1 Basic Notions - Caltech Mathematics Department

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

Since n − n ′ pm. Soncannot be prime. Let q be a prime divisor of n. Since{p1,...,pm} is the set of all primes, q must equal pj; forsomej. Then q divides n = p1 ...pm +1andp1 ...pm ⇒ q|1, a contradiction. Euler’s attempted proof. (This can be made rigorous!) Let P be the set of all primes in Z. Euler’s idea: IfP were finite, then X = p∈P 14 1 (1− 1 < ∞. ) p
Lemma. Let s be any real number > 1. Then ζ(s) = p∈P 1 (1 − 1 p 8 ) = (called the “Riemann” zeta function, though Euler studied it a century earlier). ProofofLemma. Recall: If |x| < 1, then 1 series). If s>1, 1 p s < 1. So p 1 1− 1 p s ∞ n=1 1 n s 1−x =1+x + x2 + ... (geometric =1+ 1 p s + 1 p 2s + ... Then 1+ 1 1 + + ... = ps p2s by unique factorization. Euler then argued as follows: let s → 1 from right. X=lims→1 + ∞ 1 n=1 ns → ∞ 1 n=1 , which diverges. But if P is finite, then X is a finite rational number, n a contradiction. (To make this rigorous, we need to be careful about limits and uniform convergence.) The Prime Number Theorem (PNT) For any x ≥ 2, put ∞ n=1 π(x) =#{p : prime | p ≤ x}. What does π(x) look like for x very large? The prime number theorem (PNT) says: π(x) ∼ x , as x →∞ log x In other words, the fraction of integers in [1,x] which are prime is roughly for x large. (Can’t prove it in this class.) 1 log x Twin Primes These are prime pairs (p, q) withq = p +2. Examples: (3,5), (5,7), (11, 13),... Conjecture: There exist infinitely many twin primes. Stronger conjecture: If π2(x) denotes the number of twin primes ≤ x, then π2(x) ∼ x as x →∞. (log x) 2 15 1 n s
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: Thus r ′ ∈ S and r ′ 1isprime
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 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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