777-855 826-866 Gauss and Riemann versus elementary mathematics
Problem at the 987 International Mathematical Olympiad: Given that the polynomial [ ] f (x) = x 2 + x + p yields primes for x =,, 2,..., p 3. Show that it yields primes for x =,, 2,..., p 2. Example [ ] 4 p = 4, 3 = 3. f () = 4, f () = 43, f (2) = 47, f (3) = 53. So f (4),..., f (39), i.e. 6, 7, 83, 97, 3,..., 6 are all primes. (First observed by Euler in 772.)
Theorem Let p be prime, and let f p (x) = x 2 + x + p. Then the following conditions are equivalent: p = 2, 3, 5,, 7, 4 2 f p (x) is prime for x =,, 2,..., [ p 3 ]. 3 f p (x) is prime for x =,, 2,..., p 2. 4 Q( 4p) has class number one.
Gauss ( Disquisitiones Arithmeticae, 8) stated that Q( D), where D <, has class number one if D =, 2, 3, 7,, 9, 43, 67, 63. He asked if there were other imaginary quadratic fields that had this property. It was shown independently by Baker and Stark in 966 that Gauss list was complete. p 2 3 5 7 4 4p -7 - -9-43 -67-63
Quadratic binary forms f (x, y) = ax 2 + bxy + cy 2 ; a, b, c Z Discriminant D = b 2 4ac. Example f (x, y) = x 2 + y 2 ; D = 4. Equivalent form to f (x, y): f (x, y) = f (αx + βy, γx + δy), where αδ βγ =, i.e. [ ] α β SL γ δ 2 (Z)
[ X Y ] = [ α β γ δ ] [ x y ] = [ αx + βy γx + δy ] Z 2 [ x y ] = [ δ β γ α ] [ X Y ] = [ δx βy γx + αy ] Z 2 Example f (x, y) = 3x 2 42xy + 34y 2 = f (2x 3y, 3x + 5y), where f (x, y) = x 2 + y 2.
If f (x, y) and f (x, y) are equivalent forms, they have the same discriminant. Clearly f (x, y) and f (x, y) have the same set of values as x and y run through the integers Z. Conversely, let g(x, y) and h(x, y) be two forms with the same discriminant, and assume that g(x, y) and h(x, y) take the same set of values as x and y run through Z. Then g(x, y) is equivalent to h(x, y). (We assume the forms are irreducible and primitive, i.e. the g.c.d. of the coefficients is.) (Schering (859), Chowla (966), Perlis (979))
ax 2 + bxy + cy 2 AX 2 + BXY + CY 2 (a, b, c) (A, B, C) [ ] X = Y [ α β γ δ ] [ ] x, y [ ] α β SL γ δ 2 (Z). D = b 2 4ac = B 2 4AC <.
Associate the complex numbers Then w = b + D 2a, w = B + D. 2A w = αw + β γw + δ. A form is called reduced if its associated complex number w lies in the fundamental domain for the modular group SL(2, Z). A form is reduced if (a, b, c) satisfies a < b a < c or b a = c. Every form is equivalent to a unique reduced form.
Im w = b+ D 2a 2 2 Re
Let h(d) denote the number of inequivalent forms ax 2 + bxy + cy 2 of discriminant D = b 2 4ac( or ( mod 4)). Gauss proved that h(d) is finite for all D. Conjecture The number of negative discriminants D < which have a given class number h is finite. In other words, h(d) as D.
Fact h(d) = Q( D) has class number one, i.e. the integers A D in Q( D) = {α + β D } α, β Q is a unique factorization domain A D is a PID. In our case, where D is of the form 4n, { a + b } D A D = a, b Z with same parity. 2 (A D is the set of solutions to x 2 + cx + d = ; c, d Z, that lie in Q( D).
There are 5 imaginary quadratic fields Q( D) which are Euclidean w.r.t the norm N(a + β D) = (a + β D)(a β D) = α 2 Dβ 2 : D =, 2, 3, 7,. The units of A D are ± if D <, except for { D = {±, ±i} and D = 3 ±, ±ρ, ±ρ 2 ρ = + } 3. 2 Example Example of a non-unique factorization domain: { A 5 = a + b 5 } a, b Z Q( 5). 6 = 2 3 = ( + 5)( 5).
G. Rabinovitch, Eindeutigkeit der Zerlegung in Primzahlfaktoren in quadratischer Zahlkörpern. Proceedings 5th Congress of Mathematicians, Cambridge 92. Theorem Q( 4p) has class number one f p (x) = x 2 + x + p is a prime for x =,, 2,..., p 2. [ We will prove that if f p (x) is a prime for x =,, 2,..., p 3 then Q( 4p) has class number one. ],
Proof We must show that there exists a unique reduced form ax 2 + bxy + cy 2, i.e. a < b a < c or b a = c ( ) such that the discriminant D equals 4p, i.e. D = b 2 4ac = 4p. Since D is odd, b must be odd, say b = 2k +. By ( ) we get (2k + ) 2 4ac = 4k 2 + 4k + 4ac = 4p, and so ( ) ac = k 2 + k + p = f p (k). ( ) The inequalities [ ] ( ), [ together ] with ( ), yield p 3 k p 3. Since f p ( k) = f p (k ), we get by ( ) and our assumption that ac is a prime. By ( ) we get finally that a =, b = and so by ( ), c = p. So the unique reduced form is x 2 xy + py 2. Hence the class number of Q( 4p) is one.
The historical continuity of mathematics (Felix Klein (894)) Mathematics develops and progresses as old problems are being understood and clarified by means of new methods. Simultaneously, as a better and deeper understanding of the old questions is thus obtained, new problems naturally arise.
