Divisibility and Congruence Definition. Let a ∈ Z − {0} and b ∈ Z
... 11. Cool Property 1. For all integers a, b, x, y, if a ≡ x mod m and b ≡ y mod m, then a + b ≡ x + y mod m. Proof. By hypothesis m | (x − a) and m | (y − b). This means (x − a) = mp and (y − b) = mq for some p, q ∈ Z. Adding gives ((x + y) − (a + b)) = (x − a) + (y − b) = m(p + q), and so m | (x + ...
... 11. Cool Property 1. For all integers a, b, x, y, if a ≡ x mod m and b ≡ y mod m, then a + b ≡ x + y mod m. Proof. By hypothesis m | (x − a) and m | (y − b). This means (x − a) = mp and (y − b) = mq for some p, q ∈ Z. Adding gives ((x + y) − (a + b)) = (x − a) + (y − b) = m(p + q), and so m | (x + ...
On the digital representation of integers with bounded prime factors
... We still do not know whether there are infinitely many prime numbers of the form 2n + 1 (that is, with only two nonzero binary digits) or of the form 11 . . . 11 (that is, with only the digit 1 in their decimal representation). Both questions are notorious, very difficult open problems, which at pre ...
... We still do not know whether there are infinitely many prime numbers of the form 2n + 1 (that is, with only two nonzero binary digits) or of the form 11 . . . 11 (that is, with only the digit 1 in their decimal representation). Both questions are notorious, very difficult open problems, which at pre ...
The Period and the Distribution of the Fibonacci
... 5, 8, .... Therefore it is easy to see that it has a period of 60. Theorem 1. Let an = F ( mod m) , then there exists a positive integer s such that (as ,as+1 ) = (1, 1). In particular F ( mod m) has a period. Proof. Since 0 ≤ an < m, there are only m2 possible pairs of residues, and hence there mus ...
... 5, 8, .... Therefore it is easy to see that it has a period of 60. Theorem 1. Let an = F ( mod m) , then there exists a positive integer s such that (as ,as+1 ) = (1, 1). In particular F ( mod m) has a period. Proof. Since 0 ≤ an < m, there are only m2 possible pairs of residues, and hence there mus ...
Solutions - Math@LSU
... congruence we have from Fermat’s Little Theorem implies the necessary equivalence, so we are done. 8. (Niven 2.1.25) Prove that 91 | (n12 − a12 ) for any a, n that are both coprime to 91. Give an counterexample showing that this condition is necessary. Note that 91 = 7 · 13, so we want to show that ...
... congruence we have from Fermat’s Little Theorem implies the necessary equivalence, so we are done. 8. (Niven 2.1.25) Prove that 91 | (n12 − a12 ) for any a, n that are both coprime to 91. Give an counterexample showing that this condition is necessary. Note that 91 = 7 · 13, so we want to show that ...
40(4)
... author must appear at the beginning of the paper directly under the title. Illustrations should be carefully drawn in India ink on separate sheets of bond paper or vellum, approximately twice the size they are to appear in print. Since the Fibonacci Association has adopted F{ = F2 = 1, F» +;= F« +F« ...
... author must appear at the beginning of the paper directly under the title. Illustrations should be carefully drawn in India ink on separate sheets of bond paper or vellum, approximately twice the size they are to appear in print. Since the Fibonacci Association has adopted F{ = F2 = 1, F» +;= F« +F« ...
A New 3n − 1 Conjecture Akin to Collatz Conjecture
... the Ulam conjecture (after Stanislaw Ulam), Kakutani’s problem (after Shizuo Kakutani) and so on. In this paper a new conjecture called as the 3n − 1 conjecture which is akin to the Collatz conjecture is proposed. It functions on 3n − 1, for any starting number n, its sequence eventually reaches eit ...
... the Ulam conjecture (after Stanislaw Ulam), Kakutani’s problem (after Shizuo Kakutani) and so on. In this paper a new conjecture called as the 3n − 1 conjecture which is akin to the Collatz conjecture is proposed. It functions on 3n − 1, for any starting number n, its sequence eventually reaches eit ...
Five regular or nearly-regular ternary quadratic forms
... The second type of revision is present in a variety of articles, including [5] and [6]. It deals with a binary diagonal form r2 + ks2 , k ≥ 1. We give the formulation (Theorem 9, p. 51) in the unpublished Ph.D. dissertation of Burton Jones, Univ. of Chicago, 1928. Lemma (Jones). If f = r2 + ks2 repr ...
... The second type of revision is present in a variety of articles, including [5] and [6]. It deals with a binary diagonal form r2 + ks2 , k ≥ 1. We give the formulation (Theorem 9, p. 51) in the unpublished Ph.D. dissertation of Burton Jones, Univ. of Chicago, 1928. Lemma (Jones). If f = r2 + ks2 repr ...
