Geometric and Algebraic Interpretation of Primitive Pythagorean Triples Parameters
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PYTHAGORAS in LONDON Mathematics ESO 2
PYTHAGORASPYTHAGORAS inin LONDONLONDON GuiaGuia dede treballtreball Laura, 2103. Jaume Plensa MaterialMaterial elaboratelaborat perper www.mat3.cat Maite Gorriz i Santi Vilches PYTHAGORAS in LONDON Mathematics ESO www.mat3.cat 2 AVATAR An Avatar is a personal icon in a virtual context. We will develop the capacity to program and control the avatar’s movements and we will learn how to use a powerful mathematics tool which is necessary to program an avatar. A. SPECIAL EQUATIONS A.1. Calculate without calculator and memorize the following results: a) 2², 3², 4², 5², 6², 7², 8², 9², 10², 11², 12², 13² i 14² b) √1 , √4 , √9 , √16 , √25 , √36 , √49 , √64 , √81 … A.2. Solve the following quadratic equations: a) x² = 64 b) x² – 6 = 30 c) 10 – x² = 9 d) x² = 2 e) 7² + x² = 2 f) 12 = 1,3 + x² g) 11² + x² = 13² B. A LITTLE BIT OF VOCABULARY AND BASIC CONCEPTS B.1.We need to learn specific vocabulary. Write the definitions of the following words and draw an illustrative picture. a) Equilateral triangle b) Isosceles triangle c) Escalene triangle d) Obtuse triangle e) Rectangle triangle or right triangle f) Acutangle triangle g) Hypotenuse h) Cathetus PYTHAGORAS in LONDON Mathematics ESO www.mat3.cat 3 B.2. Remember that we measure the angles from the horizontal side to the other one in the counterclockwise. Remember also that one turn is 360º (degrees). A Write the measure of each angle: B.3. Try to do the same figure in the Geogebra and add the exercise in the Moodle. (surname_B3_angles_2X.ggb) B.4. -
On Terms of Generalized Fibonacci Sequences Which Are Powers of Their Indexes
mathematics Article On Terms of Generalized Fibonacci Sequences which are Powers of their Indexes Pavel Trojovský Department of Mathematics, Faculty of Science, University of Hradec Králové, 500 03 Hradec Králové, Czech Republic, [email protected]; Tel.: +42-049-333-2860 Received: 29 June 2019; Accepted: 31 July 2019; Published: 3 August 2019 (k) Abstract: The k-generalized Fibonacci sequence (Fn )n (sometimes also called k-bonacci or k-step Fibonacci sequence), with k ≥ 2, is defined by the values 0, 0, ... , 0, 1 of starting k its terms and such way that each term afterwards is the sum of the k preceding terms. This paper is devoted to the proof of the fact that the (k) t (2) 2 (3) 2 Diophantine equation Fm = m , with t > 1 and m > k + 1, has only solutions F12 = 12 and F9 = 9 . Keywords: k-generalized Fibonacci sequence; Diophantine equation; linear form in logarithms; continued fraction MSC: Primary 11J86; Secondary 11B39. 1. Introduction The well-known Fibonacci sequence (Fn)n≥0 is given by the following recurrence of the second order Fn+2 = Fn+1 + Fn, for n ≥ 0, with the initial terms F0 = 0 and F1 = 1. Fibonacci numbers have a lot of very interesting properties (see e.g., book of Koshy [1]). One of the famous classical problems, which has attracted a attention of many mathematicians during the last thirty years of the twenty century, was the problem of finding perfect powers in the sequence of Fibonacci numbers. Finally in 2006 Bugeaud et al. [2] (Theorem 1), confirmed these expectations, as they showed that 0, 1, 8 and 144 are the only perfect powers in the sequence of Fibonacci numbers. -
Number Systems and Radix Conversion
