1 Basic Notions - Caltech Mathematics Department

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

Put B = Then (d ′ ,an) B =(0,d). Put Then aC =(aA) C = A In−2 0 0 B an/d x −d ′ ∈ SL2(Z). /d y In−2 0 0 B ∈ GLn (Z). =(0, ··· , 0,d ′ ,an) In−2 0 0 B (3) =(0, ··· , 0,d). (4) Theorem 5.1. Let a =(a1, ··· ,an) ∈ Zn −{0} with gcd equal to d. Let C be the matrix given by Lemma. Pick any N ∈ Z. Then we have: ⎛ ⎞ x = ⎜ ⎝ x1 . xn ⎟ ⎠ ∈ Z n is a solution of n i=1 ai xi = Nd if and only if ∃m1, ··· ,mn−1 ∈ Z such that n−1 x = i=1 miC i + NC n where C j denotes (∀j) the j-th column of C. Proof. Let y = x − NC n . 26
Then a · x = Nd ⇔ a·y =0 ⇕ aC(C −1 y)=(0, ··· , 0,d)(C −1 y)=0 ⇕ C −1 ⎛ ⎞ m1 ⎜ 1 ⎟ y = m = ⎜ 1 ⎟ ⎝mn−1⎠ 0 , for some mi ∈ Z, 1 ≤ i ≤ n − 1 ⇕ n−1 y = Cm = mi C i i=1 ⇕ x = Cm + NCn. Example: Find all the integral solutions of 5x +7y +11z =2. (∗) Put a =(5, 7, 11). Then the gcd of the coordinates of a is 1. By Lemma, we can find a 3 × 3 - integral matrix C of determinant ±1 such that aC = (0, 0, 1). The proof of Lemma gives a recipe for finding C. First solve 5x +7y =1. Since1=gcd(5, 7), this can be solved, and a solution (by inspection) is given by x = −4, y=3. PutC ′ 7 −4 = .Nextwehave −5 3 to solve d ′ u +11v =1,whered ′ =gcd(a1,a2) = 1. A solution is given by 11 1 u =1,v=0. LetB = . −1 0 Then the proof of Lemma says that ⎛ C = ⎝ C′ ⎞ ⎛ ⎞ 0 1 0 0 0 ⎠ ⎝ 0 ⎠ . B 0 0 1 0 27
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: n, a, N arbitrary: (general case) I
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 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 +
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1 Basic Notions - Caltech Mathematics Department

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