An identity involving the least common multiple of
... Recently, many related questions and many generalizations of the above results have been studied by several authors. The interested reader is referred to [1], [2], and [5]. In this note, using Kummer’s theorem on the p-adic valuation ¡k¢of¡binomial ...
... Recently, many related questions and many generalizations of the above results have been studied by several authors. The interested reader is referred to [1], [2], and [5]. In this note, using Kummer’s theorem on the p-adic valuation ¡k¢of¡binomial ...
Solutions to Practice Problems, Math 312 1 Find all prime numbers
... Solutions to Practice Problems, Math 312 ...
... Solutions to Practice Problems, Math 312 ...
Computing Greatest Common Divisors and Factorizations in
... ideal GCDs are computed in a certain sense, which we will precisely specify later. The algorithm has quadratic complexity for any fixed d. However, it requires some preparatory work (independent of ξ , η ), which is quite costly if d is large. It is not clear to us how to accomplish polynomial runni ...
... ideal GCDs are computed in a certain sense, which we will precisely specify later. The algorithm has quadratic complexity for any fixed d. However, it requires some preparatory work (independent of ξ , η ), which is quite costly if d is large. It is not clear to us how to accomplish polynomial runni ...
Chapter 4, Arithmetic in F[x] Polynomial arithmetic and the division
... Polynomials as functions. Let R be a commutative ring, f (x) = an xn + · · · + a0 ∈ R[x]. The function f : R → R induced by f (x) is called a polynomial function and is defined by f (r) = an rn + an−1 rn−1 + · · · + a1 r + a0 . We must be very careful in writing f (x) to differentiate between the po ...
... Polynomials as functions. Let R be a commutative ring, f (x) = an xn + · · · + a0 ∈ R[x]. The function f : R → R induced by f (x) is called a polynomial function and is defined by f (r) = an rn + an−1 rn−1 + · · · + a1 r + a0 . We must be very careful in writing f (x) to differentiate between the po ...
Chapter 2 Summary
... Determine the appropriate inequality symbols which will make each statement true (more than one symbol may apply). ...
... Determine the appropriate inequality symbols which will make each statement true (more than one symbol may apply). ...
The Pentagonal Number Theorem and All That
... publication of 1751. (This paper was written on April 6, 1741 and had no proof. Euler wrote so many papers that the publishers fell dramatically behind; they were publishing new papers many years after his death.) A typical entry, from a letter to Goldbach, reads “If these factors (1 − n)(1 − n2 )(1 ...
... publication of 1751. (This paper was written on April 6, 1741 and had no proof. Euler wrote so many papers that the publishers fell dramatically behind; they were publishing new papers many years after his death.) A typical entry, from a letter to Goldbach, reads “If these factors (1 − n)(1 − n2 )(1 ...
Math 8: Prime Factorization and Congruence
... mod p. Multiplying both sides of this equation by a yields ap ≡ a mod p. Corollary 2 Let p be a prime number. If [a] is any non-zero number in Zp , then there exists a number [b] in Zp such that [a][b] = 1. (In other words, there exists an inverse of the number [a] in Zp !) Proof: If [a] 6= [0], t ...
... mod p. Multiplying both sides of this equation by a yields ap ≡ a mod p. Corollary 2 Let p be a prime number. If [a] is any non-zero number in Zp , then there exists a number [b] in Zp such that [a][b] = 1. (In other words, there exists an inverse of the number [a] in Zp !) Proof: If [a] 6= [0], t ...
Math 319 Problem Set #6 – Solution 5 April 2002
... Factorizations starting with 2 × 3: None of these can occur, because 3 does not occur as ϕ of a prime power. Factorizations starting with 2 × 4: Since 3 cannot occur, the only such factorization is 2 × 4 × 6. This arises in several ways. We get ...
... Factorizations starting with 2 × 3: None of these can occur, because 3 does not occur as ϕ of a prime power. Factorizations starting with 2 × 4: Since 3 cannot occur, the only such factorization is 2 × 4 × 6. This arises in several ways. We get ...
Lab05MathFun0
... Extra Credit adds computing the Least Common Multiple (LCM) after the GCF is computed. The GCF makes computing the LCM very easy. Start by using Euclid's algorithm to compute the GCF. Then use the following formula to compute the LCM. Keep in mind that you must first compute an accurate GCF. LCM = N ...
... Extra Credit adds computing the Least Common Multiple (LCM) after the GCF is computed. The GCF makes computing the LCM very easy. Start by using Euclid's algorithm to compute the GCF. Then use the following formula to compute the LCM. Keep in mind that you must first compute an accurate GCF. LCM = N ...
NOTE ON 1-CROSSING PARTITIONS Given a partition π of the set
... The goal here is to generalize Bóna’s result to count 1-crossing partitions by their number of blocks, and also to examine a natural q-analogue with regard to the cyclic sieving phenomenon shown in [8] for certain q-Catalan and q-Narayana numbers. The crux is the observation that 1-crossing partiti ...
