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... so any square is congruent to 0 or 1 mod 4, hence p 3 mod 4 cannot be a sum of two squares. The remaining case of p = 2 needs to be treated separately but is trivial. In fact, we can derive a complete characterization of integers that are sums of two squares: Corollary 2.10 (Fermat's two square th ...
... so any square is congruent to 0 or 1 mod 4, hence p 3 mod 4 cannot be a sum of two squares. The remaining case of p = 2 needs to be treated separately but is trivial. In fact, we can derive a complete characterization of integers that are sums of two squares: Corollary 2.10 (Fermat's two square th ...
Answers
... (d) True - a square is a special type of rectangle. However, the converse is false, not every rectangle is a square. (e) False (the circumference of a circle is only approximately 3 times the diameter) (f) True ...
... (d) True - a square is a special type of rectangle. However, the converse is false, not every rectangle is a square. (e) False (the circumference of a circle is only approximately 3 times the diameter) (f) True ...
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... Note that G8 ≡ 0 and G9 ≡ 2, which repeats the two initial conditions and hence the sequence modulo 3 is periodic. Also note that if q = 2 and r = 1, then the resulting (q, r)−Fibonacci sequence is the classical Lucas sequence. Proposition 2: Let q, r, m be integers with m ≥ 2. (a) The (q, r)-Fibona ...
... Note that G8 ≡ 0 and G9 ≡ 2, which repeats the two initial conditions and hence the sequence modulo 3 is periodic. Also note that if q = 2 and r = 1, then the resulting (q, r)−Fibonacci sequence is the classical Lucas sequence. Proposition 2: Let q, r, m be integers with m ≥ 2. (a) The (q, r)-Fibona ...
Section 7-7 De Moivre`s Theorem
... We now come to one of the great theorems in mathematics, De Moivre’s theorem. Abraham De Moivre (1667–1754), of French birth, spent most of his life in London doing private tutoring, writing, and publishing mathematics. He belonged to many prestigious professional societies in England, Germany, and ...
... We now come to one of the great theorems in mathematics, De Moivre’s theorem. Abraham De Moivre (1667–1754), of French birth, spent most of his life in London doing private tutoring, writing, and publishing mathematics. He belonged to many prestigious professional societies in England, Germany, and ...
The Intermediate Value Theorem
... with what f (x) NEAR a. The third condition can be interpreted as saying “f (x) is continuous at a if the value of f (x) AT a is consistent with the values of f (x) NEAR a.” The Intermediate Value Theorem provides the formal justification of our intuition that a continuous function is one whose grap ...
... with what f (x) NEAR a. The third condition can be interpreted as saying “f (x) is continuous at a if the value of f (x) AT a is consistent with the values of f (x) NEAR a.” The Intermediate Value Theorem provides the formal justification of our intuition that a continuous function is one whose grap ...
Generating Elliptic Curves of Prime Order
... “anomolous” elliptic curves; these elliptic curves, of order p over fields of characteristic p, have since been shown to be insecure, [14], [16], [19]. However, the idea of the construction can be applied to quickly find non-anomolous curves as well. We present such a variant of the method to constru ...
... “anomolous” elliptic curves; these elliptic curves, of order p over fields of characteristic p, have since been shown to be insecure, [14], [16], [19]. However, the idea of the construction can be applied to quickly find non-anomolous curves as well. We present such a variant of the method to constru ...
Chapter 1 - Basics of Geometry Section 1.1
... 2. Make a conjecture. (conjecture: an unproven statement that is based on observations.) use examples to make conjecture 3. Verify the conjecture. test it (verify that it works for all cases) Counterexample: an example that shows a conjecture is false. ...
... 2. Make a conjecture. (conjecture: an unproven statement that is based on observations.) use examples to make conjecture 3. Verify the conjecture. test it (verify that it works for all cases) Counterexample: an example that shows a conjecture is false. ...
CSE 20 * Discrete Mathematics
... Assume that for some n2, all integers 2kn are divisible by a prime. WTS that n+1 is divisible by a prime. Proof by cases: Case 1: n+1 is prime. n+1 divides itself so we are done. Case 2: n+1 is composite. Then n+1=ab with 1
... Assume that for some n2, all integers 2kn are divisible by a prime. WTS that n+1 is divisible by a prime. Proof by cases: Case 1: n+1 is prime. n+1 divides itself so we are done. Case 2: n+1 is composite. Then n+1=ab with 1
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... solutions. Accordingly, we now show that e cannot be congruent if e = k + t 2 , and (k + t) is not a quadratic residue of e. For this we investigate the subsidiary-equation system (1). Say e is a prime of form (Sr + 1), represented uniquely as k + t . We have the system: x2 + y2 = ez2, x2 - y2 = w2. ...
... solutions. Accordingly, we now show that e cannot be congruent if e = k + t 2 , and (k + t) is not a quadratic residue of e. For this we investigate the subsidiary-equation system (1). Say e is a prime of form (Sr + 1), represented uniquely as k + t . We have the system: x2 + y2 = ez2, x2 - y2 = w2. ...
