Algebra and number theory: proceedings of a conference held by M. hammed Boulagouaz, Jean-Pierre Tignol, Mohammed

Number Theory

By M. hammed Boulagouaz, Jean-Pierre Tignol, Mohammed Boulagouaz

This examine demonstrates the most important manipulations surrounding Brauer teams, graded jewelry, workforce representations, excellent periods of quantity fields, p-adic differential equations, and rationality difficulties of invariant fields - exhibiting a command of the main complex tools in algebra. It describes new advancements in noncommutative valuation conception and p-adic research.

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9}, we compute ||q35 μ||. 029... 02. 01 < ||q35 μ|| − N ||q35 κ||. , log B we observe that there is no solution in the range m ∈ 22, 1015 . Thus, m ≤ 21, and n ≤ 102. However, we have assumed that n > 1000. To finish, we compute the values of all Fibonacci numbers modulo 10000 (their last four digits) and convince ourselves that there are no Fibonacci numbers with the desired pattern in the range 11 ≤ n ≤ 1000. This example stems from [25, 36]. 3 37 Simultaneous Pellian Equations The following result is due to Baker and Davenport and was historically the first example of a successful use of lower bounds for linear forms in logarithms of algebraic numbers; it actually allowed the effective computation of all common members of two binary recurrent sequences with real roots; for more details see [19, 25, 36].

Bujaˇci´c and A. 1) 1≤i≤d where αi are the conjugates of α. 10 Let α = √ p q p q is = log max{| p|, |q|}. 2. The absolute logarithmic height of α is √ √ √ 1 1 log | 2| + log | − 2| = log 2. 11 Let 1 α1 = √ √ . 3+ 5 Before calculating the absolute logarithmic height of α1 , we compute the degree of the field extension Q √3+1 √5 over Q as 1 Q √ √ 3+ 5 : Q = 4, so the degree of the algebraic number α1 = √3+1 √5 over Q is 4. The minimal polynomial of α1 is given by 1 x− √ × √ √ 3+ 5 − 3− 5 1 1 × x− √ x−√ √ √ − 3+ 5 3− 5 1 = x 4 − 4x 2 + , 4 Pα1 (x) = x−√ 1 over the rationals, and Pα1 (x) = 4x 4 − 16x 2 + 1 over the integers.

4 The largest solution of Eq. 3) is F10 = 55. Proof Suppose that n > 1000. We start by proving something weaker. Our goal is to obtain some bound on n. We rewrite Eq. 3) as αn − β n 10m − 1 . =d √ 9 5 34 S. Bujaˇci´c and A. Filipin Next we separate large and small terms on both sides of the equation. 5. 4) Our goal is to get some estimates for m in terms of n. By induction on n, it is easy to prove that αn−2 < Fn < αn−1 , n ≥ 3. Thus, αn−2 < Fn < 10m or n < m log 10 + 2, log α and 10m−1 < Fn < αn−1 .

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