... The goal here is to generalize Bóna’s result to count 1-crossing partitions by their number of blocks, and also to examine a natural q-analogue with regard to the cyclic sieving phenomenon shown in [8] for certain q-Catalan and q-Narayana numbers. The crux is the observation that 1-crossing partiti ...
Full text
... So Euler was surprised that an takes its values from two progressions, trinomials of the second degree. However, notwithstanding what he believed, one often meets in analysis series of integers that take their values from several polynomials: the arithmetic polynomials, which all have a generating r ...
... So Euler was surprised that an takes its values from two progressions, trinomials of the second degree. However, notwithstanding what he believed, one often meets in analysis series of integers that take their values from several polynomials: the arithmetic polynomials, which all have a generating r ...
THE PROBABILITY OF RELATIVELY PRIME POLYNOMIALS
... Proof. We first note that a linear factor of g(x) must have the form u(x − r), where u, r ∈ Zpk , u is a unit, and r is a root of g. Therefore, the elements h(x) ∈ Ag are exactly the polynomials h(x) = α(x − r), for some α ∈ Zpk and some root r ∈ Zpk of g. Hence, to calculate |Ag |, we need to count ...
... Proof. We first note that a linear factor of g(x) must have the form u(x − r), where u, r ∈ Zpk , u is a unit, and r is a root of g. Therefore, the elements h(x) ∈ Ag are exactly the polynomials h(x) = α(x − r), for some α ∈ Zpk and some root r ∈ Zpk of g. Hence, to calculate |Ag |, we need to count ...
Seiberg-Witten Theory and Z/2^ p actions on spin 4
... This theorem recovers as a special case a theorem of Donaldson concerning involutions on the K3 ([5] Cor. 9.1.4) and is related to a theorem of Ruberman [11]. We also remark that both possibilities in the theorem actually occur. The proof of Theorems 1.2, 1.3, 1.4, and 1.5 uses Furuta’s technique of ...
... This theorem recovers as a special case a theorem of Donaldson concerning involutions on the K3 ([5] Cor. 9.1.4) and is related to a theorem of Ruberman [11]. We also remark that both possibilities in the theorem actually occur. The proof of Theorems 1.2, 1.3, 1.4, and 1.5 uses Furuta’s technique of ...
grades 7-9
... MidMichigan Mathematical Olympiad April 25, 2009 Solutions grades 7-9 1. Arrange the whole numbers 1 through 15 in a row so that the sum of any two adjacent numbers is a perfect square. In how many ways this can be done? Solution. There are two solutions ...
... MidMichigan Mathematical Olympiad April 25, 2009 Solutions grades 7-9 1. Arrange the whole numbers 1 through 15 in a row so that the sum of any two adjacent numbers is a perfect square. In how many ways this can be done? Solution. There are two solutions ...
Lab06MathFun / Microsoft Office Word 97
... Part 6: adds computing the Least Common Multiple (LCM) after the GCF is computed. The GCF makes computing the LCM very easy. Start by using Euclid's algorithm to compute the GCF. Then use the following formula to compute and return the LCM. Keep in mind that you must first compute an accurate GCF. ...
... Part 6: adds computing the Least Common Multiple (LCM) after the GCF is computed. The GCF makes computing the LCM very easy. Start by using Euclid's algorithm to compute the GCF. Then use the following formula to compute and return the LCM. Keep in mind that you must first compute an accurate GCF. ...
FIB notes Remainder and Factor Theorems
... FIB notes Remainder and Factor Theorems 1. The degree of a polynomial indicates how many _____________ zeros the function has. 2. If the degree of a polynomial is 6 and it only crosses the x-axis 2 times then there must be ____ imaginary zeros. 3. If a polynomial is divided by a linear expression an ...
... FIB notes Remainder and Factor Theorems 1. The degree of a polynomial indicates how many _____________ zeros the function has. 2. If the degree of a polynomial is 6 and it only crosses the x-axis 2 times then there must be ____ imaginary zeros. 3. If a polynomial is divided by a linear expression an ...
Chinese Reminder Theorem
... Here the numbers mi come by factoring m = m1 m2 · · · mk where gcd(mi , mj ) = 1 whenever i 6= j. Why do this? This is answered in the text (T&W, p. 77). By breaking the problem into simultaneous congruences mod each prime factor of m, we can recombine the resulting information to obtain an answer f ...
... Here the numbers mi come by factoring m = m1 m2 · · · mk where gcd(mi , mj ) = 1 whenever i 6= j. Why do this? This is answered in the text (T&W, p. 77). By breaking the problem into simultaneous congruences mod each prime factor of m, we can recombine the resulting information to obtain an answer f ...
CMPSCI 250: Introduction to Computation
... every positive natural does. It is either prime itself or is divisible by some smaller prime, and that prime cannot be less than or equal to z. So we know that some prime greater than z must exist, though we haven’t explicitly computed it. ...
... every positive natural does. It is either prime itself or is divisible by some smaller prime, and that prime cannot be less than or equal to z. So we know that some prime greater than z must exist, though we haven’t explicitly computed it. ...