Number Systems and Radix Conversion Sanjay Rajopadhye, Colorado State University 1 Introduction These notes for CS 270 describe polynomial number systems. The material is not in the textbook, but will be required for PA1. We humans are comfortable with numbers in the decimal system where each po- sition has a weight which is a power of 10: units have a weight of 1 (100), ten’s have 10 (101), etc. Even the fractional apart after the decimal point have a weight that is a (negative) power of ten. So the number 143.25 has a value that is 1 ∗ 102 + 4 ∗ 101 + 3 ∗ 0 −1 −2 2 5 10 + 2 ∗ 10 + 5 ∗ 10 , i.e., 100 + 40 + 3 + 10 + 100 . You can think of this as a polyno- mial with coefficients 1, 4, 3, 2, and 5 (i.e., the polynomial 1x2 + 4x + 3 + 2x−1 + 5x−2+ evaluated at x = 10. There is nothing special about 10 (just that humans evolved with ten fingers). We can use any radix, r, and write a number system with digits that range from 0 to r − 1. If our radix is larger than 10, we will need to invent “new digits”. We will use the letters of the alphabet: the digit A represents 10, B is 11, K is 20, etc. 2 What is the value of a radix-r number? Mathematically, a sequence of digits (the dot to the right of d0 is called the radix point rather than the decimal point), : : : d2d1d0:d−1d−2 ::: represents a number x defined as follows X i x = dir i 2 1 0 −1 −2 = ::: + d2r + d1r + d0r + d−1r + d−2r + ::: 1 Example What is 3 in radix 3? Answer: 0.1. -
Cheat Sheet of SSE/AVX Intrinsics, for Doing Arithmetic Ll F(Int Ind, Int K) { Return Dp[Ind][K]; } on Several Numbers at Once
University of Bergen Garbage Collectors Davide Pallotti, Jan Soukup, Olav Røthe Bakken NWERC 2017 Nov 8, 2017 UiB template .bashrc .vimrc troubleshoot 1 tan v + tan w Contest (1) Any possible infinite recursion? tan(v + w) = Invalidated pointers or iterators? 1 − tan v tan w template.cpp Are you using too much memory? v + w v − w 15 lines Debug with resubmits (e.g. remapped signals, see Various). sin v + sin w = 2 sin cos #include <bits/stdc++.h> 2 2 using namespace std; Time limit exceeded: v + w v − w Do you have any possible infinite loops? cos v + cos w = 2 cos cos #define rep(i, a, b) for(int i = a; i < (b); ++i) What is the complexity of your algorithm? 2 2 #define trav(a, x) for(auto& a : x) Are you copying a lot of unnecessary data? (References) #define all(x) x.begin(), x.end() How big is the input and output? (consider scanf) (V + W ) tan(v − w)=2 = (V − W ) tan(v + w)=2 #define sz(x) (int)(x).size() Avoid vector, map. (use arrays/unordered_map) typedef long long ll; What do your team mates think about your algorithm? where V; W are lengths of sides opposite angles v; w. typedef pair<int, int> pii; typedef vector<int> vi; Memory limit exceeded: a cos x + b sin x = r cos(x − φ) What is the max amount of memory your algorithm should need? int main() { Are you clearing all datastructures between test cases? a sin x + b cos x = r sin(x + φ) cin.sync_with_stdio(0); cin.tie(0); cin.exceptions(cin.failbit); p 2 2 } Mathematics (2) where r = a + b ; φ = atan2(b; a). -
Composite Numbers That Give Valid RSA Key Pairs for Any Coprime P
information Article Composite Numbers That Give Valid RSA Key Pairs for Any Coprime p Barry Fagin ID Department of Computer Science, US Air Force Academy, Colorado Springs, CO 80840, USA; [email protected]; Tel.: +1-719-339-4514 Received: 13 August 2018; Accepted: 25 August 2018; Published: 28 August 2018 Abstract: RSA key pairs are normally generated from two large primes p and q. We consider what happens if they are generated from two integers s and r, where r is prime, but unbeknownst to the user, s is not. Under most circumstances, the correctness of encryption and decryption depends on the choice of the public and private exponents e and d. In some cases, specific (s, r) pairs can be found for which encryption and decryption will be correct for any (e, d) exponent pair. Certain s exist, however, for which encryption and decryption are correct for any odd prime r - s. We give necessary and sufficient conditions for s with this property. Keywords: cryptography; abstract algebra; RSA; computer science education; cryptography education MSC: [2010] 11Axx 11T71 1. Notation and Background Consider the RSA public-key cryptosystem and its operations of encryption and decryption [1]. Let (p, q) be primes, n = p ∗ q, f(n) = (p − 1)(q − 1) denote Euler’s totient function and (e, d) the ∗ Z encryption/decryption exponent pair chosen such that ed ≡ 1. Let n = Un be the group of units f(n) Z mod n, and let a 2 Un. Encryption and decryption operations are given by: (ae)d ≡ (aed) ≡ (a1) ≡ a mod n We consider the case of RSA encryption and decryption where at least one of (p, q) is a composite number s. -
