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Lecture Notes on Primality Testing
... by computing aR , a2R , a4R , · · · , a2 R = an−1 (all mod n). Each term in this sequence is the square of the previous one, and (assuming that a fails the Fermat Test, in which case we would be done anyway) the last term is 1. Thus if the first 1 in the sequence is preceded by a number other than − ...
... by computing aR , a2R , a4R , · · · , a2 R = an−1 (all mod n). Each term in this sequence is the square of the previous one, and (assuming that a fails the Fermat Test, in which case we would be done anyway) the last term is 1. Thus if the first 1 in the sequence is preceded by a number other than − ...
Construction of Composite Numbers by Recursively
... Conjecture 3.6. For any distinct primes p, q, there exists an r in O(log p + log q), such that hr (p) 6= hr (q). Theorem 3.7. If conjecture 3.6 is true, Factor and DMod are polynomialtime, in log n, equivalent. Proof. Given an algorithm solving DMod in polynomial-time, algorithm 3.3 solves Factor in ...
... Conjecture 3.6. For any distinct primes p, q, there exists an r in O(log p + log q), such that hr (p) 6= hr (q). Theorem 3.7. If conjecture 3.6 is true, Factor and DMod are polynomialtime, in log n, equivalent. Proof. Given an algorithm solving DMod in polynomial-time, algorithm 3.3 solves Factor in ...
Full text
... as the number of ideals in a very simple poset, usually called a fence. The purpose of this note is not to prove new theorems about the sequence {Fn}. However, we wish to demonstrate that the approach has several advantages. By attaching to each Fibonacci number a geometrical object, the number gets ...
... as the number of ideals in a very simple poset, usually called a fence. The purpose of this note is not to prove new theorems about the sequence {Fn}. However, we wish to demonstrate that the approach has several advantages. By attaching to each Fibonacci number a geometrical object, the number gets ...
Sums of triangular numbers and $t$-core partitions
... Remark 1.3. This is best possible in the sense that every partition of n < g is a g-core and when n = g there are precisely g such partitions. Studying the case when t = 2 leads to an investigation of sums of triangular numbers. Gauss showed the following famous Eureka Theorem to determine represent ...
... Remark 1.3. This is best possible in the sense that every partition of n < g is a g-core and when n = g there are precisely g such partitions. Studying the case when t = 2 leads to an investigation of sums of triangular numbers. Gauss showed the following famous Eureka Theorem to determine represent ...
9709/01 - StudyGuide.PK
... If you have been given an Answer Booklet, follow the instructions on the front cover of the Booklet. Write your Centre number, candidate number and name on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use a soft pencil for any diagrams or graphs. Do n ...
... If you have been given an Answer Booklet, follow the instructions on the front cover of the Booklet. Write your Centre number, candidate number and name on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use a soft pencil for any diagrams or graphs. Do n ...
Linear independence of continued fractions
... is a transcendental number. The generalization of transcendence is algebraic independence and there are several results concerning the algebraic independence of continued fractions. See, for instance, Bundschuh [1] or Hancl [5]. On the other hand it is a well known fact that if a positive real numbe ...
... is a transcendental number. The generalization of transcendence is algebraic independence and there are several results concerning the algebraic independence of continued fractions. See, for instance, Bundschuh [1] or Hancl [5]. On the other hand it is a well known fact that if a positive real numbe ...
Reciprocal PSLQ and the tiny Nome of Bologna David Broadhurst
... awarded by the Clay Mathematical Institute to Andrew Wiles, to honour the proof of Fermat’s last theorem, as was explained in the book Mathematics by Experiment: Plausible Reasoning in the 21st Century, by Jonathan M. Borwein and David H. Bailey. Perhaps Grigori Yakovlevich Perelman may eventually a ...
... awarded by the Clay Mathematical Institute to Andrew Wiles, to honour the proof of Fermat’s last theorem, as was explained in the book Mathematics by Experiment: Plausible Reasoning in the 21st Century, by Jonathan M. Borwein and David H. Bailey. Perhaps Grigori Yakovlevich Perelman may eventually a ...
Geometry (H) Lesson 2.1 2.1 Notes: Inductive Reasoning Lesson
... 10. Chess The small squares on a chessboard can be combined to form larger squares. For example, there are 64 1 x 1 squares and one 8 x 8 square. Use inductive reasoning to determine how many 2 x 2 squares, 3 x 3 squares, and so on, are on a chessboard. What is the total number of squares on a chess ...
... 10. Chess The small squares on a chessboard can be combined to form larger squares. For example, there are 64 1 x 1 squares and one 8 x 8 square. Use inductive reasoning to determine how many 2 x 2 squares, 3 x 3 squares, and so on, are on a chessboard. What is the total number of squares on a chess ...
