. Note that if G = n then g n = e for all g G, hence (χ(g n = so χ(g is an nth root of unity. Theorem 38. Let G be a finite abelian grou. Then (i Ĝ = G