Solutions of EGMO 2020
Solutions of EGMO 2020 Problem 1. The positive integers a0, a1, a2,..., a3030 satisfy 2an+2 = an+1 + 4an for n = 0, 1, 2,..., 3028. 2020 Prove that at least one of the numbers a0, a1, a2,..., a3030 is divisible by 2 . Problem 2. Find all lists (x1, x2, . , x2020) of non-negative real numbers such that the fol- lowing three conditions are all satisfied: (i) x1 ≤ x2 ≤ ... ≤ x2020; (ii) x2020 ≤ x1 + 1; (iii) there is a permutation (y1, y2, . , y2020) of (x1, x2, . , x2020) such that 2020 2020 X 2 X 3 (xi + 1)(yi + 1) = 8 xi . i=1 i=1 A permutation of a list is a list of the same length, with the same entries, but the entries are allowed to be in any order. For example, (2, 1, 2) is a permutation of (1, 2, 2), and they are both permutations of (2, 2, 1). Note that any list is a permutation of itself. Problem 3. Let ABCDEF be a convex hexagon such that \A = \C = \E and \B = \D = \F and the (interior) angle bisectors of \A, \C, and \E are concurrent. Prove that the (interior) angle bisectors of \B, \D, and \F must also be concurrent. Note that \A = \F AB. The other interior angles of the hexagon are similarly described. Problem 4. A permutation of the integers 1, 2,..., m is called fresh if there exists no positive integer k < m such that the first k numbers in the permutation are 1, 2,..., k in some order. Let fm be the number of fresh permutations of the integers 1, 2,..., m. -
Arxiv:1608.06086V1 [Math.NT]
POWER OF TWO AS SUMS OF THREE PELL NUMBERS JHON J. BRAVO, BERNADETTE FAYE AND FLORIAN LUCA Abstract. In this paper, we find all the solutions of the Diophantine equation a Pℓ +Pm +Pn =2 , in nonnegative integer variables (n,m,ℓ,a) where Pk is the k-th term of the Pell sequence Pn n≥0 given by P0 = 0, P1 = 1 and Pn+1 =2Pn+Pn−1 for all n 1. { } ≥ MSC: 11D45, 11B39; 11A25 Keywords: Diophantine equations, Pell numbers, Linear forms in logarithm, reduction method. 1. Introduction The Pell sequence P is the binary reccurent sequence given by P = 0, P =1 { n}n≥0 0 1 and Pn+1 = 2Pn + Pn−1 for all n 0. There are many papers in the literature dealing with Diophantine equations≥ obtained by asking that members of some fixed binary recurrence sequence be squares, factorials, triangular, or belonging to some other interesting sequence of positive integers. For example, in 2008, A. Peth˝o[18] found all the perfect powers (of exponent larger than 1) in the Pell sequence. His result is the following. Theorem 1 (A. Peth˝o, [18]). The only positive integer solutions (n, q, x) with q 2 of the Diophantine equation ≥ q Pn = x are (n, q, x) = (1, q, 1) and (7, 2, 13). That is, the only perfect powers of exponent larger than 1 in the Pell numbers are 2 P1 =1 and P7 = 13 . The case q = 2 had been treated earlier by Ljunggren [13]. Peth˝o’s result was rediscovered by J. H. -
Rapid Multiplication Modulo the Sum and Difference of Highly Composite Numbers
MATHEMATICS OF COMPUTATION Volume 72, Number 241, Pages 387{395 S 0025-5718(02)01419-9 Article electronically published on March 5, 2002 RAPID MULTIPLICATION MODULO THE SUM AND DIFFERENCE OF HIGHLY COMPOSITE NUMBERS COLIN PERCIVAL Abstract. We extend the work of Richard Crandall et al. to demonstrate how the Discrete Weighted Transform (DWT) can be applied to speed up multiplication modulo any number of the form a b where p is small. In ± p ab particular this allows rapid computation modulo numbers of thej form k 2n 1. Q · ± In addition, we prove tight bounds on the rounding errors which naturally occur in floating-point implementations of FFT and DWT multiplications. This makes it possible for FFT multiplications to be used in situations where correctness is essential, for example in computer algebra packages. 