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pythagoreantreasury[1]
... (I.47) after first proving 46 other theorems. He used some of these other theorems as building blocks to establish the proof. This proof is examined later. The Chinese may have discovered a proof sometime during the 1st millennium as a diagram similar to that shown, appears in a text called Chou pei ...
... (I.47) after first proving 46 other theorems. He used some of these other theorems as building blocks to establish the proof. This proof is examined later. The Chinese may have discovered a proof sometime during the 1st millennium as a diagram similar to that shown, appears in a text called Chou pei ...
Lectures # 7: The Class Number Formula For
... enough there is no known non-analytic proof of this fact. This also gives a very quick way of finding a give class number. Unfortunately, the method of simply finding all reduced forms actually works more quickly. Using the functional equation of the L-series, however gives a much more efficient way ...
... enough there is no known non-analytic proof of this fact. This also gives a very quick way of finding a give class number. Unfortunately, the method of simply finding all reduced forms actually works more quickly. Using the functional equation of the L-series, however gives a much more efficient way ...
Ch. 1 Review Study Guide
... You can graph a point on a coordinate plane. Use an ordered pair (x, y) to record the coordinates. The first number in the pair is the x-coordinate (left/right on the x – axis). The second number is the y-coordinate (up/down on the y – axis). To graph a point, start at the origin, O. Move horizontal ...
... You can graph a point on a coordinate plane. Use an ordered pair (x, y) to record the coordinates. The first number in the pair is the x-coordinate (left/right on the x – axis). The second number is the y-coordinate (up/down on the y – axis). To graph a point, start at the origin, O. Move horizontal ...
A65 INTEGERS 12 (2012) THE DIOPHANTINE EQUATION X4 + Y 4
... Proof. Each rational value from Lemma 5 determines a possible expression for F (z). For example, for z = 0, we find from (10) that (bz)2 −2D2 z(1−z 2 ) = F (z)Qz. Since F (z) is monic, Q = 2D2 and thus, F (z) = z 2 + (b2 /2D2 )z − 1. This polynomial also divides (a1 + b1 z)2 − 2Dz(1 + z 2 ). Long di ...
... Proof. Each rational value from Lemma 5 determines a possible expression for F (z). For example, for z = 0, we find from (10) that (bz)2 −2D2 z(1−z 2 ) = F (z)Qz. Since F (z) is monic, Q = 2D2 and thus, F (z) = z 2 + (b2 /2D2 )z − 1. This polynomial also divides (a1 + b1 z)2 − 2Dz(1 + z 2 ). Long di ...
MT 430 Intro to Number Theory MIDTERM 1 PRACTICE
... Midterm 1 will cover the material of all lectures up to and including February 14. More precisely, the following topics are covered: (1) Divisibility. Division algorithm. (2) Euclidean algorithm. Greatest common divisors. (3) Prime numbers. (4) Uniqueness of factorization. (5) Binomial coefficients. ...
... Midterm 1 will cover the material of all lectures up to and including February 14. More precisely, the following topics are covered: (1) Divisibility. Division algorithm. (2) Euclidean algorithm. Greatest common divisors. (3) Prime numbers. (4) Uniqueness of factorization. (5) Binomial coefficients. ...
Open problems in number theory
... Let f (X) and g(X) be two irreducible polynomials in Z[X]. Suppose there is no integer n such that n divides f (k) · g(k) for all k. Then there are infinitely many values of k such that f (k) and g(k) are both prime numbers. Take f (X) = X and g(X) = X + 2; we get the twin prime ...
... Let f (X) and g(X) be two irreducible polynomials in Z[X]. Suppose there is no integer n such that n divides f (k) · g(k) for all k. Then there are infinitely many values of k such that f (k) and g(k) are both prime numbers. Take f (X) = X and g(X) = X + 2; we get the twin prime ...
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[hal-00574623, v2] Averaging along Uniform Random Integers
... leading digit d ∈ {1, 2, . . . , 9} is log10 (1 + 1/d). More generally, defining the mantissa M (x) of a positive real number x as the only real number in [1, 10[ such that x = M (x)10k for some integer k, Benford’s law states that for any 1 ≤ α < β < 10, the proportion of numbers whose mantissa lie ...
... leading digit d ∈ {1, 2, . . . , 9} is log10 (1 + 1/d). More generally, defining the mantissa M (x) of a positive real number x as the only real number in [1, 10[ such that x = M (x)10k for some integer k, Benford’s law states that for any 1 ≤ α < β < 10, the proportion of numbers whose mantissa lie ...
UNIQUE FACTORIZATION IN MULTIPLICATIVE SYSTEMS
... [2, p. 21] uses positive integers of the form l+ik for the same purpose, with the numerical case 693 = 9-77 = 21-33. In this paper we examine all multiplicative systems made up of arithmetic progressions, and decide the question of unique factorization. For a fixed positive integer ra, let M be a mu ...
... [2, p. 21] uses positive integers of the form l+ik for the same purpose, with the numerical case 693 = 9-77 = 21-33. In this paper we examine all multiplicative systems made up of arithmetic progressions, and decide the question of unique factorization. For a fixed positive integer ra, let M be a mu ...
Full text
... The double bar edges leading to nodes without a superscript indicate that none of the above theorems apply. In these cases the nodes were found using (4) . Note that there are only 3 such instances. On the other hand, the 16 superscripted nodes were found easily using the theorems indicated by the s ...
