15(1)
... / = 0, 1, 2, 3, 4 but composite for / = 5, 6. It is an unsolved problem whether or not 22' + 1 has other prime values. We note in passing that, when k = 2,F6=8 = 23, and 8m ± 1 = (23 ) ^ ± 1 = (2m ) 3 ± 7 is always composite, since A 3 ± B is always factorable. It is th ought that Fg + 1 is a prime. ...
... / = 0, 1, 2, 3, 4 but composite for / = 5, 6. It is an unsolved problem whether or not 22' + 1 has other prime values. We note in passing that, when k = 2,F6=8 = 23, and 8m ± 1 = (23 ) ^ ± 1 = (2m ) 3 ± 7 is always composite, since A 3 ± B is always factorable. It is th ought that Fg + 1 is a prime. ...
Midterm #3: practice
... (b) We need to compute 2340 (mod 341). We proceed using binary exponentiation as in the previous part. The values we get modulo 341 are: 22 = 4, 24 = 16, 28 = 256, 216 = 64, 232 = 4, so that, again, values repeat. In the end, we nd that 2340 1 (mod 341). This means that 341 is a pseudoprime to th ...
... (b) We need to compute 2340 (mod 341). We proceed using binary exponentiation as in the previous part. The values we get modulo 341 are: 22 = 4, 24 = 16, 28 = 256, 216 = 64, 232 = 4, so that, again, values repeat. In the end, we nd that 2340 1 (mod 341). This means that 341 is a pseudoprime to th ...
Amazing properties of binomial coefficients
... corner in the upper right direction and from the low right corner in the upper left direction. It follows also that the triangle of the double size contains 3 copies of the initial triangle. Now let us prove by induction that there exists a bijection between T Hn and Pn , such that the vertices of t ...
... corner in the upper right direction and from the low right corner in the upper left direction. It follows also that the triangle of the double size contains 3 copies of the initial triangle. Now let us prove by induction that there exists a bijection between T Hn and Pn , such that the vertices of t ...
On integers of the forms k ± 2n and k2 n ± 1
... Erdős and Odlyzko [19] proved that the set of odd numbers k for which there exists a positive integer n with k2n + 1 being prime has positive lower asymptotic density in the set of all positive odd integers. On the other hand, Sierpiński [34] proved that there are infinitely many positive odd numb ...
... Erdős and Odlyzko [19] proved that the set of odd numbers k for which there exists a positive integer n with k2n + 1 being prime has positive lower asymptotic density in the set of all positive odd integers. On the other hand, Sierpiński [34] proved that there are infinitely many positive odd numb ...
Fermat`s Little Theorem and Chinese Remainder Theorem Solutions
... Now by Fermat’s Little Theorem, 2q−1 ≡ 1 (mod q), so d = p divides q − 1. This implies that p ≤ q − 1, so q > p. A consequence of this result is the fact that there are infinitely many prime numbers. This was known by the mathematicians of ancient Greece. 9) n = 1 : 1, 2 work n = 2 : 512 works Assum ...
... Now by Fermat’s Little Theorem, 2q−1 ≡ 1 (mod q), so d = p divides q − 1. This implies that p ≤ q − 1, so q > p. A consequence of this result is the fact that there are infinitely many prime numbers. This was known by the mathematicians of ancient Greece. 9) n = 1 : 1, 2 work n = 2 : 512 works Assum ...
Week 2
... Suppose one wants to find all prime numbers less than 200. Eratosthenes (276–194 BCE) devised a simple method to do this. Here is a brief description: List the numbers from 2 to 200. For prime numbers p = 2, 3, 5, 11, 13 in turn delete all multiples of that prime except themselves. The integers that ...
... Suppose one wants to find all prime numbers less than 200. Eratosthenes (276–194 BCE) devised a simple method to do this. Here is a brief description: List the numbers from 2 to 200. For prime numbers p = 2, 3, 5, 11, 13 in turn delete all multiples of that prime except themselves. The integers that ...
Chapter 8
... The theorem follows rather simply from some of our following work: (p – 1)! ≡ -1 (mod p) for all primes p. This result can be verified for p = 2. Now, let’s consider all odd p. Since each value 1, 2, …, p – 1 is relatively prime to p, each has an inverse mod p. We know that the inverse of 1 is 1 and ...
