Cryptology
... Exhaustive search can be exponential time (O(e(¼+o(1)) ln(p))). ERH implies: Least primitive root is O(ln(p)6). Issue: How to test if g is a primitive root of p? Selecting N. N must be large, but not easy to guess. Issue: What is a good way to choose N “randomly”? Computing y. O(ln(p)2). Polynomial ...
... Exhaustive search can be exponential time (O(e(¼+o(1)) ln(p))). ERH implies: Least primitive root is O(ln(p)6). Issue: How to test if g is a primitive root of p? Selecting N. N must be large, but not easy to guess. Issue: What is a good way to choose N “randomly”? Computing y. O(ln(p)2). Polynomial ...
Elementary Number Theory
... • Show that for integers n, if n2 is odd then n is odd • Show that for all integers n and all prime numbers p, if n2 is divisible by p, then n is divisible by p • For all integers m and n, if m+n is even then m and n are both even or m and n are both odd • The product of any non-zero rational number ...
... • Show that for integers n, if n2 is odd then n is odd • Show that for all integers n and all prime numbers p, if n2 is divisible by p, then n is divisible by p • For all integers m and n, if m+n is even then m and n are both even or m and n are both odd • The product of any non-zero rational number ...
1 The Natural Numbers
... modelled within set theory; it’s intuitively obvious that we could go on in the same way to model as many of the natural numbers as time would permit. Note that 0 ∈ 1 ∈ 2 ∈ 3 ∈ · · · and 0 ⊆ 1 ⊆ 2 ⊆ 3 · · · . Is there a set consisting of all the natural numbers? The assumptions we made in Chapter 1 ...
... modelled within set theory; it’s intuitively obvious that we could go on in the same way to model as many of the natural numbers as time would permit. Note that 0 ∈ 1 ∈ 2 ∈ 3 ∈ · · · and 0 ⊆ 1 ⊆ 2 ⊆ 3 · · · . Is there a set consisting of all the natural numbers? The assumptions we made in Chapter 1 ...
Full text
... Lemma 1 shows that Fnh is generated by a linear recurrence of order h + 1, and so Fn6 is generated by a linear recurrence of order 7. This gives an insight as to why identities (2.6)(2.8) can be considered to be special. First, there are only four terms on the right instead of a possible seven terms ...
... Lemma 1 shows that Fnh is generated by a linear recurrence of order h + 1, and so Fn6 is generated by a linear recurrence of order 7. This gives an insight as to why identities (2.6)(2.8) can be considered to be special. First, there are only four terms on the right instead of a possible seven terms ...
FUNCTIONS AND EQUATIONS 1.1. Definition of a set. A set is any
... We can reduce a fraction by finding common multipliers or factors in the numerator and denominator of a fraction. A fail safe way to do this is to factor a number into the primes of which it is made. We can factor 12 as 2 × 2 × 3. And 2 and 3 are both prime. Definition 1 (Prime Factorization). The p ...
... We can reduce a fraction by finding common multipliers or factors in the numerator and denominator of a fraction. A fail safe way to do this is to factor a number into the primes of which it is made. We can factor 12 as 2 × 2 × 3. And 2 and 3 are both prime. Definition 1 (Prime Factorization). The p ...
Here
... 2. If φ : D17 → Zm were a surjective homomorphism, then |im(φ)| = |Zm | = m divides |D17 | = 34 by the first isomorphism theorem. So we can narrow m down to 1, 2, 17, 34. If m = 34, then φ would actually be injective, and hence φ would be an isomorphism. However, D17 is not cyclic. So the only poss ...
... 2. If φ : D17 → Zm were a surjective homomorphism, then |im(φ)| = |Zm | = m divides |D17 | = 34 by the first isomorphism theorem. So we can narrow m down to 1, 2, 17, 34. If m = 34, then φ would actually be injective, and hence φ would be an isomorphism. However, D17 is not cyclic. So the only poss ...
Around the Littlewood conjecture in Diophantine approximation
... the present survey is to highlight recent results and developments on these questions. We make the choice to state more than twenty theorems and to give only a single proof. Section 1 is devoted to the Littlewood conjecture itself, while the mixed and the p-adic (a special case of the mixed) Littlew ...