#A11 INTEGERS 12 (2012) FIBONACCI VARIATIONS OF A
... Proof. We wish to build a covering C = {(ri , mi , pi )}, where the actual covering is the set of congruences {n ≡ ri (mod mi )}, and for each congruence in the covering, the corresponding prime pi is such that {Fn } has period mi modulo pi . In other words, Fn ≡ Fri (mod pi ) when n ≡ ri (mod mi ). ...
... Proof. We wish to build a covering C = {(ri , mi , pi )}, where the actual covering is the set of congruences {n ≡ ri (mod mi )}, and for each congruence in the covering, the corresponding prime pi is such that {Fn } has period mi modulo pi . In other words, Fn ≡ Fri (mod pi ) when n ≡ ri (mod mi ). ...
The Topsy-Turvy World of Continued Fractions [online]
... 47.5. Suppose that we use the recursion for pn backwards in order to define pn for negative values of n. What are the values of p−1 and p−2 ? Same question for q−1 and q−2 . 47.6. The Continued Fraction Recursion Formula (Theorem 47.1) gives a procedure for generating two lists of numbers p0 , p1 , ...
... 47.5. Suppose that we use the recursion for pn backwards in order to define pn for negative values of n. What are the values of p−1 and p−2 ? Same question for q−1 and q−2 . 47.6. The Continued Fraction Recursion Formula (Theorem 47.1) gives a procedure for generating two lists of numbers p0 , p1 , ...
QUADRATIC RESIDUES (MA2316, FOURTH WEEK) An integer a is
... The description of quadratic residues in the previous proof implies that the product of two quadratic residues is a quadratic residue, the product of a quadratic residue and a quadratic nonresidue is a quadratic nonresidue, and the product of two quadratic nonresidues is a quadratic residue. In othe ...
... The description of quadratic residues in the previous proof implies that the product of two quadratic residues is a quadratic residue, the product of a quadratic residue and a quadratic nonresidue is a quadratic nonresidue, and the product of two quadratic nonresidues is a quadratic residue. In othe ...
Polynomials for MATH136 Part A
... the roots of the polynomial, although technically we should only use that word for equations. So we can say that 2 is a zero of the polynomial x2 − 4 and it is a root of the equation x2 − 4 = 0. However not everyone adheres to this grammatical convention. Solving a polynomial equation f(x) = 0 there ...
... the roots of the polynomial, although technically we should only use that word for equations. So we can say that 2 is a zero of the polynomial x2 − 4 and it is a root of the equation x2 − 4 = 0. However not everyone adheres to this grammatical convention. Solving a polynomial equation f(x) = 0 there ...
Solution
... means that In ⊆ (a) for all n. However, a ∈ Im for some m by the definition of I; thus Im = (a) and is principal, a contradiction. (b) Since I is maximal among non-principal ideals, and Ia 6= I since a ∈ / I, we must have Ia be principal. We also know I ⊂ Ib , so all we need to check now is that Ib ...
... means that In ⊆ (a) for all n. However, a ∈ Im for some m by the definition of I; thus Im = (a) and is principal, a contradiction. (b) Since I is maximal among non-principal ideals, and Ia 6= I since a ∈ / I, we must have Ia be principal. We also know I ⊂ Ib , so all we need to check now is that Ib ...
Section 20 -- Fermat`s and Euler`s theorems
... • Suppose that a and b are not 0 nor zero divisors. We need to show that ab is neither 0 nor a zero divisor. • Since a and b are not 0 nor zero divisors, ab 6= 0. • Now suppose that (ab)c = 0. • Then a(bc) = 0. Since a is not 0 nor a zero divisors, ...
... • Suppose that a and b are not 0 nor zero divisors. We need to show that ab is neither 0 nor a zero divisor. • Since a and b are not 0 nor zero divisors, ab 6= 0. • Now suppose that (ab)c = 0. • Then a(bc) = 0. Since a is not 0 nor a zero divisors, ...
HALL-LITTLEWOOD POLYNOMIALS, ALCOVE WALKS, AND
... (Kostka-Foulkes polynomials of arbitrary type), which are certain affine Kazhdan-Lusztig polynomials [10, 23]. On the combinatorial side, we have the Lascoux-Schützenberger formula for the Kostka-Foulkes polynomials in type A [13], but no generalization of this formula to other types is known. Othe ...
... (Kostka-Foulkes polynomials of arbitrary type), which are certain affine Kazhdan-Lusztig polynomials [10, 23]. On the combinatorial side, we have the Lascoux-Schützenberger formula for the Kostka-Foulkes polynomials in type A [13], but no generalization of this formula to other types is known. Othe ...
Chinese remainder theorem
The Chinese remainder theorem is a result about congruences in number theory and its generalizations in abstract algebra. It was first published in the 3rd to 5th centuries by the Chinese mathematician Sun Tzu.In its basic form, the Chinese remainder theorem will determine a number n that, when divided by some given divisors, leaves given remainders. For example, what is the lowest number n that when divided by 3 leaves a remainder of 2, when divided by 5 leaves a remainder of 3, and when divided by 7 leaves a remainder of 2?