Evidence for the Riemann Hypothesis - Léo Agélas
... [9] J. Derbyshire : Prime Obsession: Bernhard Riemann and the Greatest Unsolved Problem in Mathematics. New York: Penguin, pp. 245-250, (2004). [10] H.M. Edwards : Riemann’s zeta function Acad. Press. New York. (1974). [11] Erdõs, P. and Kac, M.: The Gaussian Law of Errors in the Theory of Additive ...
... [9] J. Derbyshire : Prime Obsession: Bernhard Riemann and the Greatest Unsolved Problem in Mathematics. New York: Penguin, pp. 245-250, (2004). [10] H.M. Edwards : Riemann’s zeta function Acad. Press. New York. (1974). [11] Erdõs, P. and Kac, M.: The Gaussian Law of Errors in the Theory of Additive ...
FACTORING WITH CONTINUED FRACTIONS, THE PELL
... It is easy to multiply integers but, conversely, it is rather di cult to nd the prime factorization of a given large integer. This is the basis of many cryptosystems in practice. It is conjectured that factoring is an NP-problem, i.e., roughly speaking, there does not exist a fast factoring algorit ...
... It is easy to multiply integers but, conversely, it is rather di cult to nd the prime factorization of a given large integer. This is the basis of many cryptosystems in practice. It is conjectured that factoring is an NP-problem, i.e., roughly speaking, there does not exist a fast factoring algorit ...
FAMILIES OF NON-θ-CONGRUENT NUMBERS WITH
... Theorem 1.4. Let p1 , . . . , p2` be distinct primes with ` > 0 such that pi ≡ 13 (mod 24) and pji = −1 for j < i. Then the product n = 2p1 · · · p2` is a non-2π/3-congruent number. Our proofs are relatively elementary, building upon a generalization of the work of Serf [Se91] for classical congruen ...
... Theorem 1.4. Let p1 , . . . , p2` be distinct primes with ` > 0 such that pi ≡ 13 (mod 24) and pji = −1 for j < i. Then the product n = 2p1 · · · p2` is a non-2π/3-congruent number. Our proofs are relatively elementary, building upon a generalization of the work of Serf [Se91] for classical congruen ...
G30 MATH SEMINAR 1 - PROOFS BY CONTRADICTION 1
... (i) Let x, y P Z. If x and y are odd then xy is also odd. (ii) Let a, b, c P Z. If a divides b and a divides c then a divides b ` c. (iii) Let a P Z. If a2 divides a then a P t´1, 0, 1u. (iv) Let n P Z. Either a2 is divisible by 4 either a2 ´ 1 is divisible by 4. Exercise 3.3. Use a proof by contrad ...
... (i) Let x, y P Z. If x and y are odd then xy is also odd. (ii) Let a, b, c P Z. If a divides b and a divides c then a divides b ` c. (iii) Let a P Z. If a2 divides a then a P t´1, 0, 1u. (iv) Let n P Z. Either a2 is divisible by 4 either a2 ´ 1 is divisible by 4. Exercise 3.3. Use a proof by contrad ...
Theorem 4.2: W6n+k - The Fibonacci Quarterly
... Some of the results in this paper are not as "practical" as others. For example, if we put n = 10 and k = 0 in (2.13), then we seek to find W40. However, on the right-hand side, we need to know W6, Wl2, Wls,..., W60 (and many other terms) in order to find W4Q. In contrast, (2.14) is more practical s ...
... Some of the results in this paper are not as "practical" as others. For example, if we put n = 10 and k = 0 in (2.13), then we seek to find W40. However, on the right-hand side, we need to know W6, Wl2, Wls,..., W60 (and many other terms) in order to find W4Q. In contrast, (2.14) is more practical s ...
Reducing the Erdos-Moser equation 1^ n+ 2^ n+...+ k^ n=(k+ 1)^ n
... there is no solution. (r = 6, 7, 8). For the numbers K in Table 1 with r = 6, 7, 8 prime factors, conditions (i) and (ii) k require + 1 ≡ −p (mod p2 ), for p = 2, 3, 2, respectively. But the requirement is violated in each p case, and so no solution exists. (ii). Part (i) implies that the only possi ...
... there is no solution. (r = 6, 7, 8). For the numbers K in Table 1 with r = 6, 7, 8 prime factors, conditions (i) and (ii) k require + 1 ≡ −p (mod p2 ), for p = 2, 3, 2, respectively. But the requirement is violated in each p case, and so no solution exists. (ii). Part (i) implies that the only possi ...
Lecture 8 - Math TAMU
... Problem 9. Find the multiplicative order of 7 modulo 36. Problem 10. Determine the last two digits of 7303 . Problem 11. How many integers from 1 to 120 are relatively prime with 120? Problem 12. You receive a message that was encrypted using the RSA system with public key (33, 7), where 33 is the b ...
... Problem 9. Find the multiplicative order of 7 modulo 36. Problem 10. Determine the last two digits of 7303 . Problem 11. How many integers from 1 to 120 are relatively prime with 120? Problem 12. You receive a message that was encrypted using the RSA system with public key (33, 7), where 33 is the b ...
Monomials and Factoring
... Find the GCF of 48 and 60 Write the prime factorization of each number. “Factor” each number. Circle the common prime factors. The common prime factors are 2, 2, and 3. The product of the common prime factors is called the greatest common factor (GCF). ...
... Find the GCF of 48 and 60 Write the prime factorization of each number. “Factor” each number. Circle the common prime factors. The common prime factors are 2, 2, and 3. The product of the common prime factors is called the greatest common factor (GCF). ...
