1 Basic Notions - Caltech Mathematics Department

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

By part (a) this equals with ϕ(p bi i )beingpbi bi−1 − pi r {(b1,... ,br)|0≤bi≤ai,∀i} ϕ(p bi i ), (resp. 1) if bi > 0 (resp. bi = 0). Exchanging the sum and the product, and noting that r we get {(b1,... ,br)|0≤bi≤ai,∀i} ϕ(d) = d|m r i=1 ϕ(p bi i )=pai i , p ai i = m. 9 Linear congruences revisited Theorem. Fix m > 1. Let a, c ∈ Z. Put d = gcd(a, m). Then the congruence ax ≡ c (mod m) (*) has a solution x (mod m) iffd|c. Moreover, when d|c, alldsolutions are of the form x ≡ cu0 + mk (mod m), d with k ∈ Z, where(u0,v0) is a solution of au + mv = d. We already proved (*) has a solution x (mod m) iffd|c. So let d|c. Let (u0,v0) be a solution of Multiply by c, get i.e., a au + mv = d. (**) acu0 + mcv0 = cd, cu0 + m d 36 cv0 = c d
cu0 ⇒ a ≡ c (mod m) d ⇒ x ≡ cu0 (mod m) is a solution of (*). d Recall that we get all the solutions of (**) by taking (u, v) = u0 + km d ,v0− kc , d as k runs over Z. So the general solution of (*) is given by x ≡ cu0 d km + d ≡ cu0 + km d (mod m) Corollary: ax ≡ 1(modm) has a solution iff (a, m) =1. Inthiscase,∃! solution, the multiplicative inverse of a mod m, and denoted a ′ (mod m). We knew before that a has a multiplicative inverse if (a, m) =1. This corollary replaces the if by iff. Definition: (Z/m) ∗ = {a ∈ Z/m|(a, m) =1}. Note: By corollary, (Z/m) ∗ is precisely the subset of Z/m consisting of elements which have multiplicative inverses mod m. Recall: ϕ(m) =|(Z/m) ∗ |. = |{a ∈{0, 1,...,m− 1}|(a, m) =1}|. In the previous section we proved the following: Theorem: (Euler)Foranya∈Zwith (a, m) =1, a ϕ(m) ≡ 1(modm). Corollary. (Fermat’s Little Theorem) m = p (prime), p|a ⇒ a p−1 ≡ 1(modp). Remark. Fermat’s Little Theorem says that x p−1 ≡ 0(modp) 37
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: Proof of Step 1: ϕ(n1n2) =#{y ∈{
Page 39 and 40: 10 Number of solutions modulo a pri
Page 41 and 42: If j = 0 or p, then p p p p is di
Page 43 and 44: σ(n) = r j=1 σ(p ej j )= r ej +
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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