1 Basic Notions - Caltech Mathematics Department

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$MATH 5b; Solutions to Homework Set 1 January 2009 1. As K is a ...$

Lemma: x n − 1=(x − 1)(x n−1 + x n−2 + ···+ x +1) Corollary: (x − 1)|(x n − 1) So for a n − 1tobeprime,weneeda =2. Moreover, if n = md, we can apply the lemma with x = a d .Then (a d − 1)|(a n − 1) So we get the following Lemma If an − 1 is a prime, then a =2andnis prime. Definition: AMersenne prime is a prime of the form q =2 p − 1, p prime. Question: are they infinitely many Mersenne primes? Best known: The 37th Mersenne prime q is associated to p = 3021377, and this was done in 1998. One expects that p = 6972593 will give the next Mersenne prime; this is close to being proved, but not all the details have been checked. Definition: A positive integer n is perfect iff it equals the sum of all its (positive) divisors
σ(n) = r j=1 σ(p ej j )= r ej +1 p − 1 j=1 p − 1 Examples of perfect numbers: 6=1+2+3 28=1+2+4+7+14 496 8128 Questions: 1. Are there infinitely many perfect numbers? 2. Is there any odd perfect number? Note: 6=(2)(3), 28=(4)(7), 496=(16)(31), 8128=(64)(127) They all look like 2 n−1 (2 n − 1), with 2n − 1 prime (i.e., Mersenne). Theorem (Euler) Let n be a positive, even integer. Then n is perfect ⇔ n =2 p−1 (2 p − 1), for a prime p, with 2 p − 1aprime. Corollary. There exists a bijection between even perfect numbers and Mersenne primes. ProofofTheorem.(⇐) Start with n =2 p−1 q,withq =2 p − 1 a Mersenne prime. To show: n is perfect, i.e., σ(n) = 2n. Since 2 p−1 q, and since (2 p−1 ,q)=1,wehave σ(n) =σ(2 p−1 )σ(q) =(2 p − 1)(q +1)=q2 p =2n. (⇒): Let n be a even, perfect number. Since n is even, we can write . Since n is perfect, n =2 j m, with j ≥ 1, modd = n ⇒ σ(n) =σ(2 j )σ(m) =(2 j+1 − 1)σ(m) σ(n) =2n =2 j+1 m 43
Page 1 and 2: 1 Basic Notions Notation: N = {1, 2
Page 3 and 4: for any (i, j). We may assume p1
Page 5 and 6: 2 HeuristicsonPrimes Let P = {prime
Page 7 and 8: Proposition 2. a1,...,an are mutual
Page 9 and 10: So we have produced a new prime q1
Page 11 and 12: So the rational solutions of (2) ar
Page 13 and 14: Thus r ′ ∈ S and r ′ 1isprime
Page 15 and 16: Lemma. Let s be any real number > 1
Page 17 and 18: Again, take all n ≤ x and throw o
Page 19 and 20: Suppose d|Fn and d|Fn+k. Putx =22n.
Page 21 and 22: 4 Pythagorean Triples Problem: Find
Page 23 and 24: if m is a multiple of d, and that t
Page 25 and 26: n, a, N arbitrary: (general case) I
Page 27 and 28: Then a · x = Nd ⇔ a·y =0 ⇕ aC
Page 29 and 30: m=3: a r (mod 3) 0 0 1 1 2 2 3 0 4
Page 31 and 32: So the general solution of ⋆ is g
Page 33 and 34: Write, using the Euclidean algorith
Page 35 and 36: Proof of Step 1: ϕ(n1n2) =#{y ∈{
Page 37 and 38: cu0 ⇒ a ≡ c (mod m) d ⇒ x ≡
Page 39 and 40: 10 Number of solutions modulo a pri
Page 41: If j = 0 or p, then p p p p is di
Page 45 and 46: Mn =2 n − 1 Mersenne number. Defi
Page 47 and 48: 13 RSA Encryption The mathematics b
Page 49 and 50: Thus x d ≡ 1(modp) Since we are a
Page 51 and 52: 2 is a primitive root module the fo
Page 53 and 54: and T 2 = {b 2 |b ∈ T } Claim 1:
Page 55 and 56: Proposition ⇒ Corollary 1: ByEule
Page 57 and 58: Earlier we proved that there exists
Page 59 and 60: Let k =#{j ∈ S|āj ∈ S}. Then G
Page 61 and 62: p = 2: Everything is a square mod p
Page 63 and 64: ξ is called a primitive qth root o
Page 65 and 66: where a ′ a ≡ 1(modq). But
Page 67 and 68: ( a a ) because ( )=±1 andpis odd
Page 69 and 70: definition 29 ≡ (mod 43) iff 29
Page 71 and 72: where fn is a polynomial in sin 2 x
Page 73 and 74: 1=1 2 +0 2 16 = 4 2 +0 2 31 = −
Page 75 and 76: This gives: 41 = 5 2 +4 2 . Proposi
Page 77 and 78: Second case: Here we get x 2 A 2 +
Page 79 and 80: which is multiplicative, i.e., N(α
Page 81 and 82: As we have seen in the previous sec
Page 83 and 84: We also have But (1) + (3) implies
Page 85 and 86: for some C>1. Let f(X) =aX 2 + bX +

1 Basic Notions - Caltech Mathematics Department

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