Number Theory

Download e-book for kindle: A comprehensive course in number theory by Alan Baker

By Alan Baker

Built from the author's renowned textual content, A Concise creation to the idea of Numbers, this e-book presents a complete initiation to all of the significant branches of quantity conception. starting with the rudiments of the topic, the writer proceeds to extra complex subject matters, together with components of cryptography and primality checking out, an account of quantity fields within the classical vein together with houses in their devices, beliefs and excellent sessions, features of analytic quantity concept together with reviews of the Riemann zeta-function, the prime-number theorem and primes in arithmetical progressions, an outline of the Hardy-Littlewood and sieve tools from respectively additive and multiplicative quantity concept and an exposition of the mathematics of elliptic curves. The ebook comprises many labored examples, workouts and additional analyzing. Its wider insurance and flexibility make this ebook appropriate for classes extending from the user-friendly to starting graduate stories.

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Example text

Prove that the number of solutions of the congruence f (x) ≡ 0 (mod p) is 1 + dp , where d = b2 − 4ac and = 0 if p divides d. (xiii) Find the number of solutions (mod 997) of (a) x 2 + x + 1 ≡ 0, (b) x 2 + x − 2 ≡ 0, (c) x 2 + 25x − 93 ≡ 0. (xiv) With the notation of Exercise (xii), show that, if p does not divide d, then d p p x =1 f (x) p =− a . p Evaluate the sum when p divides d. (xv) Prove that if p is a prime ≡ 1 (mod 4) and if p = 2 p + 1 is a prime then 2 is a primitive root (mod p). For which primes p with p = 2 p + 1 prime is 5 a primitive root (mod p)?

We have g p−1 = 1 + py for some integer y and so, by the binomial theorem, g p−1 = 1 + pz, where z ≡ y + ( p − 1)g p−2 x (mod p). The coefficient of x is not divisible by p and so we can choose x such that (z, p) = 1. Then g has the required property. For suppose that g belongs to d (mod p j ). Then d divides φ( p j ) = p j−1 ( p − 1). But g is a primitive root (mod p) and thus p − 1 divides d. Hence d = p k ( p − 1) for some k < j. Further, since p is odd, we have k (1 + pz) p = 1 + p k+1 z k , where (z k , p) = 1.

The result follows. Let g be a primitive root (mod p). We prove now that there exists an integer x such that g = g + px is a primitive root (mod p j ) for all prime powers p j . We have g p−1 = 1 + py for some integer y and so, by the binomial theorem, g p−1 = 1 + pz, where z ≡ y + ( p − 1)g p−2 x (mod p). The coefficient of x is not divisible by p and so we can choose x such that (z, p) = 1. Then g has the required property. For suppose that g belongs to d (mod p j ). Then d divides φ( p j ) = p j−1 ( p − 1).

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