1 Basic Notions - Caltech Mathematics Department

More documents

Recommendations

Info

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

Do same with (1) times v2, (2)timesu2 to get: So a1 (u1v2 − u2v1) =(v2 − u2) −k v1 = u1 + ka2, v2 = u2 − ka1. (u1,u2) isaparticular solution which we use to generate all solutions. Conversely, for any integer k, (u1 + ka2, u2 − ka1) is a solution of a · x =1. If gcd (a1,a2) = 1, then we can solve a1x1 + a2x2 = 1 in integers. Moreover, if (u1,u2) is a particular solution, then any other solution is of the form (u1 + ka2,u2 − ka1), k∈ Z. n=2, d >1, N=1: a1x1 + a2x2 = d (*1) Since d =gcd(a1,a2), d|a1 and d|a2. Putbi = ai . Then (*) becomes d b1x1 + b2x2 =1with(b1,b2) =1. So if (u1,u2) is a particular solution, every solution is of the form u1 + k a2 d ,u2−k a1 . d This finishes the n = 2 case. We summarize the results in the following Proposition Let a1,a2 be non-zero integers, and let d be their gcd. Then the equation a1x1 + a2x2 = m is solvable in integers iff m is divisible by d. Moreover, if (u1,u2) is any particular solution, then the set of all solutions is parametrized by Z, andfor each r ∈ Z, the corresponding solution is given by x1 = u1 + r a2 d , and x2 = u2 − r a1 d . 24
n, a, N arbitrary: (general case) It will be good to understand the example at the end of the section (for n = 3). The rest of the section may be difficult and is included here for completeness. Definition: Mn(Z) ={a =(aij) : n × n − matrices with aij ∈ Z ∀i, j}. ⎛ 1 ⎜ In = ⎝ .. . ⎞ ⎟ ⎠ 1 The equation of interest is GLn(Z) ={A ∈ Mn(Z) :det(A) =±1} ⎛ x1 ⎜ (a1,...,an) ⎝ . xn ⎞ ⎟ ⎠ = Nd (*N) Lemma 1. Let a =(a1, ··· ,an) ∈ Z n −{0} with d =gcd(a1, ··· ,an). Then ∃ C ∈ GLn(Z) such that aC = den =(0, ··· , 0,d). Proof. n =1: d = |a1|, sowecantakeC =(sgn(a1)). Now let n>1, and assume Lemma by induction for m
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: if m is a multiple of d, and that t
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 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 +
show all

1 Basic Notions - Caltech Mathematics Department

Create successful ePaper yourself

Delete template?

Save as template?