... The theorem follows rather simply from some of our following work: (p – 1)! ≡ -1 (mod p) for all primes p. This result can be verified for p = 2. Now, let’s consider all odd p. Since each value 1, 2, …, p – 1 is relatively prime to p, each has an inverse mod p. We know that the inverse of 1 is 1 and ...
Link to project draft - Department of Mathematics
... prime ideals, the fundamental theorem of arithmetic have application in cryptography and other fields. 1.1. Fundamental Theorem of Arithmetic. The Fundamental Theorem of Arithmetic is an important theorem in number theory, and it is also called the unique factorization theorem or the unique prime fa ...
... prime ideals, the fundamental theorem of arithmetic have application in cryptography and other fields. 1.1. Fundamental Theorem of Arithmetic. The Fundamental Theorem of Arithmetic is an important theorem in number theory, and it is also called the unique factorization theorem or the unique prime fa ...
notes on rational and real numbers
... Then objects like 5, or sin 10◦ , or log10 7 would cease to exist, as they do not exist among rational numbers. The use of quadratic equations, or Trigonometry, or Calculus would terminate ... Well, enough of this nightmare. It is often hard to define basic mathematical notions. The rigor of such de ...
... Then objects like 5, or sin 10◦ , or log10 7 would cease to exist, as they do not exist among rational numbers. The use of quadratic equations, or Trigonometry, or Calculus would terminate ... Well, enough of this nightmare. It is often hard to define basic mathematical notions. The rigor of such de ...
A DUAL APPROACH TO TRIANGLE SEQUENCES
... periodicity of the sequence will imply algebraicity; none proves the converse. Probably there is no single such technique that will fully answer Hermite’s initial question. It is more likely that there is a whole family of techniques, each providing a periodic sequence for different classes of algebr ...
... periodicity of the sequence will imply algebraicity; none proves the converse. Probably there is no single such technique that will fully answer Hermite’s initial question. It is more likely that there is a whole family of techniques, each providing a periodic sequence for different classes of algebr ...
Number Theory
... (3) The greatest common factor of two integers a and b is the largest among the common factors of a and b. The greatest common factor of a and b is denoted by gcf(a, b). Remark 1.9. In the literature it is much more common to say f divides a (or f divides a evenly) than to say that f is a factor of ...
... (3) The greatest common factor of two integers a and b is the largest among the common factors of a and b. The greatest common factor of a and b is denoted by gcf(a, b). Remark 1.9. In the literature it is much more common to say f divides a (or f divides a evenly) than to say that f is a factor of ...
1 slide/page
... • Goldbach’s Conjecture: every even number greater than 2 is the sum of two primes. ◦ E.g., 6 = 3 + 3, 20 = 17 + 3, 28 = 17 + 11 ◦ This has been checked out to 6 × 1016 (as of 2003) ◦ Every sufficiently large integer (> 1043,000!) is the sum of four primes • Two prime numbers that differ by two are ...
... • Goldbach’s Conjecture: every even number greater than 2 is the sum of two primes. ◦ E.g., 6 = 3 + 3, 20 = 17 + 3, 28 = 17 + 11 ◦ This has been checked out to 6 × 1016 (as of 2003) ◦ Every sufficiently large integer (> 1043,000!) is the sum of four primes • Two prime numbers that differ by two are ...
4 slides/page
... ∗ Keep choosing odd numbers at random ∗ Check if they are prime (using fast randomized primality test) ∗ Keep trying until you find one ∗ Roughly 100 attempts should do it ...
... ∗ Keep choosing odd numbers at random ∗ Check if they are prime (using fast randomized primality test) ∗ Keep trying until you find one ∗ Roughly 100 attempts should do it ...
On integers n for which X n – 1 has divisors of every degree
... When considering ϕ-practical numbers n with a given squarefull part s, it is natural to consider certain “primitive” ϕ-practical numbers which have squarefull part s, which we call starters. Definition 4.1. A starter is a ϕ-practical number m such that either m/P + (m) is not ϕ-practical or P + (m)2 ...