1. Introduction In their seminal paper of 1994, Richard Crandall and Barry Fagin introduced the Discrete Weighted Transform (DWT) as a means of eliminating zero-padding when performing integer multiplication modulo Mersenne numbers [2]. While this does not give any improvement in the order of the multiplication, it nevertheless cuts the transform length (and thus time and memory) in half. For these reasons the DWT has become a fixture of the search for Mersenne primes [10]. By using the same form of irrational-base representation as is used for Mersenne numbers, we can in fact eliminate the zero-padding when working modulo a b ± provided that p ab p is sufficiently small that we have enough precision. Essen- tially, as with Mersennej numbers, we choose the base so that the reduction modulo xn 1 implicitQ in the FFT multiplication turns into a reduction modulo a b. -
A Diophantine Equation in K–Generalized Fibonacci Numbers and Repdigits
A Diophantine equation in k{generalized Fibonacci numbers and repdigits Jhon J. Bravo Departamento de Matem´aticas Universidad del Cauca Calle 5 No 4{70, Popay´an,Colombia E-mail: [email protected] Carlos Alexis G´omezRuiz Departamento de Matem´aticas Universidad del Valle 25360 Calle 13 No 100-00, Cali, Colombia E-mail: [email protected] Florian Luca School of Mathematics University of the Witwatersrand Private Bag X3 Wits 2050 Johannesburg, South Africa Max Planck Institute for Mathematics Vivatsgasse 7 53111 Bonn, Germany Department of Mathematics Faculty of Sciences University of Ostrava 30 Dubna 22 701 03 Ostrava 1, Czech Republic E-mail: [email protected] Abstract 2010 Mathematics Subject Classification: 11B39; 11J86. Key words and phrases: Generalized Fibonacci numbers, lower bounds for nonzero linear forms in logarithms of algebraic numbers, repdigits. 1 2 J. J. Bravo, C. A. G´omezand F. Luca (k) The k−generalized Fibonacci sequence fFn gn starts with the value 0;:::; 0; 1 (a total of k terms) and each term afterwards is the sum of the k preceding terms. In the present paper, we study on members of k{generalized Fibonacci sequence which are sum of two repdigts, extending a result of D´ıazand Luca [5] regarding Fibonacci numbers with the above property. 1 Introduction Given an integer k ≥ 2, we consider the k{generalized Fibonacci sequence (k) (k) or, for simplicity, the k{Fibonacci sequence F := fFn gn≥2−k given by the recurrence (k) (k) (k) (k) (1.1) Fn = Fn−1 + Fn−2 + ··· + Fn−k for all n ≥ 2; (k) (k) (k) (k) with the initial conditions F−(k−2) = F−(k−3) = ··· = F0 = 0 and F1 = 1. -
Metallic Means : Beyond the Golden Ratio, New Mathematics And
Journal of Advances in Mathematics Vol 20 (2021) ISSN: 2347-1921 https://rajpub.com/index.php/jam DOI: https://doi.org/10.24297/jam.v20i.9056 Metallic Means : Beyond the Golden Ratio, New Mathematics and Geometry of all Metallic Ratios based upon Right Triangles, The Formation of the “Triads” of Metallic Means, And their Classical Correspondence with Pythagorean Triples and p≡1(mod 4) Primes, Also the Correlation between Metallic Numbers and the Digits 3 6 9, Triangles – Triads – Triples - & 3 6 9 Dr. Chetansing Rajput M.B.B.S. Nair Hospital (Mumbai University) India, Asst. Commissioner (Govt. of Maharashtra) Lecture Link: https://youtu.be/vBfVDaFnA2k Email: [email protected] Website: https://goldenratiorajput.com/ Abstract This paper brings together the newly discovered generalised geometry of all Metallic Means and the recently published mathematical formulae those provide the precise correlations between different Metallic Ratios. The paper also puts forward the concept of the “Triads of Metallic Means”. This work also introduces the close correspondence