MC302 GRAPH THEORY Thursday, 11/21/13 (revised slides, 11/25
... Proof. First add vertices, if needed, to make both partite sets the same size. Then add edges to make all vertices have degree Δ. See next two slides! ...
... Proof. First add vertices, if needed, to make both partite sets the same size. Then add edges to make all vertices have degree Δ. See next two slides! ...
On the least common multiple of q
... where nk q denotes the q-binomial coefficient and [n]q = 1−q . Then we show that this identity 1−q is indeed a q-analogue of that of Farhi [Amer. Math. Monthly 116 (2009), 836–839]. ...
... where nk q denotes the q-binomial coefficient and [n]q = 1−q . Then we show that this identity 1−q is indeed a q-analogue of that of Farhi [Amer. Math. Monthly 116 (2009), 836–839]. ...
+ 1 - Stanford Mathematics
... But the left-hand side is, by applying Pn , equal to n2 + (2(n+ 1)−1), which equals n2 + 2n + 1 = (n + 1)2 . So Pn+1 is true. By mathematical induction, this finishes the proof. 1.6: Prove that 11n −4n is divisible by 7 when n is a natural number. The proof is by induction; Pn is the statement tha ...
... But the left-hand side is, by applying Pn , equal to n2 + (2(n+ 1)−1), which equals n2 + 2n + 1 = (n + 1)2 . So Pn+1 is true. By mathematical induction, this finishes the proof. 1.6: Prove that 11n −4n is divisible by 7 when n is a natural number. The proof is by induction; Pn is the statement tha ...
Full text
... diophantine equation are still not known, although particular instances of it have been dealt with (see, for example, [3] for the case q = 2, or [1] for the case x = 10). All solutions of the diophantine equation x2 = 2a ± 2b + 1 in positive integers (x, a, b) where found in [10], and the more gener ...
... diophantine equation are still not known, although particular instances of it have been dealt with (see, for example, [3] for the case q = 2, or [1] for the case x = 10). All solutions of the diophantine equation x2 = 2a ± 2b + 1 in positive integers (x, a, b) where found in [10], and the more gener ...
On the Erdos-Straus conjecture
... cases k = 3l + r with r ∈ {0, 1, 2}, l ≥ 0, l ∈ Z, we see that there is only one excepted case in which we get stuck: k = 3l. This because Theorem 1.1 can be used in one situation: k = 3l + 1 implies 1+(3l+1+1) ≡ 0 (mod 3). On the other hand, if k = 3l+2 we get (k+1) = (3l+2)+1 = 3(l+1) ≥ 3 and the ...
... cases k = 3l + r with r ∈ {0, 1, 2}, l ≥ 0, l ∈ Z, we see that there is only one excepted case in which we get stuck: k = 3l. This because Theorem 1.1 can be used in one situation: k = 3l + 1 implies 1+(3l+1+1) ≡ 0 (mod 3). On the other hand, if k = 3l+2 we get (k+1) = (3l+2)+1 = 3(l+1) ≥ 3 and the ...
What is Zeckendorf`s Theorem?
... Proving commutativity of ◦ over N is a simple rearrangement of finite sums. Were the circle product to always immediately yield a Zeckendorf representation of the product, associativity would be easy, but this can quickly be seen to not be the case. Donald Knuth (1938-) published his proof of assoc ...
... Proving commutativity of ◦ over N is a simple rearrangement of finite sums. Were the circle product to always immediately yield a Zeckendorf representation of the product, associativity would be easy, but this can quickly be seen to not be the case. Donald Knuth (1938-) published his proof of assoc ...
Problem Set: Proof by contradiction
... 16. Let a, b, c be integers satisfying a2 + b2 = c2 . Show that abc must be even. 17. Let P (x) = xn + an−1 xn−1 + · · · + a1 x + a0 be a polynomial with integral coefficients. Suppose that there exist four distinct integers a, b, c, d with P (a) = P (b) = P (c) = P (d) = 5. Prove that there is no i ...
... 16. Let a, b, c be integers satisfying a2 + b2 = c2 . Show that abc must be even. 17. Let P (x) = xn + an−1 xn−1 + · · · + a1 x + a0 be a polynomial with integral coefficients. Suppose that there exist four distinct integers a, b, c, d with P (a) = P (b) = P (c) = P (d) = 5. Prove that there is no i ...
pdf-file - Institut for Matematiske Fag
... partitions of n = 189 is of prime power degree greater than one. ...
... partitions of n = 189 is of prime power degree greater than one. ...