Primes of the form x2 + ny2
... can represent integers and thus also prime numbers. In particular, we will study which prime numbers are represented by the principal form x2 + ny 2 . For this, we first study the properties of quadratic forms and introduce an equivalence relation to group them together. Then, we will be able to stu ...
... can represent integers and thus also prime numbers. In particular, we will study which prime numbers are represented by the principal form x2 + ny 2 . For this, we first study the properties of quadratic forms and introduce an equivalence relation to group them together. Then, we will be able to stu ...
The Cantor Set and the Cantor Function
... by a descent and by more playing along the staircase). It is called ”L’Escalier du Diable” (i.e. ”The Devil’s Staircase”). It was composed in the 90’s by the composer György Ligeti. This composition is harmonically self-similar, like Cantor’s set. Its structure has the musical equivalent of dividin ...
... by a descent and by more playing along the staircase). It is called ”L’Escalier du Diable” (i.e. ”The Devil’s Staircase”). It was composed in the 90’s by the composer György Ligeti. This composition is harmonically self-similar, like Cantor’s set. Its structure has the musical equivalent of dividin ...
Counting degenerate polynomials of fixed degree and bounded height
... Proof. Notice that, for fixed n ≥ 2, there are only finitely many roots of unity which are ratios of two algebraic numbers of degree at most n. (See, e.g., [18] or [9, Corollary 1.3] for a more precise result asserting that deg(α/α′ ) ≤ deg α whenever α, α′ are two conjugate algebraic numbers whose ...
... Proof. Notice that, for fixed n ≥ 2, there are only finitely many roots of unity which are ratios of two algebraic numbers of degree at most n. (See, e.g., [18] or [9, Corollary 1.3] for a more precise result asserting that deg(α/α′ ) ≤ deg α whenever α, α′ are two conjugate algebraic numbers whose ...
2. Ideals in Quadratic Number Fields
... that a3 = (8). If we had unique factorization into prime ideals, this would imply a = (2). But two principal ideals are equal if and only if√their generators differ by a unit, hence we would have to conclude that 1+ 2 −3 is a unit; in fact, it is not even an element in R. Help comes from studying Fe ...
... that a3 = (8). If we had unique factorization into prime ideals, this would imply a = (2). But two principal ideals are equal if and only if√their generators differ by a unit, hence we would have to conclude that 1+ 2 −3 is a unit; in fact, it is not even an element in R. Help comes from studying Fe ...
x - El Camino College
... First, we use the Upper and Lower Bounds Theorem to find two numbers between which all the solutions must lie. • This allows us to choose a viewing rectangle that is certain to contain all the x-intercepts of P. • We use synthetic division and proceed by trial and error. ...
... First, we use the Upper and Lower Bounds Theorem to find two numbers between which all the solutions must lie. • This allows us to choose a viewing rectangle that is certain to contain all the x-intercepts of P. • We use synthetic division and proceed by trial and error. ...
Primalitv Testing and Jacobi Sums
... favorably with the older tests discussed by Williams [26], In fact, Williams never found a prime number of this size that took more than 20 minutes to prove prime on an Amdahl 470-V7 computer. On the other hand, these older tests are slower for sufficiently large n. It should also be taken into acco ...
... favorably with the older tests discussed by Williams [26], In fact, Williams never found a prime number of this size that took more than 20 minutes to prove prime on an Amdahl 470-V7 computer. On the other hand, these older tests are slower for sufficiently large n. It should also be taken into acco ...
Section 3 - The Open University
... squares of these numbers are 1, 9, 25, 49 and 81, respectively, and these are all odd. In the above example, there were only a small number of possibilities to consider, and so it was easy to prove the statement by considering each one in turn. This method of proof is known as proof by exhaustion, b ...
... squares of these numbers are 1, 9, 25, 49 and 81, respectively, and these are all odd. In the above example, there were only a small number of possibilities to consider, and so it was easy to prove the statement by considering each one in turn. This method of proof is known as proof by exhaustion, b ...
11.7 Polar Form of Complex Numbers
... time to make sure the quantities ‘modulus’, ‘argument’ and ‘principal argument’ are well-defined. Concerning the modulus, if z = 0 then the point associated with z is the origin. In this case, the only r-value which can be used here is r = 0. Hence for z = 0, |z| = 0 is well-defined. If z 6= 0, then ...
... time to make sure the quantities ‘modulus’, ‘argument’ and ‘principal argument’ are well-defined. Concerning the modulus, if z = 0 then the point associated with z is the origin. In this case, the only r-value which can be used here is r = 0. Hence for z = 0, |z| = 0 is well-defined. If z 6= 0, then ...
41(4)
... with new ideas that develop enthusiasm for number sequences or the exploration of number facts. Illustrations and tables should be wisely used to clarify the ideas of the manuscript. Unanswered questions are encouraged, and a complete list of references is absolutely necessary. ...
... with new ideas that develop enthusiasm for number sequences or the exploration of number facts. Illustrations and tables should be wisely used to clarify the ideas of the manuscript. Unanswered questions are encouraged, and a complete list of references is absolutely necessary. ...
SMOOTH CONVEX BODIES WITH PROPORTIONAL PROJECTION
... body and L a linear subspace of Rn , then K|L is the orthogonal projection of K onto L. Let G(n, i) be the Grassmannian of all i-dimensional linear subspaces of Rn . A central question in the geometric tomography of convex sets is to understand to what extent information about the projections K|L wi ...