Recurrence of incomplete quotients of continued fractions
... The measures H1 and ν are congruent. It is su cient to show this on Cn1 ,...,nk (A1 , . . . , Ak ) = {α | an1 ∈ A1 , . . . , ank ∈ Ak }, where Ai ⊆ N. By covering any such cylinder C with elementary cylinders Ik (a1 , . . . , ak ), we can verify that ν(C) ≤ H1 (C). Conversely, it follows from the de ...
... The measures H1 and ν are congruent. It is su cient to show this on Cn1 ,...,nk (A1 , . . . , Ak ) = {α | an1 ∈ A1 , . . . , ank ∈ Ak }, where Ai ⊆ N. By covering any such cylinder C with elementary cylinders Ik (a1 , . . . , ak ), we can verify that ν(C) ≤ H1 (C). Conversely, it follows from the de ...
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... The enumeration of Kekule structures for benzenoid polycyclic hydrocarbons is important because the stability and many other properties of these hydrocarbons have been found to correlate with the number of Kekule structures. Starting with the algorithm proposed by Gordon & Davison [8], many papers h ...
... The enumeration of Kekule structures for benzenoid polycyclic hydrocarbons is important because the stability and many other properties of these hydrocarbons have been found to correlate with the number of Kekule structures. Starting with the algorithm proposed by Gordon & Davison [8], many papers h ...
Triangular number
... The rightmost term in the formula, consisting of the two numbers n + 1 and 2 on top of each other within parentheses, is the standard mathematical notation for a binomial coefficient that counts the number of distinct pairs to be selected from n + 1 objects. In this form the triangular number Tn sol ...
... The rightmost term in the formula, consisting of the two numbers n + 1 and 2 on top of each other within parentheses, is the standard mathematical notation for a binomial coefficient that counts the number of distinct pairs to be selected from n + 1 objects. In this form the triangular number Tn sol ...
Homework 5 (=Exam Practice)
... Homework 5 (=Exam Practice) The exam will cover everything we did so far: induction, binomial theorem, divisibility, division algorithm, gcd, euclidean algorithm, fundamental theorem of arithmetic, prime numbers, congruences, divisibility tests, chinese remainder theorem, Fermat’s theorem, pseudopri ...
... Homework 5 (=Exam Practice) The exam will cover everything we did so far: induction, binomial theorem, divisibility, division algorithm, gcd, euclidean algorithm, fundamental theorem of arithmetic, prime numbers, congruences, divisibility tests, chinese remainder theorem, Fermat’s theorem, pseudopri ...
Newsletter – Ch 7
... 8.EE.2: Use square root and cube root symbols to represent solutions to equations of the form x2 = p and x3 = p, where p is a positive rational number. Evaluate square roots of small perfect squares and cube roots of small perfect cubes. Know that √2 is irrational. 8.G.6: Explain a proof of the Pyth ...
... 8.EE.2: Use square root and cube root symbols to represent solutions to equations of the form x2 = p and x3 = p, where p is a positive rational number. Evaluate square roots of small perfect squares and cube roots of small perfect cubes. Know that √2 is irrational. 8.G.6: Explain a proof of the Pyth ...
![[Part 1]](http://s1.studyres.com/store/data/008795780_1-2d32093eb8955eecb4220b99bc38a981-300x300.png)
[Part 1]
... tains asymptotically a proportion of the j3 f s equal to the length of the interval, and clearly the same will be true for any sub-interval (a,]3) of [ 0 , 1 ] . The classical Weyl criterion [ 1 , p. 76] states that { x.} ...
... tains asymptotically a proportion of the j3 f s equal to the length of the interval, and clearly the same will be true for any sub-interval (a,]3) of [ 0 , 1 ] . The classical Weyl criterion [ 1 , p. 76] states that { x.} ...
Maximizing the number of nonnegative subsets
... 1. Introduction. Let {x1 , · · · , xn } be a sequence of n real numbers whose sum is negative. It is natural to ask the following question: What is the maximum possible number of subsets of nonnegative Pn sum it can have? One can set x1 = n − 2 and x2 = · · · = xn = −1. This gives i=1 xi = −1 < 0 an ...
... 1. Introduction. Let {x1 , · · · , xn } be a sequence of n real numbers whose sum is negative. It is natural to ask the following question: What is the maximum possible number of subsets of nonnegative Pn sum it can have? One can set x1 = n − 2 and x2 = · · · = xn = −1. This gives i=1 xi = −1 < 0 an ...
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... teger value > p M and such that (19) is satisfied. The Lemma assures us that there are infinitely many values of p > p M such that ...
... teger value > p M and such that (19) is satisfied. The Lemma assures us that there are infinitely many values of p > p M such that ...
A Refinement of the Function $ g (m) $ on Grimm Conjecture
... Output: An m-bit prime number Due to the fact that it lies outside the scope of this paper. We omitted more details of this algorithm. As a toy example, we can generate a 32-bit prime using small primes 29, 31, 37, 41, 43, 47. Note that 231 < 29 × 31 × 37 × 41 × 43 × 47 < 232 . By an exhaustive sear ...
... Output: An m-bit prime number Due to the fact that it lies outside the scope of this paper. We omitted more details of this algorithm. As a toy example, we can generate a 32-bit prime using small primes 29, 31, 37, 41, 43, 47. Note that 231 < 29 × 31 × 37 × 41 × 43 × 47 < 232 . By an exhaustive sear ...