Number theory.doc
... especially appreciate glossaries in an open and distance education context because there is no teacher in front of them, when they encounter an unfamiliar word and cannot easily figure out its definition in the text. Module Developers should select at least the 10 most important terms to include in ...
... especially appreciate glossaries in an open and distance education context because there is no teacher in front of them, when they encounter an unfamiliar word and cannot easily figure out its definition in the text. Module Developers should select at least the 10 most important terms to include in ...
Hensel codes of square roots of p
... (a) If m = 0, then we have quadratic convergence for all the p-adic numbers which belong to the set S1 . (b) If m < 0, then the speed of convergence is faster for all the p-adic numbers which belong to the set S3 . (c) If m > 0, then the speed of convergence is slower for all the p-adic numbers whic ...
... (a) If m = 0, then we have quadratic convergence for all the p-adic numbers which belong to the set S1 . (b) If m < 0, then the speed of convergence is faster for all the p-adic numbers which belong to the set S3 . (c) If m > 0, then the speed of convergence is slower for all the p-adic numbers whic ...
Here - Math-Boise State
... Basis step: let n = 1. If m ≤ 1, then m = 1. m ∈ A, that is 1 ∈ A, is impossible because 1 is the smallest natural number and A has no smallest element: if 1 were in A, 1 would be the smallest element of A. Induction step: Suppose (inductive hypothesis) that for all m ≤ k, m 6∈ A. Our induction goa ...
... Basis step: let n = 1. If m ≤ 1, then m = 1. m ∈ A, that is 1 ∈ A, is impossible because 1 is the smallest natural number and A has no smallest element: if 1 were in A, 1 would be the smallest element of A. Induction step: Suppose (inductive hypothesis) that for all m ≤ k, m 6∈ A. Our induction goa ...
For a nonnegative integer k. the kth Fermar number Fk
... and Cunningham numbers implies that these numbers are losing their value as a yardstick to measure progress in factoring. One wonders which class of num bers will take their place. Good test numbers for factoring algorithms should meet several conditions. They should be defined a priori, to avoid th ...
... and Cunningham numbers implies that these numbers are losing their value as a yardstick to measure progress in factoring. One wonders which class of num bers will take their place. Good test numbers for factoring algorithms should meet several conditions. They should be defined a priori, to avoid th ...
Solutions CS2800 Questions selected for spring 2017
... can be formed using just three-cent and four-cent stamps. We use regular induction to prove that P (n) holds for all n ≥ 3, where P (n) says that postage of n cents cents can be formed using just three-cent and four-cent stamps. Base Case: We can form postage of three cents with a single three-cent ...
... can be formed using just three-cent and four-cent stamps. We use regular induction to prove that P (n) holds for all n ≥ 3, where P (n) says that postage of n cents cents can be formed using just three-cent and four-cent stamps. Base Case: We can form postage of three cents with a single three-cent ...
3n+1 summary - D-Scholarship@Pitt
... Mom, Dad, you have always been in my corner to help with tough situations and provide lap time when needed. I’m coming back with my shield. ...
... Mom, Dad, you have always been in my corner to help with tough situations and provide lap time when needed. I’m coming back with my shield. ...
Section 2
... numbers, like Carmichael numbers that can be used to discourage using Fermat’s Little Theorem to show that a number is prime, to discourage using the Rabin-Miller test to show a number is prime. 3. Although the test cannot be used directly to show a number is prime, its repeated use can be to verify ...
... numbers, like Carmichael numbers that can be used to discourage using Fermat’s Little Theorem to show that a number is prime, to discourage using the Rabin-Miller test to show a number is prime. 3. Although the test cannot be used directly to show a number is prime, its repeated use can be to verify ...
Elementary Number Theory, A Computational Approach
... Computing the composites in an interval is a little more complicated: sage: [n for n in range(10,30) if not is_prime(n)] ...
... Computing the composites in an interval is a little more complicated: sage: [n for n in range(10,30) if not is_prime(n)] ...
Acta Mathematica Universitatis Ostraviensis - DML-CZ
... (Rotkiewicz and Schinzel [31]). (4) There exist infinitely many square-free pseudoprimes divisible by an arbitrary given prime (Rotkiewicz [28]). (5) There exist infinitely many arithmetic progressions formed by four pseudoprimes (Rotkiewicz [34]). The smallest even pseudoprime was found by Lehmer i ...
... (Rotkiewicz and Schinzel [31]). (4) There exist infinitely many square-free pseudoprimes divisible by an arbitrary given prime (Rotkiewicz [28]). (5) There exist infinitely many arithmetic progressions formed by four pseudoprimes (Rotkiewicz [34]). The smallest even pseudoprime was found by Lehmer i ...