... body and L a linear subspace of Rn , then K|L is the orthogonal projection of K onto L. Let G(n, i) be the Grassmannian of all i-dimensional linear subspaces of Rn . A central question in the geometric tomography of convex sets is to understand to what extent information about the projections K|L wi ...
MATH 3240Q Second Midterm - Practice Problems It is impossible to
... Question 1. Find 3 primes in each category: 1. Find 3 primes p ≡ 1 mod 3. 2. Find 3 primes p ≡ 2 mod 3. 3. Find 3 primes p ≡ 1 mod 5. 4. Find 3 primes p ≡ 2 mod 5. 5. Find 3 primes p ≡ 3 mod 5. 6. Find 3 primes p ≡ 4 mod 5. 7. Are there any primes p ≡ 3 mod 21? Why? Why not? 8. Are there any primes ...
... Question 1. Find 3 primes in each category: 1. Find 3 primes p ≡ 1 mod 3. 2. Find 3 primes p ≡ 2 mod 3. 3. Find 3 primes p ≡ 1 mod 5. 4. Find 3 primes p ≡ 2 mod 5. 5. Find 3 primes p ≡ 3 mod 5. 6. Find 3 primes p ≡ 4 mod 5. 7. Are there any primes p ≡ 3 mod 21? Why? Why not? 8. Are there any primes ...
Siegel Discs
... at z. Linear maps are easier to iterate and so we ask the question whether or not a change of variable can be found for which f is conjugate to its linear part, in other words, whether or not f is linearizable. The Multiplier: If f is holomorphic and a is a p−periodic point of f , then the multiplie ...
... at z. Linear maps are easier to iterate and so we ask the question whether or not a change of variable can be found for which f is conjugate to its linear part, in other words, whether or not f is linearizable. The Multiplier: If f is holomorphic and a is a p−periodic point of f , then the multiplie ...
ABSTRACT On the Goldbach Conjecture Westin King Director: Dr
... This example is part of a proof of the Brun-Titchmarsh theorem [7]. A little more manipulation will give an explicit formula for an upper bound to π(x; k, a). Similar methods are used in proofs concerning the Goldbach conjecture, some of which use the Selberg sieve to approximate primes in arithmeti ...
... This example is part of a proof of the Brun-Titchmarsh theorem [7]. A little more manipulation will give an explicit formula for an upper bound to π(x; k, a). Similar methods are used in proofs concerning the Goldbach conjecture, some of which use the Selberg sieve to approximate primes in arithmeti ...
(pdf)
... Proof. By the definition of divisor, L = i(p − 1) = j(q − 1) + k for nonzero integers i, j, k where 1 ≤ k < q − 1. Using Fermat’s Little Theorem shows that aL = ai(p−1) = (ap−1 )i ≡ 1i ≡ 1 (mod p) and aL = aj(q−1)+k = (aq−1 )j ak ≡ 1j · ak ≡ ak (mod q). From the equations above, we see that p | aL − ...
... Proof. By the definition of divisor, L = i(p − 1) = j(q − 1) + k for nonzero integers i, j, k where 1 ≤ k < q − 1. Using Fermat’s Little Theorem shows that aL = ai(p−1) = (ap−1 )i ≡ 1i ≡ 1 (mod p) and aL = aj(q−1)+k = (aq−1 )j ak ≡ 1j · ak ≡ ak (mod q). From the equations above, we see that p | aL − ...
Math 25: Solutions to Homework # 4 (4.3 # 10) Find an integer that
... it. We know that 8 | x42y if 8 | 42y. Thus we must have y = 4 since 424 is the only number of this form divisible by 8. Now 11 | x424 only if 11 divides x − 4 + 2 − 4 = x − 6. Thus x = 6, so the total cost was 64.24, and each chicken cost 73 cents. (5.1 # 24(a)) Check the multiplication 875, 961 · 2 ...
... it. We know that 8 | x42y if 8 | 42y. Thus we must have y = 4 since 424 is the only number of this form divisible by 8. Now 11 | x424 only if 11 divides x − 4 + 2 − 4 = x − 6. Thus x = 6, so the total cost was 64.24, and each chicken cost 73 cents. (5.1 # 24(a)) Check the multiplication 875, 961 · 2 ...
Full text
... for each element xv of an 5 P , the number of U3s is equal to the total number of elements of an Sp 9 that is, M1 + M2 + M3 + MM . Besides, every U3 is a subsequence of S. As we saw In Lemma 7, Z73's are classified into four-types like Figure 2. It is easily recognized that the number of each type c ...
... for each element xv of an 5 P , the number of U3s is equal to the total number of elements of an Sp 9 that is, M1 + M2 + M3 + MM . Besides, every U3 is a subsequence of S. As we saw In Lemma 7, Z73's are classified into four-types like Figure 2. It is easily recognized that the number of each type c ...
24(2)
... In this section we stud}^ somewhat more general second-degree sequences than Pn9 and obtain necessary and sufficient conditions for certain infinite families of representations to exist. We then specialize to polygonal numbers. To this end, let P(a, 3; n) = n(an - 3)» where a, 3 are integers with (a ...
... In this section we stud}^ somewhat more general second-degree sequences than Pn9 and obtain necessary and sufficient conditions for certain infinite families of representations to exist. We then specialize to polygonal numbers. To this end, let P(a, 3; n) = n(an - 3)» where a, 3 are integers with (a ...
Algorithmic Number Theory
... 34.1 Fibonacci numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163 34.2 Chinese Remainder Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163 34.3 Wilson’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . ...
... 34.1 Fibonacci numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163 34.2 Chinese Remainder Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163 34.3 Wilson’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . ...