... the present survey is to highlight recent results and developments on these questions. We make the choice to state more than twenty theorems and to give only a single proof. Section 1 is devoted to the Littlewood conjecture itself, while the mixed and the p-adic (a special case of the mixed) Littlew ...
The Degree-Sum Theorem
... Is this also true for non-simple graphs, i.e., graphs with parallel edges and loops? ...
... Is this also true for non-simple graphs, i.e., graphs with parallel edges and loops? ...
Geometry Day One! Let`s review a little Algebra... Simplify: (what
... Conjecture: a conclusion reached by using inductive reasoning. Counterexample: an example for which a conjecture is false ...
... Conjecture: a conclusion reached by using inductive reasoning. Counterexample: an example for which a conjecture is false ...
Determine if the following conjectures are True or False. If False
... Complete the following two-column proof. NUMBER EACH STEP! 20. Given: X is the midpoint of segment AY, and Y is the midpoint of segment XB. ...
... Complete the following two-column proof. NUMBER EACH STEP! 20. Given: X is the midpoint of segment AY, and Y is the midpoint of segment XB. ...
Course Notes 5.3 Applications of Number Theory
... Fermat’s little theorem provides some help in working with prime numbers and provides the basis for many probabilistic primality tests. We will not give a proof of Fermat’s theorem, since it involves concepts from the theory of groups that would take us too far afield. An elementary proof can be fou ...
... Fermat’s little theorem provides some help in working with prime numbers and provides the basis for many probabilistic primality tests. We will not give a proof of Fermat’s theorem, since it involves concepts from the theory of groups that would take us too far afield. An elementary proof can be fou ...
Reducing the Erdos-Moser equation 1^ n+ 2^ n+...+ k^ n=(k+ 1)^ n
... Now take r = 3, so that k = p1 p2 p3 . Then q2 p2 = p1 p3 + 1 and q3 p3 = p1 p2 + 1, for some integers q2 and q3 . By (2.7) we have p1 < p2 | (p1 + q3 ) < 2p1 . Hence p2 = p1 + q3 . Substituting q3 = p2 − p1 into q3 p3 = p1 p2 + 1 yields p1 | (p2 p3 − 1). As p1 | (p2 p3 + 1), we conclude that p1 | 2 ...
... Now take r = 3, so that k = p1 p2 p3 . Then q2 p2 = p1 p3 + 1 and q3 p3 = p1 p2 + 1, for some integers q2 and q3 . By (2.7) we have p1 < p2 | (p1 + q3 ) < 2p1 . Hence p2 = p1 + q3 . Substituting q3 = p2 − p1 into q3 p3 = p1 p2 + 1 yields p1 | (p2 p3 − 1). As p1 | (p2 p3 + 1), we conclude that p1 | 2 ...
Induction
... L-shapes subject to conditions 1′ and 2′ . A 1 × 1 grid consists of only one square. This square is s, and we can leave it untiled. Since there are no other squares in the grid to tile, this gives us a valid tiling of a 1 × 1 grid. Now we prove the inductive step using a direct proof. Let n ∈ N and ...
... L-shapes subject to conditions 1′ and 2′ . A 1 × 1 grid consists of only one square. This square is s, and we can leave it untiled. Since there are no other squares in the grid to tile, this gives us a valid tiling of a 1 × 1 grid. Now we prove the inductive step using a direct proof. Let n ∈ N and ...
Searching for Large Elite Primes
... that their number is finite. For further open problems on Fermat numbers see Richard Guy’s famous book [Guy 04]. We call a prime number p elite if there is an integer index m for which all Fn with n > m are quadratic nonresidues modulo p, i.e., there is no solution to the congruence x2 ≡ Fn mod p for ...
... that their number is finite. For further open problems on Fermat numbers see Richard Guy’s famous book [Guy 04]. We call a prime number p elite if there is an integer index m for which all Fn with n > m are quadratic nonresidues modulo p, i.e., there is no solution to the congruence x2 ≡ Fn mod p for ...