= + p p s + (p 2 ) s + (p 3 + (Re(s) > ). ) s s p = ( + 2 p s + (2 2 ) s + (2 3 ) s + ) s ( + 3 s + (3 2 ) s + (3 3 ) s + ) ( + 5 s + (5 2 ) s + (5 3 ) s + ) = i <i 2 < <i k (p r i i p r i k i k ) s = n s = ζ(s) n= ζ(s) = 2 s π s sin πs Γ( s) ζ( s). 2 The functional equation for the Riemann zeta-function ζ(s).
ζ(s) = p ( p ) s = ( 2 s ) ( 3 s ) ( 5 s ) = 2 s 3 s 5 s + 6 s 7 s + s = = s n= µ(n) n s M(x)x s dx, where M(x) = n x µ(n) (see next page). if n is not square-free µ(n) = if n is square-free with an even number of distinct prime factors if n is square-free with an odd number of distinct prime factors.5 µ(n). -.5 2 3 4 5 n -
n= µ(n) n s = lim By Abel s partial summation we get: N n= µ(n) n s = N n= = M(N) N s = M(N) N s N N n= M(n) M(n ) n s N µ(n) n s. M(n){f (n + ) f (n)} n= N = M(N) + s s N N s n= N+ n+ M(n) n M(x)x s dx M(x)x s dx ( ) d dx x s dx
Theorem (Hadamard and de la Vallée Poussin (896)) Im critical line The Riemann zeta function ζ(s) has no zeros on the line Re(s) =. 2 critical strip Re π(x) x log x as x, where π(x) denotes the number of primes x. (The Prime Number Theorem)
The Riemann Hypothesis (Riemann (859)) The (non-trivial) zeros of the zeta-function ζ(s) lie on the critical line Re(s) = 2. Littlewood(92) The Riemann hypothesis is equivalent to M(x) = o(x /2+ɛ ) for all ɛ >, i.e. M(x) x /2+ɛ as x, where M(x) = µ(n). n x
Farey fractions F n of order n: { p F n = q < p } q, q n. Let A n = F n and let F n = { r < r 2 < r 3 < < r An = = }. Let δ = r A n, δ 2 = r 2 2 A n, δ 3 = r 3 3 A n,..., δ An = r An A n A n (= ). Example N = 5, F 5 = { 5, 4, 3, 2 5, 2, 3 5, 2 3, 3 4, 4 5, } =, A 5 =. r r 2 r 3 r 4 r 5 r 6 r 7 r 8 r 9 r 2 3 4 5 6 7 8 9
Theorem (Franel and Landau (924)) The Riemann hypothesis is true δ + δ 2 + + δ An = o(n /2+ɛ ) for all ɛ > as n.
Proof of Let f : [, ] C. Then A n f (r ν ) = k f ν= k= j= ( ) j ( n ) M k k ( ) Comment This is the key identity, where the rather irregular operation of summing f over the Farey fractions can be expressed more regularly using the function M and a double sum (we show this later). Formula ( ) applied to f (x) = e 2πix gives since A n ν= k j= e 2πirν = k= j= e 2πi j k = { if k if k = k ( e 2πi j n ) k M = M(n) k
Proof of (cont.) Set A = A n. Then so M(n) M(n) = = = A ν= A ν= A ν= e 2πirν e 2πi[ ν A +δν] e 2πi ν A A e 2πiδν + 2 ν= [ ] e 2πiδν + A ν= e 2πi ν A A sin (πδ ν ) 2π ν= A δ ν. ν=
Lemma With f as in the theorem A n ν= f (r ν ) = k f k= j= ( ) j ( n ) M k k ( ) Proof: Let < p q, p and q relatively prime. The term ( ) ( ) ( ) p 2p 3p f = f = f = q 2q 3q occurs on the right hand side of ( ) with the coefficient M ( ) ( ) ( ) n n n + M + M + = q 2q 3q { if q n if q > n ( ) and so we get ( ).
Proof (cont.) M(x) = k x µ(k) = k ( x ) µ(k)d k where D(x) = { for x for x <. By the Möbius inversion (see next page) we get D(x) = k M ( ) x k. Set x = n q, and we get ( ).
Theorem Möbius inversion Let f, g : R + C. f (x) = ( x ) g g(x) = µ(n)f n n x n x ( x n ) Proof: : µ(n)f n x ( x ) = µ(n) ( x ) g n mn n x m x n = ( x ) µ(n)g mn mn x = ( x ) g µ(n) r r x n r ( x ) = g = g(x)
Proof : ( x ) g = ( x ) µ(m)f n mn n x n x m x n = ( x ) µ(m)f mn mn x = ( x ) f µ(m) r r x m r ( x ) = f = f (x) We have used that { if n = µ(d) = if n >. d n
Dirichlet s L-function L(s, χ) = n= χ(n) n s = p prime ( χ(p) ) p s, Re(s) >, where χ : Z/qZ {roots of unity} is a character. L(s, χ) can be extended to a holomorphic function on C (if χ is not the trivial character). The Generalized Riemann Hypothesis (GRH) The (non-trivial) zeros of L(s, χ) lie on the line Re(s) = /2. Theorem (Dirichlet (839)) L(, χ) = 2πh(D)/(const. D ), where D = b 2 4ac < is the discriminant of an imaginary quadratic form ax 2 + bxy + cy 2, and h(d) is the class number. (χ is a primitive character mod D.)
Theorem (Hecke (98)) If GRH is true, then h(d) as D. Theorem 2 (Heilbronn (934)) If GRH is false, then h(d) as D. Corollary Gauss conjecture about the class number for imaginary quadratic forms is true.
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