between Metallic Ratios and the Pythagorean Triples as well as Pythagorean Primes. Moreover, this work illustrates the intriguing relationship between Metallic Numbers and the Digits 3 6 9. Keywords: Fibonacci sequence, Pi, Phi, Pythagoras Theorem, Divine Proportion, Silver Ratio, Golden Mean, Right Triangle, Metallic Numbers, Pell Numbers, Lucas Numbers, Golden Proportion, Metallic Ratio, Metallic Triples, 3 6 9, Pythagorean Triples, Pythagorean Primes, Golden Ratio, Metallic Mean Introduction The prime objective of this work is to synergize the following couple of newly discovered aspects of Metallic Means: 1) The Generalised Geometric Construction of all Metallic Ratios: cited by Wikipedia in its page on “Metallic Mean”[1]. -
Pythagorean Theorem
4.4. Pythagorean Theorem A right triangle is a triangle containing a right angle (90°). A triangle cannot have more than one right angle, since the sum of the two right angles plus the third angle would exceed the 180° total possessed by a triangle. The side opposite the right angle is called the hypotenuse (side c in the figure below). The sides adjacent to the right angle are called legs (or catheti, singular: cathetus). hypotenuse A c b right angle = 90° C a B Math notation for right triangle The Pythagorean Theorem states that the square of a hypotenuse is equal to the sum of the squares of the other two sides. It is one of the fundamental relations in Euclidean geometry. c2 = a2 + b2 2 Ac = c B A = a2 c a a A b C 2 Ab = b Pythagorean triangle with the squares of its sides and labels Pythagorean triples are integer values of a, b, c satisfying this equation. Math 8 * 4.4. Pythagorean Theorem 1 © ontaonta.com Fall 2021 Finding the Sides of a Right Triangle Example 1: Find the hypotenuse. A c 2:8 m C 9:6 m B 2 2 2 c = a + b ðsubstitute for a and b c2 = .9:6 m/2 + .2:8 m/2 c2 = 92:16 m2 + 7:84 m2 2 2 c = 100 m ðtake the square root of each side ù ù c2 = 100 m2 ù ù ù c2 = 100 m2 c = 10 m Example 2: Find the missing side b. 5 3 b 2 2 2 2 c = a + b ðsubtract a from each side ¨ ¨ c2 * a2 = ¨a2 + b2 ¨*¨a2 2 2 2 c * a = b ðswitch sides 2 2 2 b = c * a ðsubstitute for a and c b2 = 52 * 32 2 b = 25 * 9 = 16 ðtake the square root of each side ù ù b2 = 16 b = 4 Math 8 * 4.4. -
Arxiv:2008.10398V1 [Math.NT] 24 Aug 2020 Children He Has
JOURNAL OF THE AMERICAN MATHEMATICAL SOCIETY Volume 00, Number 0, Pages 000{000 S 0894-0347(XX)0000-0 RECURSIVELY ABUNDANT AND RECURSIVELY PERFECT NUMBERS THOMAS FINK London Institute for Mathematical Sciences, 35a South St, London W1K 2XF, UK Centre National de la Recherche Scientifique, Paris, France The divisor function σ(n) sums the divisors of n. We call n abundant when σ(n) − n > n and perfect when σ(n) − n = n. I recently introduced the recursive divisor function a(n), the recursive analog of the divisor function. It measures the extent to which a number is highly divisible into parts, such that the parts are highly divisible into subparts, so on. Just as the divisor function motivates the abundant and perfect numbers, the recursive divisor function motivates their recursive analogs, which I introduce here. A number is recursively abundant, or ample, if a(n) > n and recursively perfect, or pristine, if a(n) = n. There are striking parallels between abundant and perfect numbers and their recursive counterparts. The product of two ample numbers is ample, and ample numbers are either abundant or odd perfect numbers. Odd ample numbers exist but are rare, and I conjecture that there are such numbers not divisible by the first k primes|which is known to be true for the abundant numbers. There are infinitely many pristine numbers, but that they cannot be odd, apart from 1. Pristine numbers are the product of a power of two and odd prime solutions to certain Diophantine equations, reminiscent of how perfect numbers are the product of a power of two and a Mersenne prime.