## A Classical Introduction to Modern Number Theory by Kenneth Ireland, Michael Rosen

By Kenneth Ireland, Michael Rosen

This well-developed, obtainable textual content information the old improvement of the topic all through. It additionally offers wide-ranging insurance of vital effects with relatively straightforward proofs, a few of them new. This moment variation comprises new chapters that supply an entire evidence of the Mordel-Weil theorem for elliptic curves over the rational numbers and an outline of modern development at the mathematics of elliptic curves.

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Extra info for A Classical Introduction to Modern Number Theory

Example text

The three inequivalent solutions are 3, 8, and 13. We have two important corollaries. Corollary 1. If a and mare relatively prime, then ax one solution. == b (m) has one and only §3 The Congruence ax == b (m) 33 PROOF. In this case d = 1 so clearly d Ib, and there are d = 1 solutions. O Corollary 2. lfp is a prime and a =1= O(p), then ax == b (p) has one and only one solution. PROOF. Immediate [rom Corollary 1. O Corollaries 1 and 2 can be interpreted in terms of the ring Z/mZ. The congruence ax == b (m) is equivalent to the equation ax = li over the ring Z/mZ.

Al + 1). (b) u(n) = «p~,+l -1)/(pl _1»«p~2+l -1)/(p2 -1» .. · «pi,+l - 1)/(pl - 1». PROOF. Toprovepart (a) notice that mln iffm = p~'p~2 ... pt'andO ~ bi ~ ai for i = 1, 2, ... , 1. Thus the positive divisors of n are one-to-one correspondence with the n-tuples (b l , b2, ... , bl) with O ~ bi ~ ai for i = 1, ... ,1, and there are exact1y (al + 1)(a2 + 1) ... (al + 1) such n-tuples. To prove part (b) notice that u(n) = L p~'p~2 ... pt', where the sum is over the above set ofn-tuples. Thus, u(n) = (L~:=o P~l)(L~~=o p~2) ...

1f P is an odd prime, p -1' a, and p -1' n, then if x n == a (P) is solvable, so is x n == a (pe) for ali e ~ 1. All these congruences have the same number of solutions. If n = 1, the assertion is trivial, so we may assume n ~ 2. Suppose that x n == a (pe) is solvable. Let Xo be a solution and set Xl = Xo + bpe. A short computation shows xi == xO + nbpex"o-l (pe + 1). We wish to solve xi == a (pe+ 1). This is equivalent to finding an integer b such that nx"o-lb == «a - x"o)/pe)(p). Notice that (a - x"o)/pe is an integer and that p-l'nx"o-l.