Binary Quadratic Forms: An Algorithmic Approach (Algorithms by Johannes Buchmann

By Johannes Buchmann

The e-book offers with algorithmic difficulties with regards to binary quadratic types, reminiscent of discovering the representations of an integer by means of a sort with integer coefficients, discovering the minimal of a kind with actual coefficients and finding out equivalence of 2 types. so one can resolve these difficulties, the booklet introduces the reader to special parts of quantity idea akin to diophantine equations, relief conception of quadratic kinds, geometry of numbers and algebraic quantity thought. The publication explains functions to cryptography. It calls for in basic terms simple mathematical wisdom.

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The map that sends a solution (x, y) ∈ Z2 of the Pell equation x2 − ∆y 2 = ±4 to the matrix U (f, x, y) is a bijection between the set of those solutions and the automorphism group of f . 2. The map that sends a solution (x, y) ∈ Z2 of the Pell equation x2 −∆y 2 = 4 to the matrix U (f, x, y) is a bijection between the set of those solutions and the proper automorphism group of f . Proof. 6, the map from the theorem sends solutions of the Pell equation to automorphisms of f . The map is clearly injective.

P − 1} be a square modulo p. Then a square root of r modulo p can be computed in time O((size p)3 ). 46 3 Constructing Forms If we use this method for extracting square roots modulo p we obtain the following running time estimate for primeForm. 21. If p ≡ 3 (mod 4), then primeForm runs in time O size(p) size(∆) + (size p)3 . Now we turn to the general case. We explain the algorithm of Tonelli for extracting square roots in a finite cyclic group G of known order |G|. If we apply this algorithm to the cyclic group (Z/pZ)∗ , then we obtain an algorithm for extracting square roots modulo p.

If we use sqrtModP(∆, p) to extract the square root in primeForm, then we obtain the following result. 26. Algorithm primeForm has success probability 1/2 and running time O size(∆) size(p) + (size(p))4 . 5 The case of a prime power In this section we determine F(∆, pe ) and F ∗ (∆, pe ) for prime p and e > 1. 1. Let e ∈ N, e ≥ 1. e ∗ e 1. If ∆ p = −1, then R(∆, p ) = R (∆, p ) = 0. ∆ 2. If p = 0, e ≥ 2, and p does not divide the conductor of ∆, then R(∆, pe ) = R∗ (∆, pe ) = 0. e ∗ e 3. If ∆ p = 1, then R(∆, p ) = R (∆, p ) = 2.

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