By Kenneth Ireland, Michael Rosen

ISBN-10: 1475717792

ISBN-13: 9781475717792

ISBN-10: 1475717814

ISBN-13: 9781475717815

This well-developed, available textual content info the ancient improvement of the topic all through. It additionally offers wide-ranging assurance of important effects with relatively easy proofs, a few of them new. This moment variation comprises new chapters that supply a whole evidence of the Mordel-Weil theorem for elliptic curves over the rational numbers and an summary of contemporary growth at the mathematics of elliptic curves.

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**Extra info for A Classical Introduction to Modern Number Theory**

**Example text**

1·2··· n Clearly this integer is divisible by alI primes P, n < P < 2n. Furthermore, since (1 + 1)2n = I j=O (2~), ] Hence 22n > ( 2n) n > p< 2n Il P p>n and therefore p< 2n 2n log 2 > L log P = (}(2n) - (}(n). p>n Summing this relation for n = 1,2,4,8, ... , 2m (}(2 m) < (log 2)(2m+1 < (log 2)2m + 1. - 1 gives 2) 24 If 2m - 2 Applications of Unique Factorization 1 < x :s; 2m we obtain O(x) :s; O(2m ) < (log 2)2 m + l = (4 log 2)2m < (4 log 2)x. 1 o Corollary 1. There is a positive constant CI such that n(x) < clx/log x for x ~ 2.

Nagell [60], E. Landau [52] and Vinogradov [77]. An in tere sting discussion of the various possible ways of arranging this material can be found in P. , 13, (1967), 223-231. A more advanced discussion of congruences is given in the first chapter of Borevich and Shafarevich [9]; this book also shows how the theory of congruences is useful in determin ing whether equations can be solved in integers. We mention also the beautiful treatment by J. P. Serre [69]. Historically the notion of congruences was first introduced and used systematically in Gauss' Disquisitiones Arithmeticae.

O As usual the result for the powers of 2 is more complicated. 4. Let 21 be the highest power of2 dividing n. Suppose that a is odd and that x n == a (2 21 + 1) is solvable. Then x n == a (2 e ) is solvable for ali e ~ 21 + 1 (and consequently for aU e ~ 1). Moreover, ali these congruences ha ve the same number of solutions. We leave the proof as an exercise. One begins by assuming that xn == a (2 m ), m ~ 21 + 1, has a solution Xo. Let Xl = Xo + b2m - l • One shows, by an appropriate choice of b, that xi == a (2m + 1).

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