... The double bar edges leading to nodes without a superscript indicate that none of the above theorems apply. In these cases the nodes were found using (4) . Note that there are only 3 such instances. On the other hand, the 16 superscripted nodes were found easily using the theorems indicated by the s ...
Problem 1 Problem 2
... True/false questions. For each of the following statements, determine if it is true or false, and provide a brief justification for your claim. Credit on these questions is based on your justification. A simple true/false answer, without justification, or with an incorrect justification, won’t earn ...
... True/false questions. For each of the following statements, determine if it is true or false, and provide a brief justification for your claim. Credit on these questions is based on your justification. A simple true/false answer, without justification, or with an incorrect justification, won’t earn ...
Second Proof: Every Positive Integer is a Frobenius
... R0 (n) = p1 p2 ...pn Q0 (n) = p3n+1 . It is clear that R(n) > R0 (n) and Q0 (n) > Q(n) for all n ≥ 1. So we only need to prove that R0 (n) > Q0 (n) for n ≥ 5. It is obvious that R0 (n) > Q0 (n) for n = 5, 6, 7, 8 since R0 (5) = 2310, ...
... R0 (n) = p1 p2 ...pn Q0 (n) = p3n+1 . It is clear that R(n) > R0 (n) and Q0 (n) > Q(n) for all n ≥ 1. So we only need to prove that R0 (n) > Q0 (n) for n ≥ 5. It is obvious that R0 (n) > Q0 (n) for n = 5, 6, 7, 8 since R0 (5) = 2310, ...
Full text
... Over 23,000 problems from 42 journals and 22 contests are referenced by the site, which was developed by Stanley Rabinowitz's MathPro Press. Ample hosting space for the site was generously provided by the Department of Mathematics and Statistics at the University of Mirrouri-Rolla, through Leon M. H ...
... Over 23,000 problems from 42 journals and 22 contests are referenced by the site, which was developed by Stanley Rabinowitz's MathPro Press. Ample hosting space for the site was generously provided by the Department of Mathematics and Statistics at the University of Mirrouri-Rolla, through Leon M. H ...
Math 4600 HW 4 Due Wednesday, October 13 (1) Prove Theorem
... (b) Consider regular n-gons with the following numbers of sides and in which the radius of the circumscribed circle is 1. (i) n = 3 (ii) n = 4 (iii) n = 6 (iv) n = 100 (c) What is the length of the apothem of each polygon? (d) What is the length of each side? (e) What is the perimeter? (f) What is t ...
... (b) Consider regular n-gons with the following numbers of sides and in which the radius of the circumscribed circle is 1. (i) n = 3 (ii) n = 4 (iii) n = 6 (iv) n = 100 (c) What is the length of the apothem of each polygon? (d) What is the length of each side? (e) What is the perimeter? (f) What is t ...
Graph Colouring
... Every planar graph G has χ(G) ≤ 4. That is, a planar graph can always be coloured with four or fewer colours. The Four Colour Theorem engaged the interest of such famous mathematicians as Hamilton, De Morgan, Cayley and Birkhoff. It also prompted a famous incorrect proof that survived from 1879 to 1 ...
... Every planar graph G has χ(G) ≤ 4. That is, a planar graph can always be coloured with four or fewer colours. The Four Colour Theorem engaged the interest of such famous mathematicians as Hamilton, De Morgan, Cayley and Birkhoff. It also prompted a famous incorrect proof that survived from 1879 to 1 ...
Full text
... This is not valid, since the prime p is E 3 (mod 4). The required conclusion is now immediate. Further analysis in regard to divisibility property possessed by Lucas numbers yielded the following theorem. lh(L0K
... This is not valid, since the prime p is E 3 (mod 4). The required conclusion is now immediate. Further analysis in regard to divisibility property possessed by Lucas numbers yielded the following theorem. lh(L0K
GAUSSIAN INTEGER SOLUTIONS FOR THE FIFTH POWER
... 3. Solutions where all of a, b, c and d are Gaussian integers Our second result requires the following identity, Lemma 3.1. For all real values of a, b, c, (3.1) (a + b + ic)5 + (a − b − ic)5 − (a − b + ic)5 − (a + b − ic)5 = 80abc(a2 + b2 − c2 )i. We see that every Pythagorean triple a, b, c yields ...
... 3. Solutions where all of a, b, c and d are Gaussian integers Our second result requires the following identity, Lemma 3.1. For all real values of a, b, c, (3.1) (a + b + ic)5 + (a − b − ic)5 − (a − b + ic)5 − (a + b − ic)5 = 80abc(a2 + b2 − c2 )i. We see that every Pythagorean triple a, b, c yields ...
Full text
... 12S. ynr) + i2£ (<*„X^V^/r) < " + * < i° £y f } + 1°S (« rl £ 1 Dk / r )> and this gives the range within which the rank n of [/^ lies. For example, when r = 2, D = d, £?2 = 5, a) = -1, P 2 1 = - P 2 2 = 15 a 2 1 = 1.6, a ...
... 12S. ynr) + i2£ (<*„X^V^/r) < " + * < i° £y f } + 1°S (« rl £ 1 Dk / r )> and this gives the range within which the rank n of [/^ lies. For example, when r = 2, D = d, £?2 = 5, a) = -1, P 2 1 = - P 2 2 = 15 a 2 1 = 1.6, a ...