... When considering ϕ-practical numbers n with a given squarefull part s, it is natural to consider certain “primitive” ϕ-practical numbers which have squarefull part s, which we call starters. Definition 4.1. A starter is a ϕ-practical number m such that either m/P + (m) is not ϕ-practical or P + (m)2 ...
Smoothness of the sum and Riemann summability of double
... In the second part of the disertation we de ne two new summation methods: the Riemann summability od double trigonometric series, and Lebesgue summability of double trigonometric integrals. In the third chapter we extend the concept of the Riemann summability from single to double trigonometric seri ...
... In the second part of the disertation we de ne two new summation methods: the Riemann summability od double trigonometric series, and Lebesgue summability of double trigonometric integrals. In the third chapter we extend the concept of the Riemann summability from single to double trigonometric seri ...
Wilson`s Theorem and Fermat`s Theorem
... Proof. If k = 1, then k 2 = 1 (mod p). If k = p − 1, then k 2 = p2 − 2p + 1 = 1 (mod p) . Conversely, suppose k 2 = 1 (mod p). Then p | k 2 − 1 = (k − 1)(k + 1), and since p is prime, p | k − 1 or p | k + 1. The only number in {1, . . . , p − 1} which satisfies p | k − 1 is 1, and the only number in ...
... Proof. If k = 1, then k 2 = 1 (mod p). If k = p − 1, then k 2 = p2 − 2p + 1 = 1 (mod p) . Conversely, suppose k 2 = 1 (mod p). Then p | k 2 − 1 = (k − 1)(k + 1), and since p is prime, p | k − 1 or p | k + 1. The only number in {1, . . . , p − 1} which satisfies p | k − 1 is 1, and the only number in ...
Full text
... let Sp(n) denote the number of binomial coefficients (?) that are not divisible hyp. Then ...
... let Sp(n) denote the number of binomial coefficients (?) that are not divisible hyp. Then ...
On values taken by the largest prime factor of shifted primes
... in the open interval .0; 17=32/, such that for every integer a = 0 and real number K , the inequalities C 2 .#/ y C 1 .#/ y < ³.y; p; a/ < '. p/ log y '. p/ log y hold for all primes p ≤ y # , with at most O.y # = log K y/ exceptions, where the implied constant depends only on a, #, and K . Moreov ...
... in the open interval .0; 17=32/, such that for every integer a = 0 and real number K , the inequalities C 2 .#/ y C 1 .#/ y < ³.y; p; a/ < '. p/ log y '. p/ log y hold for all primes p ≤ y # , with at most O.y # = log K y/ exceptions, where the implied constant depends only on a, #, and K . Moreov ...
On a limit involving the product of prime numbers 1 Introduction
... log pn n n log pn Theorem 2.1 is finished, as log An → 1 implies An → e, (i.e.) relation (1.7) holds true. Now, by (2.3) of Lemma 2.2 one gets ...
... log pn n n log pn Theorem 2.1 is finished, as log An → 1 implies An → e, (i.e.) relation (1.7) holds true. Now, by (2.3) of Lemma 2.2 one gets ...
Lecture Notes for College Discrete Mathematics Szabolcs Tengely
... reaching 9 lines, he crosses them on the tenth day (thus marking them as ten). That way he groups together every ten days. Then, when he reaches ten of such groups, then he carves a big box around them. That is how he indicates hundreds. Then he circles around every ten boxes, indicating thousands, ...
... reaching 9 lines, he crosses them on the tenth day (thus marking them as ten). That way he groups together every ten days. Then, when he reaches ten of such groups, then he carves a big box around them. That is how he indicates hundreds. Then he circles around every ten boxes, indicating thousands, ...
Powers of Two as Sums of Two Lucas Numbers
... as the Fibonacci numbers, but with initial conditions L0 = 2 and L1 = 1. The study of properties of the terms of such sequences, or more generally, linear recurrence sequences, has a very long history and has generated a huge literature. For the beauty and rich applications of these numbers and thei ...
... as the Fibonacci numbers, but with initial conditions L0 = 2 and L1 = 1. The study of properties of the terms of such sequences, or more generally, linear recurrence sequences, has a very long history and has generated a huge literature. For the beauty and rich applications of these numbers and thei ...