Document Sample

21441, HW03, 09.12.2003-09.19.2003, solutions 1. Find the smallest x ≥ 0 such that x ≡ 2 (mod 3),x ≡ 1 (mod 5), and x ≡ 1 (mod 7) simultaneously. Here’s how to do it using a technique you already know. The solutions are the numbers of the form x = 3a + 2 = 5b + 1 = 7c + 1. Solving 3a − 5b = −1 and 5b − 7c = 0 simultaneously gives a = 23 + 35k, b = 14 + 21k, and c = 10 + 15k where k is arbitrary. Thus the solutions are the numbers x = 71 + 105k. So x = 71 is the smallest solution greater than or equal to zero. Here is the method given in the proof of the Chinese Remainder The- orem. To solve x ≡ ai (mod mi ) for 1 ≤ i ≤ r where the mi are pairwise relatively prime, solve Mi bi ≡ 1 (mod mi ) for 1 ≤ i ≤ r where Mi = m/mi and m = m1 m2 · · · mr . Then x ≡ r Mi bi ai i=1 (mod m). So m = 105, M1 = 35, M2 = 21, M3 = 15. We have 1 ≡ M1 b1 ≡ 35b1 ≡ 2b1 (mod 3) or b1 ≡ 2 (mod 3). We also have 1 ≡ M2 b2 ≡ 21b2 ≡ b2 (mod 5) and 1 ≡ M3 b3 ≡ 15b3 ≡ b3 (mod 7). Thus x ≡ M1 b1 a1 + M2 b2 a2 + M3 b3 a3 ≡ 35 · 2 · 2 + 21 · 1 · 1 + 15 · 1 · 1 ≡ 71 (mod 105). Thus x = 71 is the smallest solution greater than or equal to zero. 2. (a) Show that 61! + 1 ≡ 63! + 1 ≡ 0 (mod 71). The point here is to reduce these calculations to something hope- fully a lot simpler than 60 to 70 multiplications. We use the following fact: if (a, m) = 1, then ab ≡ ac (mod m) ⇔ b ≡ c (mod m). 61! + 1 ≡ 63! + 1 (mod 71) iﬀ 61! ≡ 63! (mod 71) iﬀ 1 ≡ 62 · 63 (mod 71). But 62 · 63 ≡ (−9)(−8) ≡ 72 ≡ 1 (mod 71). Since 71 is a prime we have 70! ≡ −1 (mod 71) by Wilson’s theorem. Thus 63! + 1 ≡ 0 (mod 71) iﬀ 63! ≡ 70! (mod 71) iﬀ 1 ≡ 64(65)(66)(67)(68)(69)(70) ≡ −7! (mod 71). But 7! ≡ 5040 ≡ −1 (mod 71). (b) Prove the converse of Wilson’s Theorem: if n ≥ 2 and (n − 1)! ≡ −1 (mod n), then n is prime. Suppose n is not prime. Then n = ab where 1 < a, b < n and so n = ab|(n − 1)!. Thus (n − 1)! ≡ −1 (mod n). 1 3. Show that 2730|(n13 − n) for all n. Note 2730 = 2 · 3 · 5 · 7 · 13, where all the factors are realtively prime, so n13 ≡ n (mod 2730) iﬀ n13 ≡ n (mod a) for a = 2, 3, 5, 7, and 13. By Fermat’s Little theorem, we have n13 ≡ n (mod 13), n13 ≡ n7 n6 ≡ nn6 ≡ n (mod 7), n13 ≡ (n5 )2 n3 ≡ n2 n3 ≡ n (mod 5), n13 ≡ (n3 )4 n ≡ n4 n ≡ n3 n2 ≡ nn2 ≡ n (mod 3), and n13 ≡ (n2 )6 n ≡ n6 n ≡ (n2 )3 n ≡ n3 n ≡ (n2 )2 ≡ n2 ≡ n (mod 2). 4. Show that if p is a prime and p is not 2 or 5 then p divides inﬁnitely many of the integers in the sequence: 9, 99, 999, 9999, . . .. The members of the sequence are the numbers an = 10n − 1 for n ≥ 1. Thus the claim is that for every prime p ≥ 7, 10n ≡ 1 (mod p) has inﬁnitely many solutions n ≥ 1. Indeed if n = k(p − 1) for any k ≥ 1, then 10n = (10p−1 )k ≡ 1k = 1 (mod p) by Fermat’s theorem. 5. Prove that p|(p − 1)!(1 + 1/2 + 1/3 + · · · + 1/(p − 1)) if p ≥ 3 is a prime. Here’s one solution. For every number 1 ≤ a ≤ p − 1 there is a unique number 1 ≤ a ≤ p − 1 such that aa ≡ 1 (mod p). (The equation ax ≡ 1 (mod p) has a unique solution x mod p.) Thus (p − 1)!(1 + 1/2 + · · · + 1/(p − 1)) ≡ (p − 1)!(1 + 2 + · · · + p − 1) = (p − 1)!(1 + 2 + · · · + p − 1) (mod p). But if n is odd, (n − 1)/2 is an integer and so 1 + 2 + · · · + n − 1 = n((n − 1)/2) ≡ 0 (mod n). Here is a way that mimics the proof of Wilson’s theorem using La- grange’s theorem.. Suppose p ≥ 3 is a prime, then g(x) = (x − 1)(x − 2) · · · (x − (p − 1)) = xp−1 − σ1 xp−2 + · · · + σp−3 x2 − σp−2 x + σp−1 where the σk are all integers. For example: σp−1 = (p − 1)! and σp−2 = (p − 1)!(1 + 1/2 + · · · + 1/(p − 1)). Since g(x) has the roots 1, 2, . . . , p − 1 mod p so does h(x) = (xp−1 − 1) − g(x), by Fermat’s theorem. Since deg(h) ≤ p − 2 and h has ≥ p − 1 roots mod p, h must be identically zero mod p by Lagrange’s Theorem. Thus σk ≡ 0 (mod p) for 1 ≤ k ≤ p − 2. In particular p|σp−2 as claimed. In fact for p ≥ 5, we have p2 |σp−2 (this is called Wolstenholme’s the- orem). We have (p − 1)! = g(p) = pp−1 − σ1 pp−2 + · · · + σp−3 p2 − σp−2 p + (p − 1)! or σp−2 p = pp−1 − σ1 pp−2 + · · · + σp−3 p2 Since p|σk for 1 ≤ k ≤ p − 3, p3 divides every term on the right hand side of this equation and so p3 |pσp−2 . Thus p2 |σp−2 . 2 6. Problem 2.3.14 (page 72). Let f (x) = x3 + 2x − 3. By testing all possible roots, we ﬁnd that f (x) has 3 roots mod 9 and 2 roots mod 5. In fact f (x) ≡ 0 (mod 9) iﬀ x ≡ 1, 2, or −3 (mod 9)) and f (x) ≡ 0 (mod 5) iﬀ x ≡ 1 or −2 (mod 5). Since f (x) ≡ 0 (mod 45) iﬀ f (x) ≡ 0 (mod 9) and f (x) ≡ 0 (mod 5), we see that x is a root of f (x) mod 45 iﬀ x ≡ 1, 2, or −3 (mod 9) and x ≡ 1 or −2 (mod 5). Thus by the Chinese remainder theorem, f (x) has 2 · 3 = 6 roots mod 45. f (x) ≡ 0 (mod 45) ⇔ f (x) ≡ 0 (mod 9) and f (x) ≡ 0 (mod 5) ⇔ (x ≡ 1, 2, or − 3 (mod 9)) and (x ≡ 1 or − 2 (mod 5)) ⇔ (x ≡ 1 (mod 9) and x ≡ 1 (mod 5)) or (x ≡ 1 (mod 9) and x ≡ −2 (mod 5)) or (x ≡ 2 (mod 9) and x ≡ 1 (mod 5)) or (x ≡ 2 (mod 9) and x ≡ −2 (mod 5)) or (x ≡ −3 (mod 9) and x ≡ 1 (mod 5)) or (x ≡ −3 (mod 9) and x ≡ −2 (mod 5)) ⇔ x≡1 (mod 45) or x ≡ 28 (mod 45) or x ≡ 11 (mod 45) or x ≡ 38 (mod 45) or x≡6 (mod 45) or x ≡ 33 (mod 45) So the roots of f (x) mod 45 are 1, 6, 11, 27, 33, 38. (Alternatively you could have tried all 45 residues to see which were roots, but this way may be easier.) 3

DOCUMENT INFO

Shared By:

Categories:

Tags:

Stats:

views: | 0 |

posted: | 2/22/2013 |

language: | simple |

pages: | 3 |

OTHER DOCS BY xusuqin

Docstoc is the premier online destination to start and grow small businesses. It hosts the best quality and widest selection of professional documents (over 20 million) and resources including expert videos, articles and productivity tools to make every small business better.

Search or Browse for any specific document or resource you need for your business. Or explore our curated resources for Starting a Business, Growing a Business or for Professional Development.

Feel free to Contact Us with any questions you might have.